QUANTITATIVE METHODS FOR BUSINESS 8e

QUANTITATIVE METHODS FOR BUSINESS 8e
Anderson Sweeney Williams QUANTITATIVE METHODS FOR BUSINESS 8e Slides Prepared by JOHN LOUCKS © South-Western College Publishing/Thomson Learning

Chapter 1 Introduction Body of Knowledge
Problem Solving and Decision Making Quantitative Analysis and Decision Making Quantitative Analysis Models of Cost, Revenue, and Profit Quantitative Methods in Practice The Management Scientist

Body of Knowledge Quantitative methods for business involve rational approaches to decision making based on the scientific method of problem solving. This body of knowledge is often referred to as management science, operations research or decision science. It had its early roots in World War II and is flourishing in business and industry with the aid of computers.

Problem Solving and Decision Making
7 Steps of Problem Solving (First 5 steps are the process of decision making) Identify and define the problem. Determine the set of alternative solutions. Determine the criteria for evaluating the alternatives. Evaluate the alternatives. Choose an alternative. Implement the chosen alternative. Evaluate the results.

Quantitative Analysis and Decision Making
Potential Reasons for a Quantitative Analysis Approach to Decision Making The problem is complex. The problem is very important. The problem is new. The problem is repetitive.

Quantitative Analysis
Quantitative Analysis Process Model Development Data Preparation Model Solution Report Generation

Model Development Models are representations of real objects or situations. Three forms of models are iconic, analog, and mathematical. Iconic models are physical replicas (scalar representations) of real objects. Analog models are physical in form, but do not physically resemble the object being modeled. Mathematical models represent real world problems through a system of mathematical formulas and expressions based on key assumptions, estimates, or statistical analyses.

Advantages of Models Generally, experimenting with models (compared to experimenting with the real situation): requires less time is less expensive involves less risk

Mathematical Models Cost/benefit considerations must be made in selecting an appropriate mathematical model. Frequently a less complicated (and perhaps less precise) model is more appropriate than a more complex and accurate one due to cost and ease of solution considerations.

Mathematical Models Relate decision variables (controllable inputs) with fixed or variable parameters (uncontrollable inputs). Frequently seek to maximize or minimize some objective function subject to constraints. Are said to be stochastic if any of the uncontrollable inputs is subject to variation, otherwise are said to be deterministic. Generally, stochastic models are more difficult to analyze. The values of the decision variables that provide the mathematically-best output are referred to as the optimal solution for the model.

Transforming Model Inputs into Output
Uncontrollable Inputs (Environmental Factors) Controllable Inputs (Decision Variables) Mathematical Model Output (Projected Results)

Example: Project Scheduling
Consider a construction company building a 250-unit apartment complex. The project consists of hundreds of activities involving excavating, framing, wiring, plastering, painting, landscaping, and more. Some of the activities must be done sequentially and others can be done simultaneously. Also, some of the activities can be completed faster than normal by purchasing additional resources (workers, equipment, etc.). What is the best schedule for the activities and for which activities should additional resources be purchased?

Question: How could management science be used to solve this problem? Answer: Management science can provide a structured, quantitative approach for determining the minimum project completion time based on the activities' normal times and then based on the activities' expedited (reduced) times.

Question: What would be the uncontrollable inputs? Answer: Normal and expedited activity completion times Activity expediting costs Funds available for expediting Precedence relationships of the activities

Question: What would be the decision variables of the mathematical model? The objective function? The constraints? Answer: Decision variables: which activities to expedite and by how much, and when to start each activity Objective function: minimize project completion time Constraints: do not violate any activity precedence relationships and do not expedite in excess of the funds available.

Question: Is the model deterministic or stochastic? Answer: Stochastic. Activity completion times, both normal and expedited, are uncertain and subject to variation. Activity expediting costs are uncertain. The number of activities and their precedence relationships might change before the project is completed due to a project design change.

Question: Suggest assumptions that could be made to simplify the model. Answer: Make the model deterministic by assuming normal and expedited activity times are known with certainty and are constant. The same assumption might be made about the other stochastic, uncontrollable inputs.

Data Preparation Data preparation is not a trivial step, due to the time required and the possibility of data collection errors. A model with 50 decision variables and 25 constraints could have over 1300 data elements! Often, a fairly large data base is needed. Information systems specialists might be needed.

Model Solution Involves identifying the values of the decision variables that provide the “best” output for the model. One approach is trial-and-error. might not provide the best solution inefficient (numerous calculations required) Special solution procedures have been developed for specific mathematical models. some small models/problems can be solved by hand calculations most practical applications require using a computer

Computer Software A variety of software packages are available for solving mathematical models. Spreadsheet packages such as Microsoft Excel The Management Scientist, developed by the textbook authors

Model Testing and Validation
Often, the goodness/accuracy of a model cannot be assessed until solutions are generated. Small test problems having known, or at least expected, solutions can be used for model testing and validation. If the model generates expected solutions: use the model on the full-scale problem. If inaccuracies or potential shortcomings inherent in the model are identified, take corrective action such as: collection of more-accurate input data modification of the model

Report Generation A managerial report, based on the results of the model, should be prepared. The report should be easily understood by the decision maker. The report should include: the recommended decision other pertinent information about the results (for example, how sensitive the model solution is to the assumptions and data used in the model)

Implementation and Follow-Up
Successful implementation of model results is of critical importance. Secure as much user involvement as possible throughout the modeling process. Continue to monitor the contribution of the model. It might be necessary to refine or expand the model.

Example: Austin Auto Auction
An auctioneer has developed a simple mathematical model for deciding the starting bid he will require when auctioning a used automobile. Essentially, he sets the starting bid at seventy percent of what he predicts the final winning bid will (or should) be. He predicts the winning bid by starting with the car's original selling price and making two deductions, one based on the car's age and the other based on the car's mileage. The age deduction is $800 per year and the mileage deduction is $.025 per mile.

Question: Develop the mathematical model that will give the starting bid (B ) for a car in terms of the car's original price (P ), current age (A) and mileage (M ). Answer: The expected winning bid can be expressed as: P - 800(A) (M ) The entire model is: B = .7(expected winning bid) or B = .7(P - 800(A) (M )) or B = .7(P )- 560(A) (M )

Question: Suppose a four-year old car with 60,000 miles on the odometer is up for auction. If its original price was $12,500, what starting bid should the auctioneer require? Answer: B = .7(12,500) - 560(4) (60,000) = $5460.

Question: The model is based on what assumptions? Answer: The model assumes that the only factors influencing the value of a used car are the original price, age, and mileage (not condition, rarity, or other factors). Also, it is assumed that age and mileage devalue a car in a linear manner and without limit. (Note, the starting bid for a very old car might be negative!)

Example: Iron Works, Inc.
Iron Works, Inc. (IWI) manufactures two products made from steel and just received this month's allocation of b pounds of steel. It takes a1 pounds of steel to make a unit of product 1 and it takes a2 pounds of steel to make a unit of product 2. Let x1 and x2 denote this month's production level of product 1 and product 2, respectively. Denote by p1 and p2 the unit profits for products 1 and 2, respectively. The manufacturer has a contract calling for at least m units of product 1 this month. The firm's facilities are such that at most u units of product 2 may be produced monthly.

Mathematical Model The total monthly profit = (profit per unit of product 1) x (monthly production of product 1) + (profit per unit of product 2) x (monthly production of product 2) = p1x1 + p2x2 We want to maximize total monthly profit: Max p1x1 + p2x2

Mathematical Model (continued) The total amount of steel used during monthly production = (steel required per unit of product 1) x (monthly production of product 1) + (steel required per unit of product 2) x (monthly production of product 2) = a1x1 + a2x2 This quantity must be less than or equal to the allocated b pounds of steel: a1x1 + a2x2 < b

Mathematical Model (continued) The monthly production level of product 1 must be greater than or equal to m : x1 > m The monthly production level of product 2 must be less than or equal to u : x2 < u However, the production level for product 2 cannot be negative: x2 > 0

Mathematical Model Summary Max p1x1 + p2x2 s.t a1x1 + a2x2 < b x > m x2 < u x2 > 0

Question: Suppose b = 2000, a1 = 2, a2 = 3, m = 60, u = 720, p1 = 100, p2 = Rewrite the model with these specific values for the uncontrollable inputs. Answer: Substituting, the model is: Max 100x x2 s.t x x2 < 2000 x > x2 < x2 >

Question: The optimal solution to the current model is x1 = 60 and x2 = 626 2/3. If the product were engines, explain why this is not a true optimal solution for the "real-life" problem. Answer: One cannot produce and sell 2/3 of an engine. Thus the problem is further restricted by the fact that both x1 and x2 must be integers. They could remain fractions if it is assumed these fractions are work in progress to be completed the next month.

Uncontrollable Inputs $100 profit per unit Prod. 1 $200 profit per unit Prod. 2 2 lbs. steel per unit Prod. 1 3 lbs. Steel per unit Prod. 2 2000 lbs. steel allocated 60 units minimum Prod. 1 720 units maximum Prod. 2 0 units minimum Prod. 2 60 units Prod. 1 units Prod. 2 Max 100(60) + 200(626.67) s.t. 2(60) + 3(626.67) < 2000 > 60 < 720 > Profit = $131,333.33 Steel Used = 2000 Controllable Inputs Output Mathematical Model

Example: Ponderosa Development Corp.
Ponderosa Development Corporation (PDC) is a small real estate developer operating in the Rivertree Valley. It has seven permanent employees whose monthly salaries are given in the table on the next slide. PDC leases a building for $2,000 per month. The cost of supplies, utilities, and leased equipment runs another $3,000 per month. PDC builds only one style house in the valley. Land for each house costs $55,000 and lumber, supplies, etc. run another $28,000 per house. Total labor costs are figured at $20,000 per house. The one sales representative of PDC is paid a commission of $2,000 on the sale of each house. The selling price of the house is $115,000.

Employee Monthly Salary President $10,000 VP, Development ,000 VP, Marketing ,500 Project Manager ,500 Controller ,000 Office Manager ,000 Receptionist ,000

Question: Identify all costs and denote the marginal cost and marginal revenue for each house. Answer: The monthly salaries total $35,000 and monthly office lease and supply costs total another $5,000. This $40,000 is a monthly fixed cost. The total cost of land, material, labor, and sales commission per house, $105,000, is the marginal cost for a house. The selling price of $115,000 is the marginal revenue per house.

Question: Write the monthly cost function c (x), revenue function r (x), and profit function p (x). Answer: c (x) = variable cost + fixed cost = 105,000x + 40,000 r (x) = 115,000x p (x) = r (x) - c (x) = 10,000x - 40,000

Question: What is the breakeven point for monthly sales of the houses? Answer: r (x ) = c (x ) or 115,000x = 105,000x + 40,000 Solving, x = 4. What is the monthly profit if 12 houses per month are built and sold? p (12) = 10,000(12) - 40,000 = $80,000 monthly profit

Graph of Break-Even Analysis 1200 1000 Total Revenue = 115,000x Total Cost = 40, ,000x 800 Thousands of Dollars 600 Break-Even Point = 4 Houses 400 200 1 2 3 4 5 6 7 8 9 10 Number of Houses Sold (x)

A spreadsheet software package such as Microsoft Excel can be used to perform a quantitative analysis of Ponderosa Development Corporation. We will enter the problem data in the top portion of the spreadsheet. The bottom of the spreadsheet will be used for model development.

Formula Spreadsheet A B 1 PROBLEM DATA 2 Fixed Cost 40000 3 Variable Cost Per Unit 105000 4 Selling Price Per Unit 115000 5 MODEL 6 Sales Volume 7 Total Revenue =B4*B6 8 Total Cost =B2+B3*B6 9 Total Profit (Loss) =B7-B8

Question: What is the monthly profit if 12 houses per month are built and sold? Spreadsheet Solution: A B 1 PROBLEM DATA 2 Fixed Cost 3 Variable Cost Per Unit 4 Selling Price Per Unit 5 MODEL 6 Sales Volume 7 Total Revenue 8 Total Cost 9 Total Profit (Loss) $80,000 $1,300,000 $1,380,000 12 $40,000 $105,000 $115,000

Question: What is the breakeven point for monthly sales of houses? Spreadsheet Solution: One way to determine the break-even point is to use a trial-and-error approach. We will enter trial sales quantities ranging from 1 to 7 houses in the cells D3:D9 of the spreadsheet. Then we enter the formula for profit in cell E2. Finally, we use Excel ’s Table command to compute profit (or loss) for each of the trial sales values.

Spreadsheet Solution: Trial-and-Error Approach to Determining Break-Even Point Using Excel ’s Table Command Step 1: Select cells D2:E9 Step 2: Select the Data pull-down menu Step 3: Select the Table option Step 4: Enter B6 in the Column Input Cell of the Table dialog box, and select OK

Spreadsheet Solution: Trial-and-Error Approach to Determining Break-Even Point A B C D E 1 PROBLEM DATA 2 Fixed Cost $40,000 $80,000 3 Variable Cost Per Unit $105,000 1 -$30,000 4 Selling Price Per Unit $115,000 2 -$20,000 5 MODEL 3 -$10,000 6 Sales Volume 12 4 $0 7 Total Revenue $1,380,000 5 $10,000 8 Total Cost $1,300,000 6 $20,000 9 Total Profit (Loss) $80,000 7 $30,000

Question: What is the breakeven point for monthly sales of the houses? Spreadsheet Solution: Another way to determine the break-even point using a spreadsheet is to use the Goal Seek tool. Microsoft Excel ‘s Goal Seek tool allows the user to determine the value for an input cell that will cause the output cell to equal some specified value. In our case, the goal is to set Total Profit to zero by seeking an appropriate value for Sales Volume.

Spreadsheet Solution: Goal Seek Approach to Determining Break-Even Point Using Excel ’s Goal Seek Tool Step 1: Select the Tools pull-down menu Step 2: Choose the Goal Seek option Step 3: When the Goal Seek dialog box appears: Enter B9 in the Set cell box Enter 0 in the To value box Enter B6 in the By changing cell box Select OK

Spreadsheet Solution: Goal Seek Approach to Determining Break-Even Point Completed Goal Seek Dialog Box Goal Seek ? X Set cell: B9 To value: By changing cell: B6 OK Cancel

Spreadsheet Solution: Goal Seek Approach to Determining Break-Even Point A B 1 PROBLEM DATA 2 Fixed Cost $40,000 3 Variable Cost Per Unit $105,000 4 Selling Price Per Unit $115,000 5 MODEL 6 Sales Volume 4 7 Total Revenue $460,000 8 Total Cost $460,000 9 Total Profit (Loss) $0

The End of Chapter 1

Chapter 2 Introduction to Probability
Experiments and the Sample Space Assigning Probabilities to Experimental Outcomes Events and Their Probabilities Some Basic Relationships of Probability Bayes’ Theorem

Experiments and the Sample Space
An experiment is any process that generates well-defined outcomes. The sample space for an experiment is the set of all experimental outcomes. A sample point is an element of the sample space, any one particular experimental outcome.

Example: Bradley Investments
Bradley has invested in two stocks, Markley Oil and Collins Mining. Bradley has determined that the possible outcomes of these investments three months from now are as follows. Investment Gain or Loss in 3 Months (in $000) Markley Oil Collins Mining 10 8 5 -2 -20

A Counting Rule for Multiple-Step Experiments
If an experiment consists of a sequence of k steps in which there are n1 possible results for the first step, n2 possible results for the second step, and so on, then the total number of experimental outcomes is given by (n1)(n2) (nk). Example: Bradley Investments can be viewed as a two-step experiment; it involves two stocks, each with a set of experimental outcomes. Markley Oil: n1 = 4 Collins Mining: n2 = 2 Total Number of Experimental Outcomes: n1n2 = (4)(2) = 8

Tree Diagram Markley Oil Collins Mining Experimental (Stage 1) (Stage 2) Outcomes (10, 8) Gain $18,000 (10, -2) Gain $8,000 (5, 8) Gain $13,000 (5, -2) Gain $3,000 (0, 8) Gain $8,000 (0, -2) Lose $2,000 (-20, 8) Lose $12,000 (-20, -2) Lose $22,000 Gain 8 Lose 2 Gain 10 Gain 8 Lose 2 Gain 5 Gain 8 Even Lose 2 Lose 2 Gain 8 Lose 2

Counting Rule for Combinations
Another useful counting rule enables us to count the number of experimental outcomes when n objects are to be selected from a set of N objects. The number of combinations of N objects taken n at a time is where N! = N(N - 1)(N - 2) (2)(1) n! = n(n - 1)(n - 2) (2)(1) 0! = 1

Assigning Probabilities to Experimental Outcomes
Classical Method Assigning probabilities based on the assumption of equally likely outcomes. Relative Frequency Method Assigning probabilities based on experimentation or historical data. Subjective Method Assigning probabilities based on the assignor’s judgment.

Classical Method If an experiment has n possible outcomes, this method
would assign a probability of 1/n to each outcome. Example Experiment: Rolling a die Sample Space: S = {1, 2, 3, 4, 5, 6} Probabilities: Each sample point has a 1/6 chance of occurring.

Example: Lucas Tool Rental
Relative Frequency Method Lucas would like to assign probabilities to the number of floor polishers it rents per day. Office records show the following frequencies of daily rentals for the last 40 days. Number of Number Polishers Rented of Days

Example: Lucas Tool Rental
Relative Frequency Method The probability assignments are given by dividing the number-of-days frequencies by the total frequency. Number of Number Polishers Rented of Days Probability = 4/40 = 6/40 etc.

Subjective Method When economic conditions and a company’s circumstances change rapidly it might be inappropriate to assign probabilities based solely on historical data. We can use any data available as well as our experience and intuition, but ultimately a probability value should express our degree of belief that the experimental outcome will occur. The best probability estimates often are obtained by combining the estimates from the classical or relative frequency approach with the subjective estimates.

Applying the subjective method an analyst made the following probability assignments. Exper. Outcome Net Gain/Loss Probability (10, 8) $18,000 Gain (10, -2) $8,000 Gain (5, 8) $13,000 Gain (5, -2) $3,000 Gain (0, 8) $8,000 Gain (0, -2) $2,000 Loss (-20, 8) $12,000 Loss (-20, -2) $22,000 Loss

Events and Their Probabilities
An event is a collection of sample points. The probability of any event is equal to the sum of the probabilities of the sample points in the event. Example: Event M = Markley Oil Profitable M = {(10, 8), (10, -2), (5, 8), (5, -2)} P(M) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) = = .70 Event C = Collins Mining Profitable P(C) = (found using the same logic)

Some Basic Relationships of Probability
Complement of an Event Union of Two Events Intersection of Two Events Mutually Exclusive Events

Complement of an Event The complement of event A is defined to be the event consisting of all sample points that are not in A. The complement of A is denoted by Ac. The Venn diagram below illustrates the concept of a complement. Sample Space S Event A Ac

Union of Two Events The union of events A and B is the event containing all sample points that are in A or B or both. The union is denoted by A È B . The union of A and B is illustrated below. Sample Space S Event A Event B

Union of Two Events Example: Event M = Markley Oil Profitable
Event C = Collins Mining Profitable M È C = Markley Oil Profitable or Collins Mining Profitable M È C = {(10, 8), (10, -2), (5, 8), (5, -2), (0, 8), (-20, 8)} P(M È C) = P(10, 8) + P(10, -2) + P(5, 8) + P(5, -2) + P(0, 8) + P(-20, 8) = = .82

Intersection of Two Events
The intersection of events A and B is the set of all sample points that are in both A and B. The intersection is denoted by A Ç B. The intersection of A and B is the area of overlap in the illustration below. Sample Space S Event A Event B

Intersection of Two Events
Example: Event M = Markley Oil Profitable Event C = Collins Mining Profitable M Ç C = Markley Oil Profitable and Collins Mining Profitable M Ç C = {(10, 8), (5, 8)} P(M Ç C) = P(10, 8) + P(5, 8) = = .36

Addition Law The addition law provides a way to compute the probability of event A or B or both A and B occurring. The law is written as: P(A È B) = P(A) + P(B) - P(A Ç B) Example: Markley Oil or Collins Mining Profitable We know: P(M) = .70, P(C) = .48, P(M Ç C) = .36 Thus: P(M È C) = P(M) + P(C) - P(M Ç C) = = .82 This result is the same as that obtained earlier using the definition of the probability of an event.

Addition Law for Mutually Exclusive Events
Two events are said to be mutually exclusive if the events have no sample points in common. That is, two events are mutually exclusive if, when one event occurs, the other cannot occur. Addition Law for Mutually Exclusive Events: P(A È B) = P(A) + P(B) Sample Space S Event A Event B

Conditional Probability
The probability of an event given that another event has occurred is called a conditional probability. The conditional probability of A given B is denoted by P(A|B). A conditional probability is computed as follows: Example: Collins Mining Profitable Given Markley Oil Profitable

Multiplication Law The multiplication law provides a way to compute the probability of an intersection of two events. The law is written as: P(A Ç B) = P(B)P(A|B) Example: Markley Oil and Collins Mining Profitable We know: P(M) = .70, P(C|M) = .51 Thus: P(M Ç C) = P(M)P(M|C) = (.70)(.51) = .36 This result is the same as that obtained earlier using the definition of the probability of an event.

Multiplication Law for Independent Events
Events A and B are independent if P(A|B) = P(A). Multiplication Law for Independent Events: P(A Ç B) = P(A)P(B) The multiplication law also can be used as a test to see if two events are independent. Example: Are M and C independent? Does P(M Ç C) = P(M)P(C) ? We know: P(M Ç C) = .36, P(M) = .70, P(C) = .48 But: P(M)P(C) = (.70)(.48) = .34 .34 ¹ .36, so M and C are not independent.

Bayes’ Theorem Often we begin probability analysis with initial or prior probabilities. Then, from a sample, special report, or a product test we obtain some additional information. Given this information, we calculate revised or posterior probabilities. Bayes’ theorem provides the means for revising the prior probabilities. Prior Probabilities New Information Application of Bayes’ Theorem Posterior Probabilities

Example: L. S. Clothiers A proposed shopping center will provide strong competition for downtown businesses like L. S. Clothiers. If the shopping center is built, the owner of L. S. Clothiers feels it would be best to relocate. The shopping center cannot be built unless a zoning change is approved by the town council. The planning board must first make a recommendation, for or against the zoning change, to the council. Let: A1 = town council approves the zoning change A2 = town council disapproves the change Prior Probabilities Using subjective judgment: P(A1) = .7, P(A2) = .3

Example: L. S. Clothiers New Information
The planning board has recommended against the zoning change. Let B denote the event of a negative recommendation by the planning board. Given that B has occurred, should L. S. Clothiers revise the probabilities that the town council will approve or disapprove the zoning change? Past history with the planning board and the town council indicates the following conditional probabilities: P(B|A1) = P(B|A2) = .9

Example: L. S. Clothiers Probability Tree Analysis P(B|A1) = .2
P(A1 Ç B) = .14 P(A1) = .7 P(Bc|A1) = .8 P(A1 Ç Bc) = .56 P(B|A2) = .9 P(A2 Ç B) = .27 P(A2) = .3 P(A2 Ç Bc) = .03 P(Bc|A2) = .1

Bayes’ Theorem To find the posterior probability that event Ai will occur given that event B has occurred we apply Bayes’ theorem. Bayes’ theorem is applicable when the events for which we want to compute posterior probabilities are mutually exclusive and their union is the entire sample space.

Example: L. S. Clothiers Posterior Probabilities
Given the planning board’s recommendation not to approve the zoning change, we revise the prior probabilities as follows. = .34 Conclusion The planning board’s recommendation is good news for L. S. Clothiers. The posterior probability of the town council approving the zoning change is .34 versus a prior probability of .7.

Future Use of Probability Concepts
Specific chapters and quantitative methods that make use of probability are: Chapter 3 Probability distributions Chapter 4 Decision analysis Chapter 5 Utility and decision making Chapter 12 Project scheduling: PERT/CPM Chapter 14 Inventory management Chapter 15 Waiting line models Chapter 16 Simulation

Chapter 3, Part I Discrete Probability Distributions
Random Variables Discrete Random Variables Expected Value and Variance Binomial Probability Distribution Poisson Probability Distribution

Random Variables A random variable is a numerical description of the outcome of an experiment. A random variable can be classified as being either discrete or continuous depending on the numerical values it assumes. A discrete random variable may assume either a finite number of values or an infinite sequence of values. A continuous random variable may assume any numerical value in an interval or collection of intervals.

Example: JSL Appliances
Discrete random variable with a finite number of values: Let x = number of TV sets sold at the store in one day where x can take on 5 values (0, 1, 2, 3, 4) Discrete random variable with an infinite sequence of values: Let x = number of customers arriving in one day where x can take on the values 0, 1, 2, . . . We can count the customers arriving, but there is no finite upper limit on the number that might arrive.

Discrete Probability Distributions
The probability distribution for a random variable describes how probabilities are distributed over the values of the random variable. The probability distribution is defined by a probability function, denoted by f(x), which provides the probability for each value of the random variable. The required conditions for a discrete probability function are: f(x) > 0 Sf(x) = 1 We can describe a discrete probability distribution with a table, graph, or equation.

Using past data on TV sales (below left), a tabular representation of the probability distribution for TV sales (below right) was developed. Number Units Sold of Days x f(x)

A graphical representation of the probability distribution for TV sales in one day .50 .40 Probability .30 .20 .10 Values of Random Variable x (TV sales)

Expected Value and Variance
The expected value, or mean, of a random variable is a measure of its central location. Expected value of a discrete random variable: E(x) = m = Sxf(x) The variance summarizes the variability in the values of a random variable. Variance of a discrete random variable: Var(x) = s2 = S(x - m)2f(x) The standard deviation, s, is defined as the positive square root of the variance.

Expected Value of a Discrete Random Variable x f(x) xf(x) 1.20 = E(x) The expected number of TV sets sold in a day is 1.2

Variance and Standard Deviation of a Discrete Random Variable x x - m (x - m)2 f(x) (x - m)2f(x) _____ _________ ___________ _______ _______________ 1.660 = s 2 The variance of daily sales is 1.66 TV sets squared. The standard deviation of sales is 1.29 TV sets.

Formula Spreadsheet for Computing Expected Value and Variance

Binomial Probability Distribution
Properties of a Binomial Experiment The experiment consists of a sequence of n identical trials. Two outcomes, success and failure, are possible on each trial. The probability of a success, denoted by p, does not change from trial to trial. The trials are independent.

Example: Evans Electronics
Binomial Probability Distribution Evans is concerned about a low retention rate for employees. On the basis of past experience, management has seen a turnover of 10% of the hourly employees annually. Thus, for any hourly employees chosen at random, management estimates a probability of 0.1 that the person will not be with the company next year. Choosing 3 hourly employees a random, what is the probability that 1 of them will leave the company this year? Let: p = .10, n = 3, x = 1

Binomial Probability Distribution
Binomial Probability Function where f(x) = the probability of x successes in n trials n = the number of trials p = the probability of success on any one trial

Using the Binomial Probability Function = (3)(0.1)(0.81) = .243

Using the Tables of Binomial Probabilities

Using a Tree Diagram First Worker Second Worker Third Worker Value of x Probab. L (.1) 3 .0010 Leaves (.1) 2 .0090 S (.9) Leaves (.1) L (.1) 2 .0090 Stays (.9) 1 .0810 S (.9) L (.1) 2 .0090 Leaves (.1) 1 .0810 S (.9) Stays (.9) L (.1) 1 .0810 Stays (.9) .7290 S (.9)

Using an Excel Spreadsheet Step 1: Select a cell in the worksheet where you want the binomial probabilities to appear. Step 2: Select the Insert pull-down menu. Step 3: Choose the Function option. Step 4: When the Paste Function dialog box appears: Choose Statistical from the Function Category box. Choose BINOMDIST from the Function Name box. Select OK. continued

Using an Excel Spreadsheet (continued) Step 5: When the BINOMDIST dialog box appears: Enter 1 in the Number_s box (value of x). Enter 3 in the Trials box (value of n). Enter .1 in the Probability_s box (value of p). Enter false in the Cumulative box. [Note: At this point the desired binomial probability of .243 is automatically computed and appears in the right center of the dialog box.] Select OK (and .243 will appear in the worksheet cell requested in Step 1).

The Binomial Probability Distribution
Expected Value E(x) = m = np Variance Var(x) = s 2 = np(1 - p) Standard Deviation Example: Evans Electronics E(x) = m = 3(.1) = .3 employees out of 3 Var(x) = s 2 = 3(.1)(.9) = .27

Poisson Probability Distribution
Properties of a Poisson Experiment The probability of an occurrence is the same for any two intervals of equal length. The occurrence or nonoccurrence in any interval is independent of the occurrence or nonoccurrence in any other interval.

Poisson Probability Function where f(x) = probability of x occurrences in an interval m = mean number of occurrences in an interval e =

Example: Mercy Hospital
Patients arrive at the emergency room of Mercy Hospital at the average rate of 6 per hour on weekend evenings. What is the probability of 4 arrivals in 30 minutes on a weekend evening? Using the Poisson Probability Function m = 6/hour = 3/half-hour, x = 4

Example: Mercy Hospital
Using the Tables of Poisson Probabilities

The Poisson probability distribution can be used as an approximation of the binomial probability distribution when p, the probability of success, is small and n, the number of trials, is large. Approximation is good when p < .05 and n > 20 Set m = np and use the Poisson tables.

The End of Chapter 3, Part I

Chapter 3, Part II Continuous Random Variables
Normal Probability Distribution Exponential Probability Distribution

Continuous Probability Distributions
A continuous random variable can assume any value in an interval on the real line or in a collection of intervals. It is not possible to talk about the probability of the random variable assuming a particular value. Instead, we talk about the probability of the random variable assuming a value within a given interval. The probability of the random variable assuming a value within some given interval from x1 to x2 is defined to be the area under the graph of the probability density function between x1 and x2.

Uniform Probability Distribution
A random variable is uniformly distributed whenever the probability is proportional to the length of the interval. Uniform Probability Density Function f(x) = 1/(b - a) for a < x < b = 0 elsewhere Expected Value of x E(x) = (a + b)/2 Variance of x Var(x) = (b - a)2/12 where: a = smallest value the variable can assume b = largest value the variable can assume

Example: Slater's Buffet
Slater customers are charged for the amount of salad they take. Sampling suggests that the amount of salad taken is uniformly distributed between 5 ounces and 15 ounces. Probability Density Function f(x) = 1/10 for 5 < x < 15 = 0 elsewhere where x = salad plate filling weight

Example: Slater's Buffet
What is the probability that a customer will take between 12 and 15 ounces of salad? f(x) P(12 < x < 15) = 1/10(3) = .3 1/10 x 5 10 12 15 Salad Weight (oz.)

The Normal Probability Distribution
Graph of the Normal Probability Density Function f(x) x m

Normal Probability Distribution
The Normal Curve The shape of the normal curve is often illustrated as a bell-shaped curve. The highest point on the normal curve is at the mean, which is also the median and mode of the distribution. The normal curve is symmetric. The standard deviation determines the width of the curve. The total area under the curve is 1. Probabilities for the normal random variable are given by areas under the curve.

Normal Probability Distribution
Normal Probability Density Function where m = mean s = standard deviation p = e =

Standard Normal Probability Distribution
A random variable that has a normal distribution with a mean of zero and a standard deviation of one is said to have a standard normal probability distribution. The letter z is commonly used to designate this normal random variable. Converting to the Standard Normal Distribution We can think of z as a measure of the number of standard deviations x is from m.

Example: Pep Zone Pep Zone sells auto parts and supplies including a
popular multi-grade motor oil. When the stock of this oil drops to 20 gallons, a replenishment order is placed. The store manager is concerned that sales are being lost due to stockouts while waiting for an order. It has been determined that leadtime demand is normally distributed with a mean of 15 gallons and a standard deviation of 6 gallons. The manager would like to know the probability of a stockout, P(x > 20).

Example: Pep Zone Standard Normal Distribution Area = .2967
z = (x - m)/s = ( )/6 = .83 The Standard Normal table shows an area of for the region between the z = 0 line and the z = .83 line above. The shaded tail area is = The probability of a stockout is Area = .2967 Area = .2033 Area = .5 z .83

Example: Pep Zone Using the Standard Normal Probability Table

Example: Pep Zone Using an Excel Spreadsheet
Step 1: Select a cell in the worksheet where you want the normal probability to appear. Step 2: Select the Insert pull-down menu. Step 3: Choose the Function option. Step 4: When the Paste Function dialog box appears: Choose Statistical from the Function Category box. Choose NORMDIST from the Function Name box. Select OK. continue

Example: Pep Zone Using an Excel Spreadsheet (continued)
Step 5: When the NORMDIST dialog box appears: Enter 20 in the x box. Enter 15 in the mean box. Enter 6 in the standard deviation box. Enter true in the cumulative box. Select OK. At this point, will appear in the cell selected in Step 1, indicating that the probability of demand during lead time being less than or equal to 20 gallons is The probability that demand will exceed 20 gallons is =

Example: Pep Zone If the manager of Pep Zone wants the probability of a stockout to be no more than .05, what should the reorder point be? Let z.05 represent the z value cutting the tail area of .05. Area = .05 Area = .5 Area = .45 z.05

Example: Pep Zone Using the Standard Normal Probability Table
We now look-up the area in the Standard Normal Probability table to find the corresponding z.05 value. z.05 = is a reasonable estimate.

Example: Pep Zone The corresponding value of x is given by
x = m + z.05s = (6) = 24.87 A reorder point of gallons will place the probability of a stockout during leadtime at .05. Perhaps Pep Zone should set the reorder point at 25 gallons to keep the probability under .05.

Exponential Probability Distribution
Exponential Probability Density Function for x > 0, m > 0 where m = mean e = Cumulative Exponential Distribution Function where x0 = some specific value of x

Example: Al’s Carwash The time between arrivals of cars at Al’s Carwash follows an exponential probability distribution with a mean time between arrivals of 3 minutes. Al would like to know the probability that the time between two successive arrivals will be 2 minutes or less. P(x < 2) = /3 = = .4866

Example: Al’s Carwash Graph of the Probability Density Function f(x)
.4 P(x < 2) = area = .4866 .3 .2 .1 x

Relationship Between the Poisson and Exponential Distributions
The continuous exponential probability distribution is related to the discrete Poisson distribution. The Poisson distribution provides an appropriate description of the number of occurrences per interval. The exponential distribution provides a description of the length of the interval between occurrences.

The End of Chapter 3, Part II

Chapter 4 Decision Analysis, Part A
Problem Formulation Decision Making without Probabilities Decision Making with Probabilities

Problem Formulation A decision problem is characterized by decision alternatives, states of nature, and resulting payoffs. The decision alternatives are the different possible strategies the decision maker can employ. The states of nature refer to future events, not under the control of the decision maker, which may occur. States of nature should be defined so that they are mutually exclusive and collectively exhaustive.

Influence Diagram An influence diagram is a graphical device showing the relationships among the decisions, the chance events, and the consequences. Squares or rectangles depict decision nodes. Circles or ovals depict chance nodes. Diamonds depict consequence nodes. Lines or arcs connecting the nodes show the direction of influence.

Payoff Tables The consequence resulting from a specific combination of a decision alternative and a state of nature is a payoff. A table showing payoffs for all combinations of decision alternatives and states of nature is a payoff table. Payoffs can be expressed in terms of profit, cost, time, distance or any other appropriate measure.

Decision Trees A decision tree is a chronological representation of the decision problem. Each decision tree has two types of nodes; round nodes correspond to the states of nature while square nodes correspond to the decision alternatives. The branches leaving each round node represent the different states of nature while the branches leaving each square node represent the different decision alternatives. At the end of each limb of a tree are the payoffs attained from the series of branches making up that limb.

Decision Making without Probabilities
Three commonly used criteria for decision making when probability information regarding the likelihood of the states of nature is unavailable are: the optimistic approach the conservative approach the minimax regret approach.

Optimistic Approach The optimistic approach would be used by an optimistic decision maker. The decision with the largest possible payoff is chosen. If the payoff table was in terms of costs, the decision with the lowest cost would be chosen.

Conservative Approach
The conservative approach would be used by a conservative decision maker. For each decision the minimum payoff is listed and then the decision corresponding to the maximum of these minimum payoffs is selected. (Hence, the minimum possible payoff is maximized.) If the payoff was in terms of costs, the maximum costs would be determined for each decision and then the decision corresponding to the minimum of these maximum costs is selected. (Hence, the maximum possible cost is minimized.)

Minimax Regret Approach
The minimax regret approach requires the construction of a regret table or an opportunity loss table. This is done by calculating for each state of nature the difference between each payoff and the largest payoff for that state of nature. Then, using this regret table, the maximum regret for each possible decision is listed. The decision chosen is the one corresponding to the minimum of the maximum regrets.

Example Consider the following problem with three decision alternatives and three states of nature with the following payoff table representing profits: States of Nature s s s3 d Decisions d d

Example Optimistic Approach
An optimistic decision maker would use the optimistic approach. All we really need to do is to choose the decision that has the largest single value in the payoff table. This largest value is 5, and hence the optimal decision is d3. Maximum Decision Payoff d d choose d d maximum

Example Formula Spreadsheet for Optimistic Approach

Example Solution Spreadsheet for Optimistic Approach

Example Conservative Approach
A conservative decision maker would use the conservative approach. List the minimum payoff for each decision. Choose the decision with the maximum of these minimum payoffs. Minimum Decision Payoff d choose d d maximum d

Example Formula Spreadsheet for Conservative Approach

Example Solution Spreadsheet for Conservative Approach

Example Minimax Regret Approach
For the minimax regret approach, first compute a regret table by subtracting each payoff in a column from the largest payoff in that column. In this example, in the first column subtract 4, 0, and 1 from 4; in the second column, subtract 4, 3, and 5 from 5; etc. The resulting regret table is: s1 s2 s3 d d d

Example Minimax Regret Approach (continued)
For each decision list the maximum regret. Choose the decision with the minimum of these values. Decision Maximum Regret choose d d minimum d d

Example Formula Spreadsheet for Minimax Regret Approach

Example Solution Spreadsheet for Minimax Regret Approach

Decision Making with Probabilities
Expected Value Approach If probabilistic information regarding he states of nature is available, one may use the expected value (EV) approach. Here the expected return for each decision is calculated by summing the products of the payoff under each state of nature and the probability of the respective state of nature occurring. The decision yielding the best expected return is chosen.

Expected Value of a Decision Alternative
The expected value of a decision alternative is the sum of weighted payoffs for the decision alternative. The expected value (EV) of decision alternative di is defined as: where: N = the number of states of nature P(sj ) = the probability of state of nature sj Vij = the payoff corresponding to decision alternative di and state of nature sj

Example: Burger Prince
Burger Prince Restaurant is contemplating opening a new restaurant on Main Street. It has three different models, each with a different seating capacity. Burger Prince estimates that the average number of customers per hour will be 80, 100, or The payoff table for the three models is as follows: Average Number of Customers Per Hour s1 = s2 = s3 = 120 Model A $10, $15, $14,000 Model B $ 8, $18, $12,000 Model C $ 6, $16, $21,000

Expected Value Approach Calculate the expected value for each decision. The decision tree on the next slide can assist in this calculation. Here d1, d2, d3 represent the decision alternatives of models A, B, C, and s1, s2, s3 represent the states of nature of 80, 100, and 120.

Decision Tree Payoffs .4 s1 10,000 s2 .2 2 15,000 s3 .4 d1 14,000 .4 s1 8,000 d2 1 3 s2 .2 18,000 d3 s3 .4 12,000 .4 s1 6,000 4 s2 .2 16,000 s3 .4 21,000

Expected Value For Each Decision Choose the model with largest EV, Model C. EMV = .4(10,000) + .2(15,000) + .4(14,000) = $12,600 2 d1 Model A d2 EMV = .4(8,000) + .2(18,000) + .4(12,000) = $11,600 Model B 1 3 d3 Model C EMV = .4(6,000) + .2(16,000) + .4(21,000) = $14,000 4

Formula Spreadsheet for Expected Value Approach

Solution Spreadsheet for Expected Value Approach

The End of Chapter 4, Part A

Chapter 4 Decision Analysis, Part B
Expected Value of Perfect Information Decision Analysis with Sample Information Developing a Decision Strategy Expected Value of Sample Information

Burger Prince Restaurant is contemplating opening a new restaurant on Main Street. It has three different models, each with a different seating capacity. Burger Prince estimates that the average number of customers per hour will be 80, 100, or The payoff table for the three models is as follows: Average Number of Customers Per Hour s1 = s2 = s3 = 120 Model A $10, $15, $14,000 Model B $ 8, $18, $12,000 Model C $ 6, $16, $21,000

Expected Value Approach Calculate the expected value for each decision. The decision tree on the next slide can assist in this calculation. Here d1, d2, d3 represent the decision alternatives of models A, B, C, and s1, s2, s3 represent the states of nature of 80, 100, and 120.

Decision Tree Payoffs .4 s1 10,000 s2 .2 2 15,000 s3 .4 d1 14,000 .4 s1 8,000 d2 1 3 s2 .2 18,000 d3 s3 .4 12,000 .4 s1 6,000 4 s2 .2 16,000 s3 .4 21,000

Expected Value For Each Decision Choose the model with largest EV, Model C. EMV = .4(10,000) + .2(15,000) + .4(14,000) = $12,600 2 d1 Model A d2 EMV = .4(8,000) + .2(18,000) + .4(12,000) = $11,600 Model B 1 3 d3 Model C EMV = .4(6,000) + .2(16,000) + .4(21,000) = $14,000 4

Expected Value of Perfect Information
Frequently information is available which can improve the probability estimates for the states of nature. The expected value of perfect information (EVPI) is the increase in the expected profit that would result if one knew with certainty which state of nature would occur. The EVPI provides an upper bound on the expected value of any sample or survey information.

Expected Value of Perfect Information
EVPI Calculation Step 1: Determine the optimal return corresponding to each state of nature. Step 2: Compute the expected value of these optimal returns. Step 3: Subtract the EV of the optimal decision from the amount determined in step (2).

Expected Value of Perfect Information Calculate the expected value for the optimum payoff for each state of nature and subtract the EV of the optimal decision. EVPI= .4(10,000) + .2(18,000) + .4(21,000) - 14,000 = $2,000

Spreadsheet for Expected Value of Perfect Information

Risk Analysis Risk analysis helps the decision maker recognize the difference between: the expected value of a decision alternative and the payoff that might actually occur The risk profile for a decision alternative shows the possible payoffs for the decision alternative along with their associated probabilities.

Risk Profile for the Model C Decision Alternative .50 .40 Probability .30 .20 .10 Profit ($thousands)

Sensitivity Analysis Sensitivity analysis can be used to determine how changes to the following inputs affect the recommended decision alternative: probabilities for the states of nature values of the payoffs If a small change in the value of one of the inputs causes a change in the recommended decision alternative, extra effort and care should be taken in estimating the input value.

Bayes’ Theorem and Posterior Probabilities
Knowledge of sample or survey information can be used to revise the probability estimates for the states of nature. Prior to obtaining this information, the probability estimates for the states of nature are called prior probabilities. With knowledge of conditional probabilities for the outcomes or indicators of the sample or survey information, these prior probabilities can be revised by employing Bayes' Theorem. The outcomes of this analysis are called posterior probabilities or branch probabilities for decision trees.

Computing Branch Probabilities
Branch (Posterior) Probabilities Calculation Step 1: For each state of nature, multiply the prior probability by its conditional probability for the indicator -- this gives the joint probabilities for the states and indicator. Step 2: Sum these joint probabilities over all states -- this gives the marginal probability for the indicator. Step 3: For each state, divide its joint probability by the marginal probability for the indicator -- this gives the posterior probability distribution.

Expected Value of Sample Information
The expected value of sample information (EVSI) is the additional expected profit possible through knowledge of the sample or survey information.

Expected Value of Sample Information
EVSI Calculation Step 1: Determine the optimal decision and its expected return for the possible outcomes of the sample or survey using the posterior probabilities for the states of nature. Step 2: Compute the expected value of these optimal returns. Step 3: Subtract the EV of the optimal decision obtained without using the sample information from the amount determined in step (2).

Efficiency of Sample Information
Efficiency of sample information is the ratio of EVSI to EVPI. As the EVPI provides an upper bound for the EVSI, efficiency is always a number between 0 and 1.

Sample Information Burger Prince must decide whether or not to purchase a marketing survey from Stanton Marketing for $1,000. The results of the survey are "favorable" or "unfavorable". The conditional probabilities are: P(favorable | 80 customers per hour) = .2 P(favorable | 100 customers per hour) = .5 P(favorable | 120 customers per hour) = .9 Should Burger Prince have the survey performed by Stanton Marketing?

Influence Diagram Legend: Decision Chance Consequence Market Survey Results Avg. Number of Customers Per Hour Market Survey Restaurant Size Profit

Posterior Probabilities Favorable State Prior Conditional Joint Posterior Total P(favorable) = .54

Posterior Probabilities Unfavorable State Prior Conditional Joint Posterior Total P(unfavorable) = .46

Formula Spreadsheet for Posterior Probabilities

Solution Spreadsheet for Posterior Probabilities

Decision Tree (top half) s1 (.148) $10,000 s2 (.185) 4 $15,000 d1 s3 (.667) $14,000 s1 (.148) $8,000 d2 5 s2 (.185) 2 $18,000 s3 (.667) d3 $12,000 I1 (.54) s1 (.148) $6,000 6 s2 (.185) $16,000 s3 (.667) 1 $21,000

Decision Tree (bottom half) 1 s1 (.696) $10,000 s2 (.217) 7 $15,000 I2 (.46) d1 s3 (.087) $14,000 s1 (.696) $8,000 d2 s2 (.217) 8 3 $18,000 s3 (.087) d3 $12,000 s1 (.696) $6,000 s2 (.217) 9 $16,000 s3 (.087) $21,000

EMV = .148(10,000) (15,000) + .667(14,000) = $13,593 d1 4 $17,855 d2 5 EMV = .148 (8,000) (18,000) + .667(12,000) = $12,518 2 d3 I1 (.54) 6 EMV = .148(6,000) (16,000) +.667(21,000) = $17,855 1 7 EMV = .696(10,000) (15,000) +.087(14,000)= $11,433 d1 I2 (.46) d2 8 EMV = .696(8,000) (18,000) + .087(12,000) = $10,554 3 d3 $11,433 9 EMV = .696(6,000) (16,000) +.087(21,000) = $9,475

Expected Value of Sample Information If the outcome of the survey is "favorable" choose Model C. If it is unfavorable, choose model A. EVSI = .54($17,855) + .46($11,433) - $14,000 = $900.88 Since this is less than the cost of the survey, the survey should not be purchased.

Efficiency of Sample Information The efficiency of the survey: EVSI/EVPI = ($900.88)/($2000) =

The End of Chapter 4, Part B

Chapter 5 Utility and Decision Making
The Meaning of Utility Developing Utilities for Monetary Payoffs Summary of Steps for Determining the Utility of Money Risk Avoiders versus Risk Takers Expected Monetary Value versus Utility as an Approach to Decision Making

Meaning of Utility Utilities are used when the decision criteria must be based on more than just expected monetary values. Utility is a measure of the total worth of a particular outcome, reflecting the decision maker’s attitude towards a collection of factors. Some of these factors may be profit, loss, and risk. This analysis is particularly appropriate in cases where payoffs can assume extremely high or extremely low values.

Expected Utility Approach
Once a utility function has been determined, the optimal decision can be chosen using the expected utility approach. Here, for each decision alternative, the utility corresponding to each state of nature is multiplied by the probability for that state of nature. The sum of these products for each decision alternative represents the expected utility for that alternative. The decision alternative with the highest expected utility is chosen.

Risk Avoiders vs. Risk Takers
A risk avoider will have a concave utility function when utility is measured on the vertical axis and monetary value is measured on the horizontal axis. Individuals purchasing insurance exhibit risk avoidance behavior. A risk taker, such as a gambler, pays a premium to obtain risk. His/her utility function is convex. This reflects the decision maker’s increasing marginal value of money. A risk neutral decision maker has a linear utility function. In this case, the expected value approach can be used.

Risk Avoiders vs. Risk Takers
Most individuals are risk avoiders for some amounts of money, risk neutral for other amounts of money, and risk takers for still other amounts of money. This explains why the same individual will purchase both insurance and also a lottery ticket.

Example 1 Consider the following three-state, four-decision problem with the following payoff table in dollars: s s s3 d1 +100, , ,000 d , , ,000 d , , ,000 d , , ,000 The probabilities for the three states of nature are: P(s1) = .1, P(s2) = .3, and P(s3) = .6.

Example 1 Risk-Neutral Decision Maker
If the decision maker is risk neutral the expected value approach is applicable. EV(d1) = .1(100,000) + .3(40,000) + .6(-60,000) = -$14,000 EV(d2) = .1( 50,000) + .3(20,000) + .6(-30,000) = -$ 7,000 EV(d3) = .1( 20,000) + .3(20,000) + .6(-10,000) = +$ 2,000 Note the EV for d4 need not be calculated as decision d4 is dominated by decision d2. The optimal decision is d3.

Example 1 Decision Makers with Different Utilities
Suppose two decision makers have the following utility values: Utility Amount Decision Maker I Decision Maker II $100, $50, $40, $20, -$10, -$30, -$60,

Example 1 Graph of the Two Decision Makers’ Utility Curves Utility
100 Decision Maker I 80 60 40 Decision Maker II 20 Monetary Value (in $1000’s)

Example 1 Decision Maker I
Decision Maker I has a concave utility function. He/she is a risk avoider. Decision Maker II Decision Maker II has convex utility function. He/she is a risk taker.

Example 1 Expected Utility: Decision Maker I Expected s1 s2 s3 Utility
Probability Optimal Decision: Decision Maker I The largest expected utility is Note again that d4 is dominated by d2 and hence is not considered. Decision Maker I should make decision d3.

Example 1 Expected Utility: Decision Maker II Expected
s1 s2 s Utility d d d Probability Optimal Decision: Decision Maker II The largest expected utility is Note again that d4 is dominated by d2 and hence is not considered. Decision Maker II should make decision d1.

Example 2 Suppose the probabilities for the three states of nature in Example 1 were changed to: P(s1) = .5, P(s2) = .3, and P(s3) = .2. What is the optimal decision for a risk-neutral decision maker? What is the optimal decision for Decision Maker I? . . . for Decision Maker II? What is the value of this decision problem to Decision Maker I? to Decision Maker II? What conclusion can you draw?

Example 2 Risk-Neutral Decision Maker
EV(d1) = .5(100,000) + .3(40,000) + .2(-60,000) = 50,000 EV(d2) = .5( 50,000) + .3(20,000) + .2(-30,000) = 25,000 EV(d3) = .5( 20,000) + .3(20,000) + .2(-10,000) = 14,000 The risk-neutral optimal decision is d1.

Example 2 Expected Utility: Decision Maker I
EU(d1) = .5(100) + .3(90) + .2( 0) = 77.0 EU(d2) = .5( 94) + .3(80) + .2(40) = 79.0 EU(d3) = .5( 80) + .3(80) + .2(60) = 76.0 Decision Maker I’s optimal decision is d2.

Example 2 Expected Utility: Decision Maker II
EU(d1) = .5(100) + .3(50) + .2( 0) = 65.0 EU(d2) = .5( 58) + .3(35) + .2(10) = 41.5 EU(d3) = .5( 35) + .3(35) + .2(18) = 31.6 Decision Maker II’s optimal decision is d1.

Example 2 Value of the Decision Problem: Decision Maker I
Decision Maker I’s optimal expected utility is 79. He assigned a utility of 80 to +$20,000, and a utility of 60 to -$10,000. Linearly interpolating in this range 1 point is worth $30,000/20 = $1,500. Thus a utility of 79 is worth about $20, ,500 = $18,500.

Example 2 Value of the Decision Problem: Decision Maker II
Decision Maker II’s optimal expected utility is 65. He assigned a utility of 100 to $100,000, and a utility of 58 to $50,000. In this range, 1 point is worth $50,000/42 = $1190. Thus a utility of 65 is worth about $50, (1190) = $58,330. The decision problem is worth more to Decision Maker II (since $58,330 > $18,500).

Chapter 6 Forecasting Quantitative Approaches to Forecasting
The Components of a Time Series Measures of Forecast Accuracy Using Smoothing Methods in Forecasting Using Trend Projection in Forecasting Using Trend and Seasonal Components in Forecasting Using Regression Analysis in Forecasting Qualitative Approaches to Forecasting

Quantitative Approaches to Forecasting
Quantitative methods are based on an analysis of historical data concerning one or more time series. A time series is a set of observations measured at successive points in time or over successive periods of time. If the historical data used are restricted to past values of the series that we are trying to forecast, the procedure is called a time series method. If the historical data used involve other time series that are believed to be related to the time series that we are trying to forecast, the procedure is called a causal method.

Time Series Methods Three time series methods are: smoothing
trend projection trend projection adjusted for seasonal influence

Components of a Time Series
The trend component accounts for the gradual shifting of the time series over a long period of time. Any regular pattern of sequences of values above and below the trend line is attributable to the cyclical component of the series. The seasonal component of the series accounts for regular patterns of variability within certain time periods, such as over a year. The irregular component of the series is caused by short-term, unanticipated and non-recurring factors that affect the values of the time series. One cannot attempt to predict its impact on the time series in advance.

Measures of Forecast Accuracy
Mean Squared Error The average of the squared forecast errors for the historical data is calculated. The forecasting method or parameter(s) which minimize this mean squared error is then selected. Mean Absolute Deviation The mean of the absolute values of all forecast errors is calculated, and the forecasting method or parameter(s) which minimize this measure is selected. The mean absolute deviation measure is less sensitive to individual large forecast errors than the mean squared error measure.

Smoothing Methods In cases in which the time series is fairly stable and has no significant trend, seasonal, or cyclical effects, one can use smoothing methods to average out the irregular components of the time series. Four common smoothing methods are: Moving averages Centered moving averages Weighted moving averages Exponential smoothing

Smoothing Methods Moving Average Method
The moving average method consists of computing an average of the most recent n data values for the series and using this average for forecasting the value of the time series for the next period.

Example: Robert’s Drugs
During the past ten weeks, sales of cases of Comfort brand headache medicine at Robert's Drugs have been as follows: Week Sales Week Sales If Robert's uses a 3-period moving average to forecast sales, what is the forecast for Week 11?

Excel Spreadsheet Showing Input Data

Steps to Moving Average Using Excel Step 1: Select the Tools pull-down menu. Step 2: Select the Data Analysis option. Step 3: When the Data Analysis Tools dialog appears, choose Moving Average. Step 4: When the Moving Average dialog box appears: Enter B4:B13 in the Input Range box. Enter 3 in the Interval box. Enter C4 in the Output Range box. Select OK.

Spreadsheet Showing Results Using n = 3

Smoothing Methods Centered Moving Average Method
The centered moving average method consists of computing an average of n periods' data and associating it with the midpoint of the periods. For example, the average for periods 5, 6, and 7 is associated with period 6. This methodology is useful in the process of computing season indexes.

Smoothing Methods Weighted Moving Average Method
In the weighted moving average method for computing the average of the most recent n periods, the more recent observations are typically given more weight than older observations. For convenience, the weights usually sum to 1.

Smoothing Methods Exponential Smoothing
Using exponential smoothing, the forecast for the next period is equal to the forecast for the current period plus a proportion () of the forecast error in the current period. Using exponential smoothing, the forecast is calculated by: [the actual value for the current period] + (1- )[the forecasted value for the current period], where the smoothing constant,  , is a number between 0 and 1.

Trend Projection If a time series exhibits a linear trend, the method of least squares may be used to determine a trend line (projection) for future forecasts. Least squares, also used in regression analysis, determines the unique trend line forecast which minimizes the mean square error between the trend line forecasts and the actual observed values for the time series. The independent variable is the time period and the dependent variable is the actual observed value in the time series.

Trend Projection Using the method of least squares, the formula for the trend projection is: Tt = b0 + b1t. where: Tt = trend forecast for time period t b1 = slope of the trend line b0 = trend line projection for time 0 b1 = ntYt - t Yt b0 = Y - b1t nt 2 - (t )2 where: Yt = observed value of the time series at time period t Y = average of the observed values for Yt t = average time period for the n observations

During the past ten weeks, sales of cases of Comfort brand headache medicine at Robert's Drugs have been as follows: Week Sales Week Sales If Robert's uses exponential smoothing to forecast sales, which value for the smoothing constant ,  = .1 or  = .8, gives better forecasts?

Exponential Smoothing To evaluate the two smoothing constants, determine how the forecasted values would compare with the actual historical values in each case. Let: Yt = actual sales in week t Ft = forecasted sales in week t F1 = Y1 = 110 For other weeks, Ft+1 = .1Yt + .9Ft

Exponential Smoothing For  = .1, 1 -  = .9 F = 110 F2 = .1Y1 + .9F1 = .1(110) + .9(110) = 110 F3 = .1Y2 + .9F2 = .1(115) + .9(110) = 110.5 F4 = .1Y3 + .9F3 = .1(125) + .9(110.5) = F5 = .1Y4 + .9F4 = .1(120) + .9(111.95) = F6 = .1Y5 + .9F5 = .1(125) + .9(112.76) = F7 = .1Y6 + .9F6 = .1(120) + .9(113.98) = F8 = .1Y7 + .9F7 = .1(130) + .9(114.58) = F9 = .1Y8 + .9F8 = .1(115) + .9(116.12) = F10= .1Y9 + .9F9 = .1(110) + .9(116.01) =

Exponential Smoothing For  = .8, 1 -  = .2 F = 110 F2 = .8(110) + .2(110) = 110 F3 = .8(115) + .2(110) = 114 F4 = .8(125) + .2(114) = F5 = .8(120) + .2(122.80) = F6 = .8(125) + .2(120.56) = F7 = .8(120) + .2(124.11) = F8 = .8(130) + .2(120.82) = F9 = .8(115) + .2(128.16) = F10= .8(110) + .2(117.63) =

Mean Squared Error In order to determine which smoothing constant gives the better performance, calculate, for each, the mean squared error for the nine weeks of forecasts, weeks 2 through 10 by: [(Y2-F2)2 + (Y3-F3)2 + (Y4-F4) (Y10-F10)2]/9

 =  = .8 Week Yt Ft (Yt - Ft) Ft (Yt - Ft)2 Sum Sum MSE Sum/ Sum/9 108.25 94.17

Steps to Exponential Smoothing Using Excel Step 1: Select the Tools pull-down menu. Step 2: Select the Data Analysis option. Step 3: When the Data Analysis Tools dialog appears, choose Exponential Smoothing. Step 4: When the Exponential Smoothing dialog box appears: Enter B4:B13 in the Input Range box. Enter 0.9 (for a = 0.1) in Damping Factor box. Enter C4 in the Output Range box. Select OK.

Spreadsheet Showing Results Using a = 0.1

Repeating the Process for a = 0.8 Step 4: When the Exponential Smoothing dialog box appears: Enter B4:B13 in the Input Range box. Enter 0.2 (for a = 0.8) in Damping Factor box. Enter D4 in the Output Range box. Select OK.

Spreadsheet Showing Results Using a = 0.1 and a = 0.8

Example: Auger’s Plumbing Service
The number of plumbing repair jobs performed by Auger's Plumbing Service in each of the last nine months are listed below. Month Jobs Month Jobs Month Jobs March June September April July October May August November Forecast the number of repair jobs Auger's will perform in December using the least squares method.

Trend Projection (month) t Yt tYt t 2 (Mar.) (Apr.) (May) (June) (July) (Aug.) (Sep.) (Oct.) (Nov.) Sum

Trend Projection (continued) t = 45/9 = Y = 3480/9 = ntYt - t Yt (9)(17844) - (45)(3480) b1 = = = 7.4 nt 2 - (t) (9)(285) - (45)2 b0 = Y - b1t = (5) = T10 = (7.4)(10) =

Steps to Trend Projection Using Excel Step 1: Select an empty cell (B13) in the worksheet. Step 2: Select the Insert pull-down menu. Step 3: Choose the Function option. Step 4: When the Paste Function dialog box appears: Choose Statistical in Function Category box. Choose Forecast in the Function Name box. Select OK. more

Steps to Trend Projecting Using Excel (continued) Step 5: When the Forecast dialog box appears: Enter 10 in the x box (for month 10). Enter B4:B12 in the Known y’s box. Enter A4:A12 in the Known x’s box. Select OK.

Spreadsheet Showing Trend Projection for Month 10

Example: Auger’s Plumbing Service (B)
Forecast for December (Month 10) using a three-period (n = 3) weighted moving average with weights of .6, .3, and .1. Then, compare this Month 10 weighted moving average forecast with the Month 10 trend projection forecast.

Three-Month Weighted Moving Average The forecast for December will be the weighted average of the preceding three months: September, October, and November. F10 = .1YSep. + .3YOct. + .6YNov. = .1(399) + .3(412) + .6(408) = Trend Projection F10 = (from earlier slide) 408.3

Conclusion Due to the positive trend component in the time series, the trend projection produced a forecast that is more in tune with the trend that exists. The weighted moving average, even with heavy (.6) placed on the current period, produced a forecast that is lagging behind the changing data.

Forecasting with Trend and Seasonal Components
Steps of Multiplicative Time Series Model 1. Calculate the centered moving averages (CMAs). 2. Center the CMAs on integer-valued periods. 3. Determine the seasonal and irregular factors (StIt ). 4. Determine the average seasonal factors. 5. Scale the seasonal factors (St ). 6. Determine the deseasonalized data. 7. Determine a trend line of the deseasonalized data. 8. Determine the deseasonalized predictions. 9. Take into account the seasonality.

Example: Terry’s Tie Shop
Business at Terry's Tie Shop can be viewed as falling into three distinct seasons: (1) Christmas (November-December); (2) Father's Day (late May - mid-June); and (3) all other times. Average weekly sales (in $'s) during each of these three seasons during the past four years has been as follows: Year Season Determine a forecast for the average weekly sales in year 5 for each of the three seasons.

Dollar Moving Scaled Year Season Sales (Yt) Average StIt St Yt/St

1. Calculate the centered moving averages. There are three distinct seasons in each year. Hence, take a three season moving average to eliminate seasonal and irregular factors. For example the first moving average is: ( )/3 = 2. Center the CMAs on integer-valued periods. The first moving average computed in step 1 ( ) will be centered on season 2 of year 1. Note that the moving averages from step 1 center themselves on integer-valued periods because n is an odd number.

3. Determine the seasonal and irregular factors (St ,It ). Isolate the trend and cyclical components. For each period t, this is given by: Yt /(Moving Average for period t ). 4. Determine the average seasonal factors. Averaging all St It values corresponding to that season: Season 1: ( ) / = 1.180 Season 2: ( ) /4 = 1.238 Season 3: ( ) / =

5. Scale the seasonal factors (St ). Divide each seasonal factor by the average of the seasonal factors. Then average the seasonal factors = ( )/3 = Season 1: /1.002 = 1.178 Season 2: /1.002 = 1.236 Season 3: /1.002 = Total = 3.000

6. Determine the deseasonalized data. Divide the data point values, Yt , by St . 7. Determine a trend line of the deseasonalized data. Using the least squares method for t = 1, 2, ..., 12, gives: Tt = t

8. Determine the deseasonalized predictions. Substitute t = 13, 14, and 15 into the least squares equation: T13 = (33.96)(13) = 2022 T14 = (33.96)(14) = 2056 T15 = (33.96)(15) = 2090 9. Take into account the seasonality. Multiply each deseasonalized prediction by its seasonal factor to give the following forecasts for year 5: Season 1: (1.178)(2022) = Season 2: (1.236)(2056) = Season 3: ( .586)(2090) = 2382 2541 1225

Qualitative Approaches to Forecasting
Delphi Approach A panel of experts, each of whom is physically separated from the others and is anonymous, is asked to respond to a sequential series of questionnaires. After each questionnaire, the responses are tabulated and the information and opinions of the entire group are made known to each of the other panel members so that they may revise their previous forecast response. The process continues until some degree of consensus is achieved.

Scenario Writing Scenario writing consists of developing a conceptual scenario of the future based on a well defined set of assumptions. After several different scenarios have been developed, the decision maker determines which is most likely to occur in the future and makes decisions accordingly.

Subjective or Interactive Approaches These techniques are often used by committees or panels seeking to develop new ideas or solve complex problems. They often involve "brainstorming sessions". It is important in such sessions that any ideas or opinions be permitted to be presented without regard to its relevancy and without fear of criticism.

Chapter 7 An Introduction to Linear Programming
Linear Programming Problem Problem Formulation A Maximization Problem Graphical Solution Procedure Computer Solutions A Minimization Problem Special Cases

Linear Programming (LP) Problem
A mathematical programming problem is one that seeks to maximize an objective function subject to constraints. If both the objective function and the constraints are linear, the problem is referred to as a linear programming problem. Linear functions are functions in which each variable appears in a separate term raised to the first power and is multiplied by a constant (which could be 0). Linear constraints are linear functions that are restricted to be "less than or equal to", "equal to", or "greater than or equal to" a constant.

LP Solutions The maximization or minimization of some quantity is the objective in all linear programming problems. A feasible solution satisfies all the problem's constraints. An optimal solution is a feasible solution that results in the largest possible objective function value when maximizing (or smallest when minimizing). A graphical solution method can be used to solve a linear program with two variables.

Problem Formulation Problem formulation or modeling is the process of translating a verbal statement of a problem into a mathematical statement.

Guidelines for Model Formulation
Understand the problem thoroughly. Write a verbal description of the objective. Write a verbal description of each constraint. Define the decision variables. Write the objective in terms of the decision variables. Write the constraints in terms of the decision variables.

Example 1: A Maximization Problem
LP Formulation Max z = 5x1 + 7x2 s.t x < 6 2x1 + 3x2 < 19 x1 + x2 < 8 x1, x2 > 0

Example 1: Graphical Solution
Constraint #1 Graphed x2 8 7 6 5 4 3 2 1 x1 < 6 (6, 0) x1

Constraint #2 Graphed x2 8 7 6 5 4 3 2 1 (0, 6 1/3) 2x1 + 3x2 < 19 (9 1/2, 0) x1

Constraint #3 Graphed x2 (0, 8) 8 7 6 5 4 3 2 1 x1 + x2 < 8 (8, 0) x1

Combined-Constraint Graph x2 8 7 6 5 4 3 2 1 x1 + x2 < 8 x1 < 6 2x1 + 3x2 < 19 x1

Feasible Solution Region x2 8 7 6 5 4 3 2 1 Feasible Region x1

Objective Function Line x2 8 7 6 5 4 3 2 1 (0, 5) 5x1 + 7x2 = 35 (7, 0) x1

Optimal Solution x2 8 7 6 5 4 3 2 1 5x1 + 7x2 = 46 Optimal Solution x1

Summary of the Graphical Solution Procedure for Maximization Problems
Prepare a graph of the feasible solutions for each of the constraints. Determine the feasible region that satisfies all the constraints simultaneously.. Draw an objective function line. Move parallel objective function lines toward larger objective function values without entirely leaving the feasible region. Any feasible solution on the objective function line with the largest value is an optimal solution.

Slack and Surplus Variables
A linear program in which all the variables are non-negative and all the constraints are equalities is said to be in standard form. Standard form is attained by adding slack variables to "less than or equal to" constraints, and by subtracting surplus variables from "greater than or equal to" constraints. Slack and surplus variables represent the difference between the left and right sides of the constraints. Slack and surplus variables have objective function coefficients equal to 0.

Example 1 Standard Form Max z = 5x1 + 7x2 + 0s1 + 0s2 + 0s3
s.t x s = 6 2x1 + 3x s2 = 19 x1 + x s3 = 8 x1, x2 , s1 , s2 , s3 > 0

Extreme Points and the Optimal Solution
The corners or vertices of the feasible region are referred to as the extreme points. An optimal solution to an LP problem can be found at an extreme point of the feasible region. When looking for the optimal solution, you do not have to evaluate all feasible solution points. You have to consider only the extreme points of the feasible region.

The Five Extreme Points 8 7 6 5 4 3 2 1 5 4 Feasible Region 3 1 2 x1

Computer Solutions Computer programs designed to solve LP problems are now widely available. Most large LP problems can be solved with just a few minutes of computer time. Small LP problems usually require only a few seconds. Linear programming solvers are now part of many spreadsheet packages, such as Microsoft Excel.

Interpretation of Computer Output
In this chapter we will discuss the following output: objective function value values of the decision variables reduced costs slack/surplus In Chapter 8 we will discuss how an optimal solution is affected by a change in: a coefficient of the objective function the right-hand side value of a constraint

Example 1: Spreadsheet Solution
Partial Spreadsheet Showing Problem Data

Partial Spreadsheet Showing Solution

Interpretation of Computer Output We see from the previous slide that: Objective Function Value = 46 Decision Variable #1 (x1) = 5 Decision Variable #2 (x2) = 3 Slack in Constraint # = 1 (= 6 - 5) Slack in Constraint # = 0 (= ) Slack in Constraint # = 0 (= 8 - 8)

Reduced Cost The reduced cost for a decision variable whose value is 0 in the optimal solution is the amount the variable's objective function coefficient would have to improve (increase for maximization problems, decrease for minimization problems) before this variable could assume a positive value. The reduced cost for a decision variable with a positive value is 0.

Reduced Costs

Example 2: A Minimization Problem
LP Formulation Min z = 5x1 + 2x2 s.t x1 + 5x2 > 10 4x x2 > 12 x1 + x2 > 4 x1, x2 > 0

Graph the Constraints Constraint 1: When x1 = 0, then x2 = 2; when x2 = 0, then x1 = 5. Connect (5,0) and (0,2). The ">" side is above this line. Constraint 2: When x2 = 0, then x1 = 3. But setting x1 to 0 will yield x2 = -12, which is not on the graph. Thus, to get a second point on this line, set x1 to any number larger than 3 and solve for x2: when x1 = 5, then x2 = 8. Connect (3,0) and (5,8). The ">" side is to the right. Constraint 3: When x1 = 0, then x2 = 4; when x2 = 0, then x1 = 4. Connect (4,0) and (0,4). The ">" side is above this line.

Constraints Graphed x2 5 4 3 2 1 Feasible Region 4x1 - x2 > 12 x1 + x2 > 4 2x1 + 5x2 > 10 x1

Graph the Objective Function Set the objective function equal to an arbitrary constant (say 20) and graph it. For 5x1 + 2x2 = 20, when x1 = 0, then x2 = 10; when x2= 0, then x1 = 4. Connect (4,0) and (0,10). Move the Objective Function Line Toward Optimality Move it in the direction which lowers its value (down), since we are minimizing, until it touches the last point of the feasible region, determined by the last two constraints.

Objective Function Graphed x2 Min z = 5x1 + 2x2 4x1 - x2 > 12 x1 + x2 > 4 5 4 3 2 1 2x1 + 5x2 > 10 x1

Solve for the Extreme Point at the Intersection of the Two Binding Constraints 4x1 - x2 = 12 x1+ x2 = 4 Adding these two equations gives: 5x1 = 16 or x1 = 16/5. Substituting this into x1 + x2 = 4 gives: x2 = 4/5 Solve for the Optimal Value of the Objective Function Solve for z = 5x1 + 2x2 = 5(16/5) + 2(4/5) = 88/5. Thus the optimal solution is x1 = 16/5; x2 = 4/5; z = 88/5

Optimal Solution x2 Min z = 5x1 + 2x2 4x1 - x2 > 12 x1 + x2 > 4 5 4 3 2 1 2x1 + 5x2 > 10 Optimal: x1 = 16/5 x2 = 4/5 x1

Partial Spreadsheet Showing Formulas

Feasible Region The feasible region for a two-variable linear programming problem can be nonexistent, a single point, a line, a polygon, or an unbounded area. Any linear program falls in one of three categories: is infeasible has a unique optimal solution or alternate optimal solutions has an objective function that can be increased without bound A feasible region may be unbounded and yet there may be optimal solutions. This is common in minimization problems and is possible in maximization problems.

Special Cases Alternative Optimal Solutions
In the graphical method, if the objective function line is parallel to a boundary constraint in the direction of optimization, there are alternate optimal solutions, with all points on this line segment being optimal. Infeasibility A linear program which is overconstrained so that no point satisfies all the constraints is said to be infeasible. Unbounded (See example on upcoming slide.)

Example: Infeasible Problem
Solve graphically for the optimal solution: Max z = 2x1 + 6x2 s.t x1 + 3x2 < 12 2x1 + x2 > 8 x1, x2 > 0

Example: Infeasible Problem
There are no points that satisfy both constraints, hence this problem has no feasible region, and no optimal solution. x2 2x1 + x2 > 8 8 4x1 + 3x2 < 12 4 x1 3 4

Example: Unbounded Problem
Solve graphically for the optimal solution: Max z = 3x1 + 4x2 s.t x1 + x2 > 5 3x1 + x2 > 8 x1, x2 > 0

Example: Unbounded Problem
The feasible region is unbounded and the objective function line can be moved parallel to itself without bound so that z can be increased infinitely. x2 3x1 + x2 > 8 8 Max 3x1 + 4x2 5 x1 + x2 > 5 x1 2.67 5

Chapter 8 Linear Programming: Sensitivity Analysis and Interpretation of Solution
Introduction to Sensitivity Analysis Graphical Sensitivity Analysis Sensitivity Analysis: Computer Solution Simultaneous Changes

Standard Computer Output
Software packages such as The Management Scientist and Microsoft Excel provide the following LP information: Information about the objective function: its optimal value coefficient ranges (ranges of optimality) Information about the decision variables: their optimal values their reduced costs Information about the constraints: the amount of slack or surplus the dual prices right-hand side ranges (ranges of feasibility)

Standard Computer Output
In Chapter 7 we discussed: objective function value values of the decision variables reduced costs slack/surplus In this chapter we will discuss: changes in the coefficients of the objective function changes in the right-hand side value of a constraint

Sensitivity Analysis Sensitivity analysis (or post-optimality analysis) is used to determine how the optimal solution is affected by changes, within specified ranges, in: the objective function coefficients the right-hand side (RHS) values Sensitivity analysis is important to the manager who must operate in a dynamic environment with imprecise estimates of the coefficients. Sensitivity analysis allows him to ask certain what-if questions about the problem.

Example 1 LP Formulation Max z = 5x1 + 7x2 s.t. x1 < 6

Example 1 Graphical Solution x2 x1 + x2 < 8 Max 5x1 + 7x2 x1 < 6
4 3 2 1 x1 + x2 < 8 Max 5x1 + 7x2 x1 < 6 Optimal: x1 = 5, x2 = 3, z = 46 2x1 + 3x2 < 19 x1

Objective Function Coefficients
Let us consider how changes in the objective function coefficients might affect the optimal solution. The range of optimality for each coefficient provides the range of values over which the current solution will remain optimal. Managers should focus on those objective coefficients that have a narrow range of optimality and coefficients near the endpoints of the range.

Example 1 Changing Slope of Objective Function Feasible Region x1 5 4
8 7 6 5 4 3 2 1 5 Feasible Region 4 3 1 2 x1

Range of Optimality Graphically, the limits of a range of optimality are found by changing the slope of the objective function line within the limits of the slopes of the binding constraint lines. The slope of an objective function line, Max c1x1 + c2x2, is -c1/c2, and the slope of a constraint, a1x1 + a2x2 = b, is -a1/a2.

Example 1 Range of Optimality for c1
The slope of the objective function line is -c1/c2. The slope of the first binding constraint, x1 + x2 = 8, is -1 and the slope of the second binding constraint, x1 + 3x2 = 19, is -2/3. Find the range of values for c1 (with c2 staying 7) such that the objective function line slope lies between that of the two binding constraints: -1 < -c1/7 < -2/3 Multiplying through by -7 (and reversing the inequalities): 14/3 < c1 < 7

Example 1 Range of Optimality for c2
Find the range of values for c2 ( with c1 staying 5) such that the objective function line slope lies between that of the two binding constraints: -1 < -5/c2 < -2/3 Multiplying by -1: > 5/c2 > 2/3 Inverting, < c2/5 < 3/2 Multiplying by 5: < c2 < 15/2

Example 1 Range of Optimality for c1 and c2

Right-Hand Sides Let us consider how a change in the right-hand side for a constraint might affect the feasible region and perhaps cause a change in the optimal solution. The improvement in the value of the optimal solution per unit increase in the right-hand side is called the dual price. The range of feasibility is the range over which the dual price is applicable. As the RHS increases, other constraints will become binding and limit the change in the value of the objective function.

Dual Price Graphically, a dual price is determined by adding +1 to the right hand side value in question and then resolving for the optimal solution in terms of the same two binding constraints. The dual price is equal to the difference in the values of the objective functions between the new and original problems. The dual price for a nonbinding constraint is 0. A negative dual price indicates that the objective function will not improve if the RHS is increased.

Relevant Cost and Sunk Cost
A resource cost is a relevant cost if the amount paid for it is dependent upon the amount of the resource used by the decision variables. Relevant costs are reflected in the objective function coefficients. A resource cost is a sunk cost if it must be paid regardless of the amount of the resource actually used by the decision variables. Sunk resource costs are not reflected in the objective function coefficients.

A Cautionary Note on the Interpretation of Dual Prices
Resource cost is sunk The dual price is the maximum amount you should be willing to pay for one additional unit of the resource. Resource cost is relevant The dual price is the maximum premium over the normal cost that you should be willing to pay for one unit of the resource.

Example 1 Dual Prices Constraint 1: Since x1 < 6 is not a binding constraint, its dual price is 0. Constraint 2: Change the RHS value of the second constraint to 20 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 20 and x1 + x2 = 8. The solution is x1 = 4, x2 = 4, z = 48. Hence, the dual price = znew - zold = = 2.

Example 1 Dual Prices Constraint 3: Change the RHS value of the third constraint to 9 and resolve for the optimal point determined by the last two constraints: 2x1 + 3x2 = 19 and x1 + x2 = 9. The solution is: x1 = 8, x2 = 1, z = 47. Hence, the dual price is znew - zold = = 1.

Example 1 Dual Prices Adjustable Cells Final Reduced Objective
Allowable Cell Name Value Cost Coefficient Increase Decrease $B$8 X1 5.0 0.0 5 2 $C$8 X2 3.0 7 0.5 Constraints Shadow Constraint Price R.H. Side $B$13 #1 6 1E+30 1 $B$14 #2 19 $B$15 #3 8

Range of Feasibility The range of feasibility for a change in the right hand side value is the range of values for this coefficient in which the original dual price remains constant. Graphically, the range of feasibility is determined by finding the values of a right hand side coefficient such that the same two lines that determined the original optimal solution continue to determine the optimal solution for the problem.

Example 1 Range of Feasibility Adjustable Cells Final Reduced
Objective Allowable Cell Name Value Cost Coefficient Increase Decrease $B$8 X1 5.0 0.0 5 2 $C$8 X2 3.0 7 0.5 Constraints Shadow Constraint Price R.H. Side $B$13 #1 6 1E+30 1 $B$14 #2 19 $B$15 #3 8

Example 2: Olympic Bike Co.
Olympic Bike is introducing two new lightweight bicycle frames, the Deluxe and the Professional, to be made from special aluminum and steel alloys. The anticipated unit profits are $10 for the Deluxe and $15 for the Professional. The number of pounds of each alloy needed per frame is summarized below. A supplier delivers 100 pounds of the aluminum alloy and 80 pounds of the steel alloy weekly. How many Deluxe and Professional frames should Olympic produce each week? Aluminum Alloy Steel Alloy Deluxe Professional

Model Formulation Verbal Statement of the Objective Function Maximize total weekly profit. Verbal Statement of the Constraints Total weekly usage of aluminum alloy < 100 pounds. Total weekly usage of steel alloy < 80 pounds. Definition of the Decision Variables x1 = number of Deluxe frames produced weekly. x2 = number of Professional frames produced weekly.

Model Formulation (Continued) Max 10x1 + 15x2 (Total Weekly Profit) s.t x x2 < (Aluminum Available) 3x x2 < (Steel Available) x1, x2 > 0

Optimal Solution According to the output: x1 (Deluxe frames) = 15 x2 (Professional frames) = 17.5 Objective function value = $412.50

Range of Optimality Question Suppose the profit on deluxe frames is increased to $20. Is the above solution still optimal? What is the value of the objective function when this unit profit is increased to $20?

Sensitivity Report

Range of Optimality Answer The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and Since 20 is within this range, the optimal solution will not change. The optimal profit will change: 20x1 + 15x2 = 20(15) + 15(17.5) = $

Range of Optimality Question If the unit profit on deluxe frames were $6 instead of $10, would the optimal solution change?

Range of Optimality

Range of Optimality Answer The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 7.5 and Since 6 is outside this range, the optimal solution would change.

Range of Optimality and 100% Rule
The 100% rule states that simultaneous changes in objective function coefficients will not change the optimal solution as long as the sum of the percentages of the change divided by the corresponding maximum allowable change in the range of optimality for each coefficient does not exceed 100%.

Range of Optimality and 100% Rule Question If simultaneously the profit on Deluxe frames was raised to $16 and the profit on Professional frames was raised to $17, would the current solution be optimal? Answer If c1 = 16, the amount c1 changed is = 6 . The maximum allowable increase is = 12.5, so this is a 6/12.5 = 48% change. If c2 = 17, the amount that c2 changed is = 2. The maximum allowable increase is = 5 so this is a 2/5 = 40% change. The sum of the change percentages is 88%. Since this does not exceed 100%, the optimal solution would not change.

Range of Feasibility and 100% Rule
The 100% rule states that simultaneous changes in right-hand sides will not change the dual prices as long as the sum of the percentages of the changes divided by the corresponding maximum allowable change in the range of feasibility for each right-hand side does not exceed 100%.

Range of Feasibility and Sunk Costs Question Given that aluminum is a sunk cost, what is the maximum amount the company should pay for 50 extra pounds of aluminum?

Range of Feasibility and Sunk Costs

Range of Feasibility and Sunk Costs Answer Since the cost for aluminum is a sunk cost, the shadow price provides the value of extra aluminum. The shadow price for aluminum is the same as its dual price (for a maximization problem). The shadow price for aluminum is $3.125 per pound and the maximum allowable increase is 60 pounds. Since 50 is in this range, then the $3.125 is valid. Thus, the value of 50 additional pounds is = 50($3.125) = $

Range of Feasibility and Relevant Costs Question If aluminum were a relevant cost, what is the maximum amount the company should pay for 50 extra pounds of aluminum? Answer If aluminum were a relevant cost, the shadow price would be the amount above the normal price of aluminum the company would be willing to pay. Thus if initially aluminum cost $4 per pound, then additional units in the range of feasibility would be worth $4 + $3.125 = $7.125 per pound.

Example 3 Consider the following linear program:
Min 6x x2 ($ cost) s.t x x2 < 8 10x x2 > 30 x2 > 2 x1, x2 > 0

Example 3 The Management Scientist Output
OBJECTIVE FUNCTION VALUE = Variable Value Reduced Cost x x Constraint Slack/Surplus Dual Price

Example 3 The Management Scientist Output (Continued)
OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit x x No Upper Limit RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit No Upper Limit

Example 3 Optimal Solution According to the output: x1 = 1.5 x2 = 2.0
Objective function value =

Example 3 Range of Optimality Question
Suppose the unit cost of x1 is decreased to $4. Is the current solution still optimal? What is the value of the objective function when this unit cost is decreased to $4?

Example 3 OBJECTIVE COEFFICIENT RANGES
Variable Lower Limit Current Value Upper Limit x x No Upper Limit RIGHTHAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit No Upper Limit

Example 3 Range of Optimality Answer
The output states that the solution remains optimal as long as the objective function coefficient of x1 is between 0 and 12. Since 4 is within this range, the optimal solution will not change. However, the optimal total cost will be affected: 6x1 + 9x2 = 4(1.5) + 9(2.0) = $24.00.

Example 3 Range of Optimality Question
How much can the unit cost of x2 be decreased without concern for the optimal solution changing?

Example 3 Range of Optimality Answer
The output states that the solution remains optimal as long as the objective function coefficient of x2 does not fall below 4.5.

Example 3 Range of Optimality and 100% Rule Question
If simultaneously the cost of x1 was raised to $7.5 and the cost of x2 was reduced to $6, would the current solution remain optimal? Answer If c1 = 7.5, the amount c1 changed is = The maximum allowable increase is = 6, so this is a 1.5/6 = 25% change. If c2 = 6, the amount that c2 changed is = 3. The maximum allowable decrease is = 4.5, so this is a 3/4.5 = 66.7% change. The sum of the change percentages is 25% % = 91.7%. Since this does not exceed 100% the optimal solution would not change.

Example 3 Range of Feasibility Question
If the right-hand side of constraint 3 is increased by 1, what will be the effect on the optimal solution?

Example 3 Range of Feasibility Answer
A dual price represents the improvement in the objective function value per unit increase in the right-hand side. A negative dual price indicates a deterioration (negative improvement) in the objective, which in this problem means an increase in total cost because we're minimizing. Since the right-hand side remains within the range of feasibility, there is no change in the optimal solution. However, the objective function value increases by $4.50.

Chapter 9 Linear Programming Applications
Blending Problem Portfolio Planning Problem Product Mix Problem Transportation Problem

Blending Problem Frederick's Feed Company receives four raw grains from which it blends its dry pet food. The pet food advertises that each 8-ounce can meets the minimum daily requirements for vitamin C, protein and iron. The cost of each raw grain as well as the vitamin C, protein, and iron units per pound of each grain are summarized on the next slide. Frederick's is interested in producing the 8-ounce mixture at minimum cost while meeting the minimum daily requirements of 6 units of vitamin C, 5 units of protein, and 5 units of iron.

Blending Problem Vitamin C Protein Iron
Grain Units/lb Units/lb Units/lb Cost/lb

Blending Problem Define the decision variables
xj = the pounds of grain j (j = 1,2,3,4) used in the 8-ounce mixture Define the objective function Minimize the total cost for an 8-ounce mixture: MIN .75x x x x4

Blending Problem Define the constraints
Total weight of the mix is 8-ounces (.5 pounds): (1) x1 + x2 + x3 + x4 = .5 Total amount of Vitamin C in the mix is at least 6 units: (2) 9x1 + 16x2 + 8x3 + 10x4 > 6 Total amount of protein in the mix is at least 5 units: (3) 12x1 + 10x2 + 10x3 + 8x4 > 5 Total amount of iron in the mix is at least 5 units: (4) 14x2 + 15x3 + 7x4 > 5 Nonnegativity of variables: xj > 0 for all j

Blending Problem The Management Scientist Output
OBJECTIVE FUNCTION VALUE = VARIABLE VALUE REDUCED COSTS X X X X Thus, the optimal blend is about .10 lb. of grain 1, .21 lb. of grain 2, .09 lb. of grain 3, and .10 lb. of grain 4. The mixture costs Frederick’s 40.6 cents.

Portfolio Planning Problem
Winslow Savings has $20 million available for investment. It wishes to invest over the next four months in such a way that it will maximize the total interest earned over the four month period as well as have at least $10 million available at the start of the fifth month for a high rise building venture in which it will be participating.

For the time being, Winslow wishes to invest only in 2-month government bonds (earning 2% over the 2-month period) and 3-month construction loans (earning 6% over the 3-month period). Each of these is available each month for investment. Funds not invested in these two investments are liquid and earn 3/4 of 1% per month when invested locally.

Formulate a linear program that will help Winslow Savings determine how to invest over the next four months if at no time does it wish to have more than $8 million in either government bonds or construction loans.

Define the decision variables gj = amount of new investment in government bonds in month j cj = amount of new investment in construction loans in month j lj = amount invested locally in month j, where j = 1,2,3,4

Define the objective function Maximize total interest earned over the 4-month period. MAX (interest rate on investment)(amount invested) MAX .02g g g g4 + .06c c c c4 l l l l4

Define the constraints Month 1's total investment limited to $20 million: (1) g1 + c1 + l1 = 20,000,000 Month 2's total investment limited to principle and interest invested locally in Month 1: (2) g2 + c2 + l2 = l1 or g2 + c l1 + l2 = 0

Define the constraints (continued) Month 3's total investment amount limited to principle and interest invested in government bonds in Month 1 and locally invested in Month 2: (3) g3 + c3 + l3 = 1.02g l2 or g1 + g3 + c l2 + l3 = 0

Define the constraints (continued) Month 4's total investment limited to principle and interest invested in construction loans in Month 1, goverment bonds in Month 2, and locally invested in Month 3: (4) g4 + c4 + l4 = 1.06c g l3 or g2 + g c1 + c l3 + l4 = 0 $10 million must be available at start of Month 5: (5) c g l4 > 10,000,000

Define the constraints (continued) No more than $8 million in government bonds at any time: (6) g < 8,000,000 (7) g1 + g2 < 8,000,000 (8) g2 + g3 < 8,000,000 (9) g3 + g4 < 8,000,000

Define the constraints (continued) No more than $8 million in construction loans at any time: (10) c1 < 8,000,000 (11) c1 + c2 < 8,000,000 (12) c1 + c2 + c3 < 8,000,000 (13) c2 + c3 + c4 < 8,000,000 Nonnegativity: gj, cj, lj > 0 for j = 1,2,3,4

Problem: Floataway Tours
Floataway Tours has $420,000 that may be used to purchase new rental boats for hire during the summer. The boats can be purchased from two different manufacturers. Floataway Tours would like to purchase at least 50 boats and would like to purchase the same number from Sleekboat as from Racer to maintain goodwill. At the same time, Floataway Tours wishes to have a total seating capacity of at least 200. Pertinent data concerning the boats are summarized on the next slide. Formulate this problem as a linear program.

Data Maximum Expected Boat Builder Cost Seating Daily Profit Speedhawk Sleekboat $ $ 70 Silverbird Sleekboat $ $ 80 Catman Racer $ $ 50 Classy Racer $ $110

Define the decision variables x1 = number of Speedhawks ordered x2 = number of Silverbirds ordered x3 = number of Catmans ordered x4 = number of Classys ordered Define the objective function Maximize total expected daily profit: Max: (Expected daily profit per unit) x (Number of units) Max: 70x1 + 80x2 + 50x x4

Define the constraints (1) Spend no more than $420,000: 6000x x x x4 < 420,000 (2) Purchase at least 50 boats: x1 + x2 + x3 + x4 > 50 (3) Number of boats from Sleekboat equals number of boats from Racer: x1 + x2 = x3 + x4 or x1 + x2 - x3 - x4 = 0

Define the constraints (continued) (4) Capacity at least 200: 3x1 + 5x2 + 2x3 + 6x4 > 200 Nonnegativity of variables: xj > 0, for j = 1,2,3,4

Complete Formulation Max 70x1 + 80x2 + 50x x4 s.t. 6000x x x x4 < 420,000 x1 + x2 + x3 + x4 > 50 x1 + x2 - x3 - x4 = 0 3x1 + 5x2 + 2x3 + 6x4 > 200 xj > 0, for j = 1,2,3,4

The Management Science Output OBJECTIVE FUNCTION VALUE = Variable Value Reduced Cost x x x x Constraint Slack/Surplus Dual Price

Solution Summary Purchase 28 Speedhawks from Sleekboat. Purchase 28 Classy’s from Racer. Total expected daily profit is $5, The minimum number of boats was exceeded by 6 (surplus for constraint #2). The minimum seating capacity was exceeded by 52 (surplus for constraint #4).

Sensitivity Report

Problem: U.S. Navy The Navy has 9,000 pounds of material in Albany, Georgia which it wishes to ship to three installations: San Diego, Norfolk, and Pensacola. They require 4,000, 2,500, and 2,500 pounds, respectively. Government regulations require equal distribution of shipping among the three carriers. The shipping costs per pound for truck, railroad, and airplane transit are shown on the next slide. Formulate and solve a linear program to determine the shipping arrangements (mode, destination, and quantity) that will minimize the total shipping cost.

Problem: U.S. Navy Data Destination Mode San Diego Norfolk Pensacola
Truck $ $ $ 5 Railroad Airplane

Problem: U.S. Navy Define the Decision Variables
We want to determine the pounds of material, xij , to be shipped by mode i to destination j. The following table summarizes the decision variables: San Diego Norfolk Pensacola Truck x x x13 Railroad x x x23 Airplane x x x33

Problem: U.S. Navy Define the Objective Function
Minimize the total shipping cost. Min: (shipping cost per pound for each mode per destination pairing) x (number of pounds shipped by mode per destination pairing). Min: 12x11 + 6x12 + 5x x x22 + 9x23 + 30x x x33

Problem: U.S. Navy Define the Constraints
Equal use of transportation modes: (1) x11 + x12 + x13 = 3000 (2) x21 + x22 + x23 = 3000 (3) x31 + x32 + x33 = 3000 Destination material requirements: (4) x11 + x21 + x31 = 4000 (5) x12 + x22 + x32 = 2500 (6) x13 + x23 + x33 = 2500 Nonnegativity of variables: xij > 0, i = 1,2,3 and j = 1,2,3

Problem: U.S. Navy Partial Spreadsheet Showing Problem Data

Problem: U.S. Navy Partial Spreadsheet Showing Solution

Problem: U.S. Navy The Management Scientist Output
OBJECTIVE FUNCTION VALUE = Variable Value Reduced Cost x x x x x x x x x

Problem: U.S. Navy Solution Summary
San Diego will receive 1000 lbs. by truck and 3000 lbs. by airplane. Norfolk will receive 2000 lbs. by truck and 500 lbs. by railroad. Pensacola will receive 2500 lbs. by railroad. The total shipping cost will be $142,000.

Chapter 10 Transportation, Assignment, and Transshipment Problems
The Transportation Problem: The Network Model and a Linear Programming Formulation The Assignment Problem: The Network Model and a Linear Programming Formulation The Transshipment Problem: The Network Model and a Linear Programming Formulation

Transportation, Assignment, and Transshipment Problems
A network model is one which can be represented by a set of nodes, a set of arcs, and functions (e.g. costs, supplies, demands, etc.) associated with the arcs and/or nodes. Transportation, assignment, and transshipment problems of this chapter, as well as PERT/CPM problems (Chapter 12), are all examples of network problems.

Transportation, Assignment, and Transshipment Problems
Each of the three models of this chapter can be formulated as linear programs and solved by general purpose linear programming codes. For each of the three models, if the right-hand side of the linear programming formulations are all integers, the optimal solution will be in terms of integer values for the decision variables. However, there are many computer packages (including The Management Scientist) which contain separate computer codes for these models which take advantage of their network structure.

Transportation Problem
The transportation problem seeks to minimize the total shipping costs of transporting goods from m origins (each with a supply si ) to n destinations (each with a demand dj ), when the unit shipping cost from an origin, i, to a destination, j, is cij. The network representation for a transportation problem with two sources and three destinations is given on the next slide.

Network Representation 1 d1 c11 1 s1 c12 c13 2 d2 c21 c22 2 s2 c23 3 d3 SOURCES DESTINATIONS

LP Formulation The linear programming formulation in terms of the amounts shipped from the origins to the destinations, xij , can be written as: Min cijxij i j s.t. xij < si for each origin i j xij = dj for each destination j i xij > 0 for all i and j

LP Formulation Special Cases The following special-case modifications to the linear programming formulation can be made: Minimum shipping guarantee from i to j: xij > Lij Maximum route capacity from i to j: xij < Lij Unacceptable route: Remove the corresponding decision variable.

Example: BBC Building Brick Company (BBC) has orders for 80 tons of bricks at three suburban locations as follows: Northwood tons, Westwood tons, and Eastwood tons. BBC has two plants, each of which can produce 50 tons per week. How should end of week shipments be made to fill the above orders given the following delivery cost per ton: Northwood Westwood Eastwood Plant Plant

Example: BBC Partial Spreadsheet Showing Problem Data

Example: BBC Partial Spreadsheet Showing Optimal Solution

Example: BBC Optimal Solution From To Amount Cost
Plant 1 Northwood Plant 1 Westwood ,350 Plant 2 Northwood Plant 2 Eastwood Total Cost = $2,490

Example: BBC Partial Sensitivity Report (first half)

Example: BBC Partial Sensitivity Report (second half)

Assignment Problem An assignment problem seeks to minimize the total cost assignment of m agents to m tasks, given that the cost of agent i performing task j is cij. It assumes all agents are assigned and each task is performed. An assignment problem is a special case of a transportation problem in which all supplies and all demands are equal to 1; hence assignment problems may be solved as linear programs. The network representation of an assignment problem with three agents and three tasks is shown on the next slide.

Assignment Problem Network Representation c11 c12 c13 c21 c22 c23 c32
AGENTS TASKS

Assignment Problem Linear Programming Formulation Min cijxij i j
s.t. xij = for each agent i j xij = for each task j i xij = 0 or 1 for all i and j.

Assignment Problem LP Formulation Special Cases
Number of agents exceeds the number of tasks: xij < for each agent i j Number of tasks exceeds the number of agents: Add enough dummy agents to equalize the number of agents and the number of tasks. The objective function coefficients for these new variable would be zero.

Assignment Problem LP Formulation Special Cases (continued)
The assignment alternatives are evaluated in terms of revenue or profit: Solve as a maximization problem. An assignment is unacceptable: Remove the corresponding decision variable. An agent is permitted to work a tasks: xij < a for each agent i j

Example: Hungry Owner A contractor pays his subcontractors a fixed fee plus mileage for work performed. On a given day the contractor is faced with three electrical jobs associated with various projects. Given below are the distances between the subcontractors and the projects. Projects A B C Westside Subcontractors Federated Goliath Universal How should the contractors be assigned to minimize total costs?

Example: Hungry Owner Network Representation Subcontractors Projects
West. 50 A 36 16 28 Fed. 30 B 18 Subcontractors Projects 35 Gol. 32 C 20 25 25 Univ. 14

Example: Hungry Owner Linear Programming Formulation
Min 50x11+36x12+16x13+28x21+30x22+18x23 +35x31+32x32+20x33+25x41+25x42+14x43 s.t. x11+x12+x13 < 1 x21+x22+x23 < 1 x31+x32+x33 < 1 x41+x42+x43 < 1 x11+x21+x31+x41 = 1 x12+x22+x32+x42 = 1 x13+x23+x33+x43 = 1 xij = 0 or 1 for all i and j Agents Tasks

Example: Hungry Owner Optimal Assignment
Subcontractor Project Distance Westside C Federated A Goliath (unassigned) Universal B Total Distance = 69 miles

Transshipment Problem
Transshipment problems are transportation problems in which a shipment may move through intermediate nodes (transshipment nodes)before reaching a particular destination node. Transshipment problems can be converted to larger transportation problems and solved by a special transportation program. Transshipment problems can also be solved by general purpose linear programming codes. The network representation for a transshipment problem with two sources, three intermediate nodes, and two destinations is shown on the next slide.

Network Representation 3 c36 c13 c37 1 6 s1 d1 c14 c15 c46 4 c47 c23 c24 7 2 c56 d2 s2 c25 5 c57 SOURCES INTERMEDIATE NODES DESTINATIONS

Linear Programming Formulation xij represents the shipment from node i to node j Min cijxij i j s.t. xij < si for each origin i j xik - xkj = 0 for each intermediate i j node k xij = dj for each destination j i xij > for all i and j

Example: Transshipping
Thomas Industries and Washburn Corporation supply three firms (Zrox, Hewes, Rockwright) with customized shelving for its offices. They both order shelving from the same two manufacturers, Arnold Manufacturers and Supershelf, Inc. Currently weekly demands by the users are 50 for Zrox, 60 for Hewes, and 40 for Rockwright. Both Arnold and Supershelf can supply at most 75 units to its customers. Additional data is shown on the next slide.

Because of long standing contracts based on past orders, unit costs from the manufacturers to the suppliers are: Thomas Washburn Arnold Supershelf The cost to install the shelving at the various locations are: Zrox Hewes Rockwright Thomas Washburn

Network Representation Zrox ZROX 50 1 Arnold 5 Thomas 75 ARNOLD 5 8 8 Hewes 60 HEWES 3 4 7 Super Shelf Wash- Burn 75 WASH BURN 4 4 Rock- Wright 40

Linear Programming Formulation Decision Variables Defined xij = amount shipped from manufacturer i to supplier j xjk = amount shipped from supplier j to customer k where i = 1 (Arnold), 2 (Supershelf) j = 3 (Thomas), 4 (Washburn) k = 5 (Zrox), 6 (Hewes), 7 (Rockwright) Objective Function Defined Minimize Overall Shipping Costs: Min 5x13 + 8x14 + 7x23 + 4x24 + 1x35 + 5x36 + 8x37 + 3x45 + 4x46 + 4x47

Constraints Defined Amount Out of Arnold: x13 + x14 < 75 Amount Out of Supershelf: x23 + x24 < 75 Amount Through Thomas: x13 + x23 - x35 - x36 - x37 = 0 Amount Through Washburn: x14 + x24 - x45 - x46 - x47 = 0 Amount Into Zrox: x35 + x45 = 50 Amount Into Hewes: x36 + x46 = 60 Amount Into Rockwright: x37 + x47 = 40 Non-negativity of Variables: xij > 0, for all i and j.

Optimal Solution (from Management Scientist ) Objective Function Value = Variable Value Reduced Costs X X X X X X X X X X

Optimal Solution Zrox ZROX 50 50 75 1 Arnold 5 Thomas 75 ARNOLD 5 25 8 8 Hewes 60 35 HEWES 3 4 7 Super Shelf Wash- Burn 40 75 WASH BURN 4 75 4 Rock- Wright 40

Optimal Solution (continued) Constraint Slack/Surplus Dual Prices

Optimal Solution (continued) OBJECTIVE COEFFICIENT RANGES Variable Lower Limit Current Value Upper Limit X X No Limit X No Limit X No Limit X No Limit X X No Limit X No Limit X X No Limit

Optimal Solution (continued) RIGHT HAND SIDE RANGES Constraint Lower Limit Current Value Upper Limit No Limit

Chapter 11 Integer Linear Programming
Types of Integer Linear Programming Models Graphical and Computer Solutions for an All-Integer Linear Program Applications Involving 0-1 Variables Modeling Flexibility Provided by 0-1 Variables

Types of Integer Programming Models
A linear program in which all the variables are restricted to be integers is called an all-integer linear program (ILP). The linear program that results from dropping the integer requirements is called the LP Relaxation of the ILP. If only a subset of the variables are restricted to be integers, the problem is called a mixed-integer linear program (MILP). Binary variables are variables whose values are restricted to be 0 or 1. If all variables are restricted to be 0 or 1, the problem is called a 0-1 or binary integer linear program.

Example: All-Integer LP
Consider the following all-integer linear program: Max 3x1 + 2x2 s.t x1 + x2 < 9 x1 + 3x2 < 7 -x1 + x2 < 1 x1, x2 > 0 and integer

LP Relaxation Solving the problem as a linear program ignoring the integer constraints, the optimal solution to the linear program gives fractional values for both x1 and x2. From the graph on the next slide, we see that the optimal solution to the linear program is: x1 = 2.5, x2 = 1.5, z = 10.5

LP Relaxation x2 5 -x1 + x2 < 1 3x1 + x2 < 9 4 Max 3x1 + 2x2 3 LP Optimal (2.5, 1.5) 2 x1 + 3x2 < 7 1 x1

Rounding Up If we round up the fractional solution (x1 = 2.5, x2 = 1.5) to the LP relaxation problem, we get x1 = 3 and x2 = 2. From the graph on the next page, we see that this point lies outside the feasible region, making this solution infeasible.

Rounded Up Solution x2 5 -x1 + x2 < 1 3x1 + x2 < 9 4 Max 3x1 + 2x2 3 ILP Infeasible (3, 2) 2 LP Optimal (2.5, 1.5) x1 + 3x2 < 7 1 x1

Rounding Down By rounding the optimal solution down to x1 = 2, x2 = 1, we see that this solution indeed is an integer solution within the feasible region, and substituting in the objective function, it gives z = 8. We have found a feasible all-integer solution, but have we found the OPTIMAL all-integer solution? The answer is NO! The optimal solution is x1 = 3 and x2 = 0 giving z = 9, as evidenced in the next two slides.

Complete Enumeration of Feasible ILP Solutions There are eight feasible integer solutions to this problem: x x z optimal solution

5 -x1 + x2 < 1 3x1 + x2 < 9 4 Max 3x1 + 2x2 3 ILP Optimal (3, 0) 2 x1 + 3x2 < 7 1 x1

Partial Spreadsheet Showing Formulas

Partial Spreadsheet Showing Optimal Solution

Special 0-1 Constraints When xi and xj represent binary variables designating whether projects i and j have been completed, the following special constraints may be formulated: At most k out of n projects will be completed: xj < k j Project j is conditional on project i: xj - xi < 0 Project i is a corequisite for project j: xj - xi = 0 Projects i and j are mutually exclusive: xi + xj < 1

Example: Metropolitan Microwaves
Metropolitan Microwaves, Inc. is planning to expand its operations into other electronic appliances. The company has identified seven new product lines it can carry. Relevant information about each line follows: Initial Floor Space Exp. Rate Product Line Invest (Sq.Ft.) of Return 1. Black & White TVs $ 6, % 2. Color TVs , 3. Large Screen TVs , 4. VHS VCRs , 5. Beta VCRs , 6. Video Games , 7. Home Computers ,

Metropolitan has decided that they should not stock large screen TVs unless they stock either B&W or color TVs. Also, they will not stock both types of VCRs, and they will stock video games if they stock color TVs. Finally, the company wishes to introduce at least three new product lines. If the company has $45,000 to invest and 420 sq. ft. of floor space available, formulate an integer linear program for Metropolitan to maximize its overall expected rate of return.

Define the Decision Variables xj = 1 if product line j is introduced; = 0 otherwise. Define the Objective Function Maximize total overall expected return: Max (6000)x (12000)x (20000)x3 + .102(14000)x (15000)x (2000)x6 + .132(32000)x7

Define the Constraints 1) Money: 6x1 + 12x2 + 20x3 + 14x4 + 15x5 + 2x6 + 32x7 < 45 2) Space: 125x1 +150x2 +200x3 +40x4 +40x5 +20x6 +100x7 < 420 3) Stock large screen TVs only if stock B&W or color: x1 + x2 > x3 or x1 + x2 - x3 > 0

Define the Constraints (continued) 4) Do not stock both types of VCRs: x4 + x5 < 1 5) Stock video games if they stock color TV's: x2 - x6 > 0 6) At least 3 new lines: x1 + x2 + x3 + x4 + x5 + x6 + x7 > 3 7) Variables are 0 or 1: xj = 0 or 1 for j = 1, , , 7

Partial Spreadsheet Showing Example Formulas

Solver Parameters Dialog Box Solver Parameters ? X Set Target Cell: Solve Equal To: Max Min Value of: Close By Changing Cells: $B$13:$H$13 Guess Options Subject to the Constraints: Add $B$13:$H$13 = binary $D$17:$D$18 <= $H$17:$H$18 $D$19 >= $H$19 $D$20 <= $H$20 Reset All Change Delete Help

Solver Options Dialog Box Select Assume Linear Model Select Assume Non-negative Set Tolerance equal to 0

Example: Tom’s Tailoring
Tom's Tailoring has five idle tailors and four custom garments to make. The estimated time (in hours) it would take each tailor to make each garment is listed below. (An 'X' in the table indicates an unacceptable tailor-garment assignment.) Tailor Garment Wedding gown Clown costume X Admiral's uniform X 9 Bullfighter's outfit X

Formulate an integer program for determining the tailor-garment assignments that minimize the total estimated time spent making the four garments. No tailor is to be assigned more than one garment and each garment is to be worked on by only one tailor. This problem can be formulated as a 0-1 integer program. The LP solution to this problem will automatically be integer (0-1).

Define the decision variables xij = 1 if garment i is assigned to tailor j = 0 otherwise. Number of decision variables = [(number of garments)(number of tailors)] - (number of unacceptable assignments) = [4(5)] - 3 = 17 Define the objective function Minimize total time spent making garments: Min 19x x x x x x21 + 14x x x x x x33 + 9x x x x x45

Define the Constraints Exactly one tailor per garment: 1) x11 + x12 + x13 + x14 + x15 = 1 2) x21 + x22 + x24 + x25 = 1 3) x31 + x32 + x33 + x35 = 1 4) x42 + x43 + x44 + x45 = 1

Define the Constraints (continued) No more than one garment per tailor: 5) x11 + x21 + x31 < 1 6) x21 + x22 + x23 + x24 < 1 7) x31 + x33 + x34 < 1 8) x41 + x42 + x44 < 1 9) x51 + x52 + x53 + x54 < 1 Nonnegativity: xij > 0 for i = 1, . . ,4 and j = 1, . . ,5

Chapter 12 Project Scheduling: PERT/CPM
Project Scheduling with Known Activity Times Project Scheduling with Uncertain Activity Times Considering Time-Cost Trade-Offs

PERT/CPM PERT stands for Program Evaluation Review Technique.
CPM stands for Critical Path Method. PERT/CPM is used to plan the scheduling of individual activities that make up a project. PERT/CPM can be used to determine the earliest/latest start and finish times for each activity, the entire project completion time and the slack time for each activity.

Project Network A project network can be constructed to model the precedence of the activities. The nodes of the network represent the activities. The arcs of the network reflect the precedence relationships of the activities. A critical path for the network is a path consisting of activities with zero slack.

Determining the Critical Path
Step 1: Make a forward pass through the network as follows: For each activity i beginning at the Start node, compute: Earliest Start Time = the maximum of the earliest finish times of all activities immediately preceding activity i. (This is 0 for an activity with no predecessors.) Earliest Finish Time = (Earliest Start Time) + (Time to complete activity i. The project completion time is the maximum of the Earliest Finish Times at the Finish node.

Step 2: Make a backwards pass through the network as follows: Move sequentially backwards from the Finish node to the Start node. At a given node, j, consider all activities ending at node j. For each of these activities, (i,j), compute: Latest Finish Time = the minimum of the latest start times beginning at node j. (For node N, this is the project completion time.) Latest Start Time = (Latest Finish Time) - (Time to complete activity (i,j)).

Step 3: Calculate the slack time for each activity by: Slack = (Latest Start) - (Earliest Start), or = (Latest Finish) - (Earliest Finish). A critical path is a path of activities, from the Start node to the Finish node, with 0 slack times.

Uncertain Activity Times
In the three-time estimate approach, the time to complete an activity is assumed to follow a Beta distribution. An activity’s mean completion time is: t = (a + 4m + b)/6 An activity’s completion time variance is: 2 = ((b-a)/6)2 a = the optimistic completion time estimate b = the pessimistic completion time estimate m = the most likely completion time estimate

Uncertain Activity Times
In the three-time estimate approach, the critical path is determined as if the mean times for the activities were fixed times. The overall project completion time is assumed to have a normal distribution with mean equal to the sum of the means along the critical path and variance equal to the sum of the variances along the critical path.

Example: ABC Associates
Consider the following project: Immed. Optimistic Most Likely Pessimistic Activity Predec. Time (Hr.) Time (Hr.) Time (Hr.) A B C A D A E A F B,C G B,C H E,F I E,F J D,H K G,I

PERT Network Representation

Activity Expected Time and Variances t = (a + 4m + b)/6 2 = ((b-a)/6)2 Activity Expected Time Variance A /9 B /9 C D /9 E /36 F /9 G /9 H /9 I J /9 K /9

Earliest/Latest Times Activity ES EF LS LF Slack A *critical B C * D E F * G H I * J K * Estimated Project Completion Time: Max EF = 23

Critical Path (A-C-F-I-K) 6 4 3 5 2 1 6 11 15 20 19 22 20 23 13 19 14 20 0 6 6 7 12 13 13 18 6 9 9 13 18 23 0 4 5 9 9 11 16 18

Probability the project will be completed within 24 hrs 2 = 2A + 2C + 2F + 2H + 2K = 4/ / /9 = 2  = z = ( )/(24-23)/1.414 = .71 From the Standard Normal Distribution table: P(z < .71) = =

PERT/Cost PERT/Cost is a technique for monitoring costs during a project. Work packages (groups of related activities) with estimated budgets and completion times are evaluated. A cost status report may be calculated by determining the cost overrun or underrun for each work package. Cost overrun or underrun is calculated by subtracting the budgeted cost from the actual cost of the work package. For work in progress, overrun or underrun may be determined by subtracting the prorated budget cost from the actual cost to date.

PERT/Cost The overall project cost overrun or underrun at a particular time during a project is determined by summing the individual cost overruns and underruns to date of the work packages.

Example: How Are We Doing?
Consider the following PERT network:

Earliest/Latest Times Activity ES EF LS LF Slack A B C D E F G H I J

Activity Status (end of eleventh week) Activity Actual Cost % Complete A $6, B , C , D E , F , G , H I J

Cost Status Report (Assuming a budgeted cost of $6000 for each activity) Activity Actual Cost Value Difference A $6, (1.00)x6000 = $200 B , (1.00)x6000 = C , (.90)x6000 = D E , (.25)x6000 = F , (.75)x6000 = G , (.50)x6000 = H I J Totals $25, $26, $- 900

PERT Diagram at End of Week 11 The activity completion times are the times remaining for each activity.

Corrective Action Note that the project is currently experiencing a $900 cost underrun, but the overall completion time is now 25.5 weeks or a .5 week delay. Management should consider using some of the $900 cost savings and apply it to activity G to assist in a more rapid completion of this activity (and hence the entire project).

Critical Path Method In the Critical Path Method (CPM) approach to project scheduling, it is assumed that the normal time to complete an activity, tj , which can be met at a normal cost, cj , can be crashed to a reduced time, tj’, under maximum crashing for an increased cost, cj’. Using CPM, activity j's maximum time reduction, Mj , may be calculated by: Mj = tj - tj'. It is assumed that its cost per unit reduction, Kj , is linear and can be calculated by: Kj = (cj' - cj)/Mj.

Chapter 13 Inventory Management: Independent-Demand
Economic Order Quantity (EOQ) Model Economic Production Lot Size Model An Inventory Model with Planned Shortages Quantity Discounts for the EOQ Model A Single-Period Inventory Model with Probabilistic Demand An Order-Quantity, Reorder-Point Model with Probabilistic Demand A Periodic-Review Model with Probabilistic Demand

Inventory Models The study of inventory models is concerned with two basic questions: How much should be ordered each time When should the reordering occur The objective is to minimize total variable cost over a specified time period (assumed to be annual in the following review).

Inventory Costs Ordering cost -- salaries and expenses of processing an order, regardless of the order quantity Holding cost -- usually a percentage of the value of the item assessed for keeping an item in inventory (including finance costs, insurance, security costs, taxes, warehouse overhead, and other related variable expenses) Backorder cost -- costs associated with being out of stock when an item is demanded (including lost goodwill) Purchase cost -- the actual price of the items Other costs

Deterministic Models The simplest inventory models assume demand and the other parameters of the problem to be deterministic and constant. The deterministic models covered in this chapter are: Economic order quantity (EOQ) Economic production lot size EOQ with planned shortages EOQ with quantity discounts

Economic Order Quantity (EOQ)
The most basic of the deterministic inventory models is the economic order quantity (EOQ). The variable costs in this model are annual holding cost and annual ordering cost. For the EOQ, annual holding and ordering costs are equal.

Economic Order Quantity
Assumptions Demand is constant throughout the year at D items per year. Ordering cost is $Co per order. Holding cost is $Ch per item in inventory per year. Purchase cost per unit is constant (no quantity discount). Delivery time (lead time) is constant. Planned shortages are not permitted.

Economic Order Quantity
Formulas Optimal order quantity: Q* = 2DCo/Ch Number of orders per year: D/Q* Time between orders (cycle time): Q*/D years Total annual cost: [(1/2)Q*Ch] + [DCo/Q*] (holding + ordering)

Example: Bart’s Barometer Business
Economic Order Quantity Model Bart's Barometer Business (BBB) is a retail outlet which deals exclusively with weather equipment. Currently BBB is trying to decide on an inventory and reorder policy for home barometers. Barometers cost BBB $50 each and demand is about 500 per year distributed fairly evenly throughout the year. Reordering costs are $80 per order and holding costs are figured at 20% of the cost of the item. BBB is open 300 days a year (6 days a week and closed two weeks in August). Lead time is 60 working days.

Total Variable Cost Model Total Costs = (Holding Cost) + (Ordering Cost) TC = [Ch(Q/2)] + [Co(D/Q)] = [.2(50)(Q/2)] + [80(500/Q)] = 5Q + (40,000/Q)

Optimal Reorder Quantity Q * = 2DCo /Ch = 2(500)(80)/10 =  90 Optimal Reorder Point Lead time is m = 60 days and daily demand is d = 500/300 or1.667. Thus the reorder point r = (1.667)(60) = Bart should reorder 90 barometers when his inventory position reaches 100 (that is 10 on hand and one outstanding order).

Number of Orders Per Year Number of reorder times per year = (500/90) = 5.56 or once every (300/5.56) = 54 working days (about every 9 weeks). Total Annual Variable Cost TC = 5(90) + (40,000/90) = = $894.

We’ll now use a spreadsheet to implement the Economic Order Quantity model. We’ll confirm our earlier calculations for Bart’s problem and perform some sensitivity analysis. This spreadsheet can be modified to accommodate other inventory models presented in this chapter.

Partial Spreadsheet with Input Data

Partial Spreadsheet Showing Formulas for Output

Partial Spreadsheet Showing Output

Summary of Spreadsheet Results A 16.15% negative deviation from the EOQ resulted in only a 1.55% increase in the Total Annual Cost. Annual Holding Cost and Annual Ordering Cost are no longer equal. The Reorder Point is not affected, in this model, by a change in the Order Quantity.

Economic Production Lot Size
The economic production lot size model is a variation of the basic EOQ model. A replenishment order is not received in one lump sum as it is in the basic EOQ model. Inventory is replenished gradually as the order is produced (which requires the production rate to be greater than the demand rate). This model's variable costs are annual holding cost and annual set-up cost (equivalent to ordering cost). For the optimal lot size, annual holding and set-up costs are equal.

Assumptions Demand occurs at a constant rate of D items per year. Production rate is P items per year (and P >D). Set-up cost: $Co per run. Holding cost: $Ch per item in inventory per year. Purchase cost per unit is constant (no quantity discount). Set-up time (lead time) is constant. Planned shortages are not permitted.

Formulas Optimal production lot-size: Q * = 2DCo /[(1-D/P )Ch] Number of production runs per year: D/Q * Time between set-ups (cycle time): Q */D years Total annual cost: [(1/2)(1-D/P )Q *Ch] + [DCo/Q *] (holding + ordering)

Example: Non-Slip Tile Co.
Economic Production Lot Size Model Non-Slip Tile Company (NST) has been using production runs of 100,000 tiles, 10 times per year to meet the demand of 1,000,000 tiles annually. The set-up cost is $5,000 per run and holding cost is estimated at 10% of the manufacturing cost of $1 per tile. The production capacity of the machine is 500,000 tiles per month. The factory is open 365 days per year.

Total Annual Variable Cost Model This is an economic production lot size problem with D = 1,000,000, P = 6,000,000, Ch = .10, Co = 5,000 TC = (Holding Costs) + (Set-Up Costs) = [Ch(Q/2)(1 - D/P )] + [DCo/Q] = Q + 5,000,000,000/Q

Optimal Production Lot Size Q * = 2DCo/[(1 -D/P )Ch] = 2(1,000,000)(5,000) /[(.1)(1 - 1/6)] = 346,410 Number of Production Runs Per Year The number of runs per year = D/Q * = 2.89 times per year.

Total Annual Variable Cost How much is NST losing annually by using their present production schedule? Optimal TC = (346,410) + 5,000,000,000/346,410 = $28,868 Current TC = (100,000) + 5,000,000,000/100,000 = $54,167 Difference = 54, ,868 = $25,299

Idle Time Between Production Runs There are 2.89 cycles per year. Thus, each cycle lasts (365/2.89) = days. The time to produce 346,410 per run = (346,410/6,000,000)365 = 21.1 days. Thus, the machine is idle for = days between runs.

Maximum Inventory Current Policy: maximum inventory = (1-D/P )Q * = (1-1/6)100,000  83,333 Optimal Policy: maximum inventory = (1-1/6)346,410 = 288,675. Machine Utilization The machine is producing tiles D/P = 1/6 of the time.

EOQ with Planned Shortages
With the EOQ with planned shortages model, a replenishment order does not arrive at or before the inventory position drops to zero. Shortages occur until a predetermined backorder quantity is reached, at which time the replenishment order arrives. The variable costs in this model are annual holding, backorder, and ordering. For the optimal order and backorder quantity combination, the sum of the annual holding and backordering costs equals the annual ordering cost.

Assumptions Demand occurs at a constant rate of D items per year. Ordering cost: $Co per order. Holding cost: $Ch per item in inventory per year. Backorder cost: $Cb per item backordered per year. Purchase cost per unit is constant (no quantity discount). Set-up time (lead time) is constant. Planned shortages are permitted (backordered demand units are withdrawn from a replenishment order when it is delivered).

Formulas Optimal order quantity: Q * = 2DCo/Ch (Ch+Cb )/Cb Maximum number of backorders: S * = Q *(Ch/(Ch+Cb)) Number of orders per year: D/Q * Time between orders (cycle time): Q */D years Total annual cost: [Ch(Q *-S *)2/2Q *] + [DCo/Q *] + [S *2Cb/2Q *] (holding + ordering + backordering)

Example: Hervis Rent-a-Car
EOQ with Planned Shortages Model Hervis Rent-a-Car has a fleet of 2,500 Rockets serving the Los Angeles area. All Rockets are maintained at a central garage. On the average, eight Rockets per month require a new engine. Engines cost $850 each. There is also a $120 order cost (independent of the number of engines ordered). Hervis has an annual holding cost rate of 30% on engines. It takes two weeks to obtain the engines after they are ordered. For each week a car is out of service, Hervis loses $40 profit.

Optimal Order Policy D = 8 x 12 = 96; Co = $120; Ch = .30(850) = $255; Cb = 40 x 52 = $2080 Q * = 2DCo/Ch (Ch + Cb)/Cb = 2(96)(120)/255 x ( )/2080 =  10 S * = Q *(Ch/(Ch+Cb)) = 10(255/( )) = 1.09  1

Optimal Order Policy (continued) Demand is 8 per month or 2 per week. Since lead time is 2 weeks, lead time demand is 4. Thus, since the optimal policy is to order 10 to arrive when there is one backorder, the order should be placed when there are 3 engines remaining in inventory.

Stockout: When and How Long How many days after receiving an order does Hervis run out of engines? How long is Hervis without any engines per cycle? Inventory exists for Cb/(Cb+Ch) = 2080/( ) = of the order cycle. (Note, (Q*-S*)/Q* = also, before Q * and S * are rounded.) An order cycle is Q */D = years = 38.3 days. Thus, Hervis runs out of engines .8908(38.3) = 34 days after receiving an order. Hervis is out of stock for approximately = 4 days.

EOQ with Quantity Discounts
The EOQ with quantity discounts model is applicable where a supplier offers a lower purchase cost when an item is ordered in larger quantities. This model's variable costs are annual holding, ordering and purchase costs. For the optimal order quantity, the annual holding and ordering costs are not necessarily equal.

Assumptions Demand occurs at a constant rate of D items per year. Ordering Cost is $Co per order. Holding Cost is $Ch = $CiI per item in inventory per year (note holding cost is based on the cost of the item, Ci). Purchase Cost is $C1 per item if the quantity ordered is between 0 and x1, $C2 if the order quantity is between x1 and x2 , etc. Delivery time (lead time) is constant. Planned shortages are not permitted.

Formulas Optimal order quantity: the procedure for determining Q * will be demonstrated Number of orders per year: D/Q * Time between orders (cycle time): Q */D years Total annual cost: [(1/2)Q *Ch] + [DCo/Q *] + DC (holding + ordering + purchase)

Example: Nick's Camera Shop
EOQ with Quantity Discounts Model Nick's Camera Shop carries Zodiac instant print film. The film normally costs Nick $3.20 per roll, and he sells it for $ Zodiac film has a shelf life of 18 months. Nick's average sales are 21 rolls per week. His annual inventory holding cost rate is 25% and it costs Nick $20 to place an order with Zodiac. If Zodiac offers a 7% discount on orders of 400 rolls or more, a 10% discount for 900 rolls or more, and a 15% discount for 2000 rolls or more, determine Nick's optimal order quantity. D = 21(52) = 1092; Ch = .25(Ci); Co = 20

Unit-Prices’ Economical, Feasible Order Quantities For C4 = .85(3.20) = $2.72 To receive a 15% discount Nick must order at least 2,000 rolls. Unfortunately, the film's shelf life is 18 months. The demand in 18 months (78 weeks) is 78 X 21 = 1638 rolls of film. If he ordered 2,000 rolls he would have to scrap 372 of them. This would cost more than the 15% discount would save.

Unit-Prices’ Economical, Feasible Order Quantities For C3 = .90(3.20) = $2.88 Q3* = 2DCo/Ch = 2(1092)(20)/[.25(2.88)] = (not feasible) The most economical, feasible quantity for C3 is 900. For C2 = .93(3.20) = $2.976 Q2* = 2DCo/Ch = 2(1092)(20)/[.25(2.976)] = (not feasible) The most economical, feasible quantity for C2 is 400.

Unit-Prices’ Economical, Feasible Order Quantities For C1 = 1.00(3.20) = $3.20 Q1* = 2DCo/Ch = 2(1092)(20)/.25(3.20) = (feasible) When we reach a computed Q that is feasible we stop computing Q's. (In this problem we have no more to compute anyway.)

Total Cost Comparison Compute the total cost for the most economical, feasible order quantity in each price category for which a Q * was computed. TCi = (1/2)(Qi*Ch) + (DCo/Qi*) + DCi TC3 = (1/2)(900)(.72) +((1092)(20)/900)+(1092)(2.88) = 3493 TC2 = (1/2)(400)(.744)+((1092)(20)/400)+(1092)(2.976) = 3453 TC1 = (1/2)(234)(.80) +((1092)(20)/234)+(1092)(3.20) = 3681 Comparing the total costs for 234, 400 and 900, the lowest total annual cost is $ Nick should order 400 rolls at a time.

Probabilistic Models In many cases demand (or some other factor) is not known with a high degree of certainty and a probabilistic inventory model should actually be used. These models tend to be more complex than deterministic models. The probabilistic models covered in this chapter are: single-period order quantity reorder-point quantity periodic-review order quantity

Single-Period Order Quantity
A single-period order quantity model (sometimes called the newsboy problem) deals with a situation in which only one order is placed for the item and the demand is probabilistic. If the period's demand exceeds the order quantity, the demand is not backordered and revenue (profit) will be lost. If demand is less than the order quantity, the surplus stock is sold at the end of the period (usually for less than the original purchase price).

Assumptions Period demand follows a known probability distribution: normal: mean is µ, standard deviation is  uniform: minimum is a, maximum is b Cost of overestimating demand: $co Cost of underestimating demand: $cu Shortages are not backordered. Period-end stock is sold for salvage (not held in inventory).

Formulas Optimal probability of no shortage: P(demand < Q *) = cu/(cu+co) Optimal probability of shortage: P(demand > Q *) = 1 - cu/(cu+co) Optimal order quantity, based on demand distribution: normal: Q * = µ + z uniform: Q * = a + P(demand < Q *)(b-a)

Example: McHardee Press
Single-Period Order Quantity Model McHardee Press publishes the Fast Food Restaurant Menu Book and wishes to determine how many copies to print. There is a fixed cost of $5,000 to produce the book and the incremental profit per copy is $.45. Any unsold copies of the book can be sold at salvage at a $.55 loss. Sales for this edition are estimated to be normally distributed. The most likely sales volume is 12,000 copies and they believe there is a 5% chance that sales will exceed 20,000. How many copies should be printed?

Single-Period Order Quantity m = 12,000. To find  note that z = 1.65 corresponds to a 5% tail probability. Therefore, (20, ,000) = 1.65 or  = 4848 Using incremental analysis with Co = .55 and Cu = .45, (Cu/(Cu+Co)) = .45/( ) = .45 Find Q * such that P(D < Q *) = The probability of 0.45 corresponds to z = Thus, Q * = 12, (4848) = 11,418 books

Single-Period Order Quantity (revised) If any unsold copies of the book can be sold at salvage at a $.65 loss, how many copies should be printed? Co = .65, (Cu/(Cu + Co)) = .45/( ) = .4091 Find Q * such that P(D < Q *) = z = -.23 gives this probability. Thus, Q * = 12, (4848) = 10,885 books However, since this is less than the breakeven volume of 11,111 books (= 5000/.45), no copies should be printed because if the company produced only 10,885 copies it will not recoup its $5,000 fixed cost of producing the book.

Reorder Point Quantity
A firm's inventory position consists of the on-hand inventory plus on-order inventory (all amounts previously ordered but not yet received). An inventory item is reordered when the item's inventory position reaches a predetermined value, referred to as the reorder point. The reorder point represents the quantity available to meet demand during lead time. Lead time is the time span starting when the replenishment order is placed and ending when the order arrives.

Reorder Point Quantity
Under deterministic conditions, when both demand and lead time are constant, the reorder point associated with EOQ-based models is set equal to lead time demand. Under probabilistic conditions, when demand and/or lead time varies, the reorder point often includes safety stock. Safety stock is the amount by which the reorder point exceeds the expected (average) lead time demand.

Safety Stock and Service Level
The amount of safety stock in a reorder point determines the chance of a stockout during lead time. The complement of this chance is called the service level. Service level, in this context, is defined as the probability of not incurring a stockout during any one lead time. Service level, in this context, also is the long-run proportion of lead times in which no stockouts occur.

Reorder Point Assumptions Lead-time demand is normally distributed
with mean µ and standard deviation . Approximate optimal order quantity: EOQ Service level is defined in terms of the probability of no stockouts during lead time and is reflected in z. Shortages are not backordered. Inventory position is reviewed continuously.

Reorder Point Formulas Reorder point: r = µ + z Safety stock: z
Average inventory: 1/2(Q) + z Total annual cost: [(1/2)Q *Ch] + [z Ch] + [DCo/Q *] (holding(normal) + holding(safety) + ordering)

Example: Robert’s Drug
Reorder Point Model Robert's Drugs is a drug wholesaler supplying 55 independent drug stores. Roberts wishes to determine an optimal inventory policy for Comfort brand headache remedy. Sales of Comfort are relatively constant as the past 10 weeks of data indicate: Week Sales (cases) Week Sales (cases)

Each case of Comfort costs Roberts $10 and Roberts uses a 14% annual holding cost rate for its inventory. The cost to prepare a purchase order for Comfort is $12. What is Roberts’ optimal order quantity?

Optimal Order Quantity The average weekly sales over the 10 week period is 120 cases. Hence D = 120 X 52 = 6,240 cases per year; Ch = (.14)(10) = 1.40; Co = 12. Q * = 2DCo/Ch = (2)(6240)(12)/1.40 = 327

The lead time for a delivery of Comfort has averaged four working days. Lead time has therefore been estimated as having a normal distribution with a mean of 80 cases and a standard deviation of 10 cases. Roberts wants at most a 2% probability of selling out of Comfort during this lead time. What reorder point should Roberts use?

Optimal Reorder Point Lead time demand is normally distributed with m = 80,  = 10. Since Roberts wants at most a 2% probability of selling out of Comfort, the corresponding z value is That is, P(z > 2.06) = (about .02). Hence Roberts should reorder Comfort when supply reaches m + z = (10) = 101 cases. The safety stock is 21 cases.

Total Annual Inventory Cost Ordering: (DCo/Q *) = ((6240)(12)/327) = $ Holding-Normal: (1/2)Q *Co = (1/2)(327)(1.40) = Holding-Safety Stock: Ch(21) = (1.40)(21) = TOTAL = $487

Periodic Review System
A periodic review system is one in which the inventory level is checked and reordering is done only at specified points in time (at fixed intervals usually). Assuming the demand rate varies, the order quantity will vary from one review period to another. (This is in contrast to the continuous review system in which inventory is monitored continuously and a fixed-quantity order can be placed whenever the reorder point is reached.) At the time a periodic-review order quantity is being decided, the concern is that the on-hand inventory and the quantity being ordered is enough to satisfy demand from the time this order is placed until the next order is received (not placed).

Periodic Review Order Quantity
Assumptions Inventory position is reviewed at constant intervals (periods). Demand during review period plus lead time period is normally distributed with mean µ and standard deviation . Service level is defined in terms of the probability of no stockouts during a review period plus lead time period and is reflected in z. On-hand inventory at ordering time: I Shortages are not backordered. Lead time is less than the length of the review period.

Periodic Review Order Quantity
Formulas Replenishment level: M = µ + z Order quantity: Q = M - I

Example: Ace Brush Periodic Review Order Quantity Model
Joe Walsh is a salesman for the Ace Brush Company. Every three weeks he contacts Dollar Department Store so that they may place an order to replenish their stock. Weekly demand for Ace brushes at Dollar approximately follows a normal distribution with a mean of 60 brushes and a standard deviation of 9 brushes. Once Joe submits an order, the lead time until Dollar receives the brushes is one week. Dollar would like at most a 2% chance of running out of stock during any replenishment period. If Dollar has 75 brushes in stock when Joe contacts them, how many should they order?

Example: Ace Brush Demand During Uncertainty Period
The review period plus the following lead time totals 4 weeks. This is the amount of time that will elapse before the next shipment of brushes will arrive. Weekly demand is normally distributed with: Mean weekly demand, µ = 60 Weekly standard deviation,  = 9 Weekly variance,  = 81 Demand for 4 weeks is normally distributed with: Mean demand over 4 weeks, µ = 4x = 240 Variance of demand over 4 weeks,  2 = 4x = 324 Standard deviation over 4 weeks,  = (324)1/2 = 18

Example: Ace Brush Replenishment Level
M = µ + z where z is determined by the desired stockout probability. For a 2% stockout probability (2% tail area), z = Thus, M = (18) = 277 brushes As the store currently has 75 brushes in stock, Dollar should order: = 202 brushes The safety stock is: z = (2.05)(18) = 37 brushes

Chapter 14 Inventory Management: Dependent-Demand
Objectives of MRP Inputs to MRP Gross and Net Requirements Lot Sizing and Safety Stock MRP and a Time-phased Production Schedule Implementing an MRP System Just-in-Time

Material Requirements Planning
Material requirements planning (MRP) is used to control a manufacturing inventory system. The major function of an MRP system is to translate the demand for finished goods into detailed inventory requirements for all their components. MRP serves as part of a data processing system whose function is to monitor and control the status of production and perform inventory control. One goal of MRP is to minimize the investment in manufacturing inventories. Another goal is to ensure all new materials, parts and subassemblies are available when needed, thus preventing production delays from occurring.

MRP Inputs One input to an MRP system is the master production schedule (MPS) which summarizes requirements and deadlines for finished goods over the planning horizon. Another input is the bill of materials (BOM) which is a structured parts list detailing the sequencing of the assembly of the product. A third input is an MRP inventory record for each raw material, part or subassembly showing everything affecting the inventory level over the planning horizon.

Bill of Materials One graphical representation of a bill of materials is known as a product structure tree. In this representation, the finished product (end item) is shown at the top of the hierarchy (level 0). At the level below this (level 1) are the subassemblies or parts making up the finished product. In the next level (level 2) are the subassemblies or parts making up the subassemblies of level 1. The parts or subassemblies shown at each level of the hierarchy are said to be the parent of the parts or subassemblies directly below them in the hierarchy. Each item in the bill of materials, except for the end product, has a parent.

Inventory Record Information contained in the inventory record can be classified as either inventory transaction information or planning information. Inventory transaction information includes events such as the receipt of goods from a supplier, the disbursement of items from inventory to satisfy production, the occurrence of scrap, etc. Planning information includes lead time, safety stock, lot sizing method, etc.

MRP Calculations In manufacturing there is often a dependent demand between different components. In this case, (Net Component Requirements) = (Gross Component Requirements) - (Number of Components in Inventory). The process of generating net requirements for components from the master production schedule for an end item is called a BOM explosion. The gross component requirement is the quantity of the component necessary to support production at the next higher level of assembly.

MRP Calculations The net requirements calculation enables one to determine the required additions to inventory. An order placed during a previous planning period but scheduled to arrive during the current planning period is called a scheduled receipt. An order placed during the current planning period is referred to as a planned order release. An order’s arrival is known as a planned order receipt. Items are normally assumed to arrive at the beginning of a time period. Both net and gross requirements are assumed to refer to requirements at the beginning of a time period.

Lot Sizing Two often used methods for determining the amount of inventory to order during the planning period are the fixed-order-size and the lot-for-lot rules. The fixed order size rule is appropriate for purchased components when such a lot size is necessary to take advantage of quantity discounts, when an economic order quantity (EOQ) rule is being used, or when batches must be of a certain size due to equipment capacities. The lot for lot rule has the lot size equal to the net requirement for the period in which the lot will become available. In this case, the planned order receipts are identical to the net requirements for a time period.

Safety Stock Safety stock can be used to buffer the production system against uncertainty. This includes uncertainty about gross requirements, lead times, defective parts, pilferage, etc. Safety stock is indicated in the MRP planning worksheet as a projected balance in each time bucket. In calculating net requirements, it should be included in the total for the gross requirements.

Time Phasing MRP determines the date the net requirements are needed by a procedure called time phasing. In time phasing, a production plan for components is developed by working backwards from the desired completion date of the finished product through the various manufacturing stages. Time phasing is used to determine the appropriate planning horizon for each subassembly or parent corresponding to the planning horizon for the end product. An MRP worksheet can then be prepared for the part or subassembly over this planning horizon.

MRP Record Updating Two approaches used to update MRP records are the regeneration approach and the net change approach. In the regeneration approach, the records for all items are updated periodically. In the net change approach, the MRP system recalculates net requirements whenever changes make it necessary; however, only the records affected by the new or revised information are updated.

Example: Columbia Mopeds
Columbia Mopeds is a manufacturer of off-road mopeds. The following product structure tree represents the bill of materials for its dual carburetor model 621 moped. MOPED ENGINE ASSEMBLY GAS TANK WHEEL ASSEMBLY (2) FRAME MOTOR CARBURETOR (2) HUB ASSEMBLY TIRE

Component Lead Times Component Lead Time (Weeks) Engine Assembly Motor Carburetor Gas Tank Wheel Assembly Tire Hub Assembly Frame

The company is currently planning production for weeks 10 through 16. Based on existing orders and demand forecasts, the master production schedule is as follows: a) Determine how many units of each component will be needed to support the production of 1000 mopeds in week 10. b) Determine the planned order release date for each component to support the production of 1000 mopeds in week 10. 11 800 1400 900 10 16 15 14 13 12 1000 1300 WEEK PROD. QTY.

To find the gross requirements of each component to support the production of 1000 mopeds in week 10: Each moped requires one engine assembly, one gas tank, two wheel assemblies and one frame. Thus, 1000 engine assemblies, 1000 gas tanks, 2000 ( = 2 x 1000) wheel assemblies, and 1000 frames will be needed to support production. Each engine assembly requires one motor and two carburetors. Hence the 1000 engine assemblies require 1000 motors and 2000 carburetors. Since each wheel assembly requires one tire and one hub assembly, 2000 tires and 2000 hub assemblies will be required to produce 2000 wheel assemblies.

Gross Requirements Item Units Moped Engine Assembly Motor Carburetor Gas Tank Wheel Assembly Tire Hub Assembly Frame

To determine the planned order release date for each of the components required to support the production of 1000 mopeds in week 10: Note that each of the four level 1 components (engine assembly, gas tank, wheel assembly, and frame) must be available in week 10. Given that the lead time for the engine assembly is one week, its planned order release date should be week 9 (= 10 -1). Similarly, as the lead time for gas tanks is two weeks, its planned release date should be week 8 (= 10- 2). Following the same reasoning, the planned order release date for the wheel assembly should be week 9 (= ) and for the frame, week 7 (= ).

Considering the level 2 components, observe that in order to have engine assemblies ready for assembly in week 9 (their planned release date), the motor and carburetor must be available by week 9. Given that the motors have a one-week lead time, this means they should have a planned release date of week 8 (= 9 -1). Similarly, the planned release date for the carburetors should be week 7 (= 9 - 2). Repeating this process for the level 2 components of the wheel assemblies, the planned release date for the tires should be week 6 (= 9 - 3), and the planned release date for the hubs should be week 8 (= 9 - 1).

Planned Order Release Dates Item Planned Order Release Date (Week) Moped Engine Assembly Motor Carburetor Gas Tank Wheel Assembly Tire Hub Assembly Frame

MRP Worksheet for Gas Tanks for Weeks , assuming: There is a projected balance of 400 gas tanks at the beginning of week 10. There is a scheduled receipt of 900 tanks in week 10. Columbia Mopeds uses a lot-for-lot rule for ordering gas tanks. A safety stock of 200 gas tanks is desired. Lead time for gas tanks is two weeks. The following gross requirements, by week: 11 800 1400 900 10 16 15 14 13 12 1000 1300 WEEK GR. REQ.

MRP Worksheet for Gas Tank Initially GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM:: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 10 1300 11 12 13 800 15 14 1400 16 WEEK 400

MRP Worksheet for Gas Tank Week 10 To complete the worksheet, begin at week 10. Since there is a projected balance of 400 gas tanks at the beginning of week 10 and scheduled receipts of 900 gas tanks in week 10 (assumed to arrive at the beginning of the week), a total of 1300 (= ) gas tanks are available at the beginning of week 10. As the gross requirements in week 10 are for 1000 gas tanks, this leaves a projected balance of 300 (= ) at the end of week 10 (i.e. at the beginning of week 11). Since 300 gas tanks is greater than the desired safety stock of 200 gas tanks, there are no net requirements in week 10 and therefore no planned order receipts.

MRP Worksheet for Gas Tank Week 10 The worksheet now looks as follows: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM:: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 10 1300 11 12 13 800 15 14 1400 16 WEEK 400

MRP Worksheet for Gas Tank Week 11 There are no gross requirements for week 11. Thus, the projected balance at the end of week 11 (the beginning of week 12) remains at 300 and there are no net requirements for week 11. GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 10 1300 11 12 13 800 15 14 1400 16 WEEK 400

MRP Worksheet for Gas Tank Week 12 In week 12 there are gross requirements for 1300 gas tanks. In addition, 200 gas tanks must remain in inventory as safety stock. Hence, the total requirement for week 12 is 1500 (= ) gas tanks. Since there is a projected balance of 300 at the beginning of week 12, the net requirement for week 12 is 1200 (= ) gas tanks. As a lot-for-lot rule is being used, this means that there should be a planned order receipt of 1200 gas tanks in week 12 (at the beginning of the week). Since the lead time for gas tanks is two weeks, this means that there should be a planned order release for the 1200 gas tanks in week 10 (= ).

MRP Worksheet for Gas Tank Week 12 The MRP worksheet would now look as follows: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 1200 10 200 1300 11 12 13 800 15 14 1400 16 WEEK 400

MRP Worksheet for Gas Tank Week 13 The same approach is used to calculate the net requirements for week 13. The formula for determining the net requirements using the lot-for-lot rule is: (Net Requirements) = (Gross Requirements) + (Projected Balance) - (Desired Safety Stock) The projected balance at the end of the week will equal the desired safety stock. Hence net requirements for week 13 will be 800 (= ), and the projected balance at the end of the week will be 200. Because a lot for lot rule is being used, this implies there should be a planned order receipt of 800 gas tanks in week 13. Given the two week lead time, the planned order release of these 800 gas tanks would be week 11.

MRP Worksheet for Gas Tank Week 13 Now the MRP worksheet would be: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 10 1200 200 1300 11 12 13 800 15 14 1400 16 WEEK 400

MRP Worksheet for Gas Tank Weeks 14 & 15 In week 14, since the gross requirements are 0, the net requirements will also be 0 and there will be no planned order receipt. The projected balance at the end of week 14 will remain at the safety stock level of 200 units. For week 15, 1400 gas tanks are required. Hence, the net requirements for this week will be 1400 (= ). Thus, there should be a planned order receipt of 1400 units in week 15. These should have a planned order release date of week 13 (= ). The projected balance at the end of week 15 will be 200 units.

MRP Worksheet for Gas Tank Weeks 14 & 15 Now, the MRP worksheet is: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 1200 10 800 200 1300 11 12 13 1400 15 14 16 WEEK 400

MRP Worksheet for Gas Tank Week 16 For week 16, the gross requirements of 900 translates into net requirements of 900 (= ), and a planned order receipt of The planned order release date for this order will be week 14 (= ). The projected balance at the end of week 16 will then be equal to the desired safety stock of 200 units.

MRP Worksheet for Gas Tank Week 16 Thus the completed MRP worksheet for weeks 10 through 16 is: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: GAS TANK LEAD TIME: 2 LOT SIZE: L-F-L SAFETY STK: 200 1000 900 300 1200 10 800 200 1300 11 12 13 1400 15 14 16 WEEK 400

MRP Worksheet for Frames Week 10 Assume there is a projected balance of 600 frames at the beginning of week 10 and a scheduled receipt of 2000 frames in week 10. The firm uses a fixed order size of 2000 frames and desires no safety stock of frames. Lead time for frames is 3 weeks. The following chart gives the weekly gross requirements for frames: 11 800 1400 900 10 16 15 14 13 12 1000 1300 WEEK GR. REQ.

MRP Worksheet for Frames Initially GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 10 1300 11 12 13 800 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames Week 10 Note that there are 2600 frames (the 600 projected balance plus the 2000 scheduled receipt) available for week 10, but only 1000 frames are required. Hence the projected balance at the end of week 10 (the beginning of week 11) is 1600 (= ). Thus, there are no net requirements and no planned order receipts are necessary.

MRP Worksheet for Frames Week 10 GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 1600 10 1300 11 12 13 800 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames Weeks 11 & 12 Since the gross requirements in week 11 are 0, the projected balance at the end of week 11 remains at 1600 units. Hence, there are no net requirements and no planned order receipts for week 11. In week 12, the gross requirements of 1300 are less than the projected balance of 1600 units. Therefore, the net requirements and the planned order receipts for week 12 are 0. The projected balance is now 300 (= ).

MRP Worksheet for Frames Weeks 11 & 12 The MRP worksheet is now: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 1600 10 300 1300 11 12 13 800 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames Week 13 In week 13, the gross requirements of 800 frames exceed the projected balance of 300 at the beginning of the week. The net requirements for week 13 would then be 500 (= ). Hence there must be a planned order receipt. As the lot size is 2000, this is the planned order receipt. Since the lead time is 3 weeks, the planned order release date for this order must be week 10 (= ). This order will leave a projected balance of 1500 (= ) at the end of week 13.

MRP Worksheet for Frames Week 13 Now the MRP worksheet is: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 1600 10 300 1300 11 12 13 800 500 1500 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames Weeks 14 & 15 Since the gross requirements for week 14 are 0, there are no net requirements and no planned order receipt and the projected balance at the end of week 14 remains at 1500. In week 15, the gross requirements of 1400 units is less than the projected balance of 1500 units at the beginning of the week. Thus the net requirements and the planned order receipts are both 0. However, the projected balance at the end of week 15 is now 100 (= ).

MRP Worksheet for Frames Weeks 14 & 15 The MRP worksheet is now: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 1600 10 300 1300 11 12 13 800 500 100 1500 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames Week 16 In week 16, the gross requirements of 900 units exceed that of the projected balance of 100 units. Hence the net requirements in week 16 are 800 units (= ). Thus, there must be a planned order receipt in week 16 of the fixed order size of 2000. Given the 3 week lead time, the planned order release of these 2000 units must be week 13 (= ). The projected balance at the end of week 16 will be 1200 units (= ).

MRP Worksheet for Frames Week 16 The complete MRP worksheet is: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STK: 0 1000 2000 1600 10 300 1300 11 12 13 800 500 100 1500 1200 15 14 900 1400 16 WEEK 600

MRP Worksheet for Frames With Safety Stock Now consider a desired safety stock of 300 for frames. Note in the completed worksheet for frames that the projected balance exceeds the desired safety stock level of 300 in weeks 10 through 14. In week 13, however, the true net requirements are 800 (= gross requirements (800) + safety stock (300) - projected balance at the end of week 12 (300)). In week 15, the completed worksheet shows a projected balance of only This is less than the desired safety stock of Hence the new net requirements for week 15 is 200 units (= gross requirements (1400) + safety stock (300) - projected balance at the end of week 14 (1500)).

MRP Worksheet for Frames With Safety Stock There should be a planned order receipt of the fixed size of 2000 in week 15. Since there is a 3 week lead time, this means a planned order release date of week 12 (= ). In week 16, the gross requirements of 500 plus the desired safety stock of 300 does not exceed the projected balance of 2100 at the beginning of week 16 (end of week 15). Hence, the net requirements are 0 and there is no planned order receipt. Then the projected balance at the end of week 16 will be 1200 units (= ).

MRP Worksheet for Frames With Safety Stock Thus the completed MRP worksheet when a safety stock of 300 is desired is as follows: GROSS REQUIREMENTS SCHEDULED RECEIPTS PROJECTED BALANCE NET REQUIREMENTS PLANNED ORDER RECEIPTS PLANNED ORDER RELEASES ITEM: FRAME LEAD TIME: 3 LOT SIZE: SAFETY STOCK: 300 1000 2000 1600 10 300 1300 11 12 13 800 2100 1500 1200 15 14 900 1400 16 WEEK 600 200

Chapter 15 Waiting Line Models
The Structure of a Waiting Line System Queuing Systems Queuing System Input Characteristics Queuing System Operating Characteristics Analytical Formulas The Single-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Times The Multiple-Channel Waiting Line Model with Poisson Arrivals and Exponential Service Times Economic Analysis of Waiting Lines

Structure of a Waiting Line System
Queuing theory is the study of waiting lines. Four characteristics of a queuing system are: the manner in which customers arrive the time required for service the priority determining the order of service the number and configuration of servers in the system.

Structure of a Waiting Line System
In general, the arrival of customers into the system is a random event. Frequently the arrival pattern is modeled as a Poisson process. Service time is also usually a random variable. A distribution commonly used to describe service time is the exponential distribution. The most common queue discipline is first come, first served (FCFS). An elevator is an example of last come, first served (LCFS) queue discipline.

Queuing Systems A three part code of the form A/B/s is used to describe various queuing systems. A identifies the arrival distribution, B the service (departure) distribution and s the number of servers for the system. Frequently used symbols for the arrival and service processes are: M - Markov distributions (Poisson/exponential), D - Deterministic (constant) and G - General distribution (with a known mean and variance). For example, M/M/k refers to a system in which arrivals occur according to a Poisson distribution, service times follow an exponential distribution and there are k servers working at identical service rates.

Queuing System Input Characteristics
 = the average arrival rate 1/ = the average time between arrivals µ = the average service rate for each server 1/µ = the average service time  = the standard deviation of the service time

Queuing System Operating Characteristics
P0 = probability the service facility is idle Pn = probability of n units in the system Pw = probability an arriving unit must wait for service Lq = average number of units in the queue awaiting service L = average number of units in the system Wq = average time a unit spends in the queue awaiting service W = average time a unit spends in the system

Analytical Formulas For nearly all queuing systems, there is a relationship between the average time a unit spends in the system or queue and the average number of units in the system or queue. These relationships, known as Little's flow equations are: L = W and Lq = Wq When the queue discipline is FCFS, analytical formulas have been derived for several different queuing models including the following: M/M/1, M/M/k, M/G/1, M/G/k with blocked customers cleared, and M/M/1 with a finite calling population. Analytical formulas are not available for all possible queuing systems. In this event, insights may be gained through a simulation of the system.

Example: SJJT, Inc. (A) M/M/1 Queuing System
Joe Ferris is a stock trader on the floor of the New York Stock Exchange for the firm of Smith, Jones, Johnson, and Thomas, Inc. Stock transactions arrive at a mean rate of 20 per hour. Each order received by Joe requires an average of two minutes to process. Orders arrive at a mean rate of 20 per hour or one order every 3 minutes. Therefore, in a 15 minute interval the average number of orders arriving will be  = 15/3 = 5.

Example: SJJT, Inc. (A) Arrival Rate Distribution Question
What is the probability that no orders are received within a 15-minute period? Answer P (x = 0) = (50e -5)/0! = e -5 = .0067

What is the probability that exactly 3 orders are received within a 15-minute period? Answer P (x = 3) = (53e -5)/3! = 125(.0067)/6 = .1396

What is the probability that more than 6 orders arrive within a 15-minute period? Answer P (x > 6) = 1 - P (x = 0) - P (x = 1) - P (x = 2) - P (x = 3) - P (x = 4) - P (x = 5) - P (x = 6) = = .238

Example: SJJT, Inc. (A) Service Rate Distribution Question
What is the mean service rate per hour? Answer Since Joe Ferris can process an order in an average time of 2 minutes (= 2/60 hr.), then the mean service rate, µ, is µ = 1/(mean service time), or 60/2 = 30/hr.

Example: SJJT, Inc. (A) Service Time Distribution Question
What percentage of the orders will take less than one minute to process? Answer Since the units are expressed in hours, P (T < 1 minute) = P (T < 1/60 hour). Using the exponential distribution, P (T < t ) = 1 - e-µt. Hence, P (T < 1/60) = 1 - e-30(1/60) = = .3935

What percentage of the orders will be processed in exactly 3 minutes? Answer Since the exponential distribution is a continuous distribution, the probability a service time exactly equals any specific value is 0.

What percentage of the orders will require more than 3 minutes to process? Answer The percentage of orders requiring more than 3 minutes to process is: P (T > 3/60) = e-30(3/60) = e =

Example: SJJT, Inc. (A) Average Time in the System Question
What is the average time an order must wait from the time Joe receives the order until it is finished being processed (i.e. its turnaround time)? Answer This is an M/M/1 queue with  = 20 per hour and  = 30 per hour. The average time an order waits in the system is: W = 1/(µ -  ) = 1/( ) = 1/10 hour or 6 minutes

Example: SJJT, Inc. (A) Average Length of Queue Question
What is the average number of orders Joe has waiting to be processed? Answer The average number of orders waiting in the queue is: Lq = 2/[µ(µ - )] = (20)2/[(30)(30-20)] = 400/300 = 4/3

Example: SJJT, Inc. (A) Utilization Factor Question
What percentage of the time is Joe processing orders? Answer The percentage of time Joe is processing orders is equivalent to the utilization factor, /. Thus, the percentage of time he is processing orders is: / = 20/30 = 2/3 or %

Example: SJJT, Inc. (A) Formula Spreadsheet

Example: SJJT, Inc. (A) Spreadsheet Solution

Example: SJJT, Inc. (B) M/M/2 Queuing System
Smith, Jones, Johnson, and Thomas, Inc. has begun a major advertising campaign which it believes will increase its business 50%. To handle the increased volume, the company has hired an additional floor trader, Fred Hanson, who works at the same speed as Joe Ferris. Note that the new arrival rate of orders,  , is 50% higher than that of problem (A). Thus,  = 1.5(20) = 30 per hour.

Example: SJJT, Inc. (B) Sufficient Service Rate Question
Why will Joe Ferris alone not be able to handle the increase in orders? Answer Since Joe Ferris processes orders at a mean rate of µ = 30 per hour, then  = µ = 30 and the utilization factor is 1. This implies the queue of orders will grow infinitely large. Hence, Joe alone cannot handle this increase in demand.

Example: SJJT, Inc. (B) Probability of n Units in System Question
What is the probability that neither Joe nor Fred will be working on an order at any point in time? Answer Given that  = 30, µ = 30, k = 2 and ( /µ) = 1, the probability that neither Joe nor Fred will be working is: = 1/[(1 + (1/1!)(30/30)1] + [(1/2!)(1)2][2(30)/(2(30)-30)] = 1/( ) = 1/3

Example: SJJT, Inc. (B) Average Time in System Question
What is the average turnaround time for an order with both Joe and Fred working? Answer The average turnaround time is the average waiting time in the system, W. µ( /µ)k (30)(30)(30/30)2 Lq = P0 = (1/3) = 1/3 (k-1)!(kµ -  )2 (1!)((2)(30)-30))2 L = Lq + ( /µ) = 1/3 + (30/30) = 4/3 W = L/(4/3)/30 = 4/90 hr. = 2.67 min.

Example: SJJT, Inc. (B) Average Length of Queue Question
What is the average number of orders waiting to be filled with both Joe and Fred working? Answer The average number of orders waiting to be filled is Lq. This was calculated earlier as 1/3.

Example: SJJT, Inc. (B) Formula Spreadsheet

Example: SJJT, Inc. (B) Spreadsheet Solution

Example: SJJT, Inc. (B) Creating Special Excel Function to Compute P0
Select the Tools pull-down menu Select the Macro option Choose the Visual Basic Editor When the Visual Basic Editor appears Select the Insert pull-down menu Choose the Module option When the Module sheet appears Enter Function Po (k,lamda,mu) Enter Visual Basic program (on next slide) Select the File pull-down menu Choose the Close and Return to MS Excel option

Example: SJJT, Inc. (B) Visual Basic Module for P0 Function
Function Po(k, lamda, mu) Sum = 0 For n = 0 to k - 1 Sum = Sum + (lamda / mu) ^ n / Application.Fact(n) Next Po = 1/(Sum+(lamda/mu)^k/Application.Fact(k))* (k*mu/(k*mu-lamda))) End Function

Example: SJJT, Inc. (C) Economic Analysis of Queuing Systems
The advertising campaign of Smith, Jones, Johnson and Thomas, Inc. (see problems (A) and (B)) was so successful that business actually doubled. The mean rate of stock orders arriving at the exchange is now 40 per hour and the company must decide how many floor traders to employ. Each floor trader hired can process an order in an average time of 2 minutes. Based on a number of factors the brokerage firm has determined the average waiting cost per minute for an order to be $.50. Floor traders hired will earn $20 per hour in wages and benefits. Using this information compare the total hourly cost of hiring 2 traders with that of hiring 3 traders.

Example: SJJT, Inc. (C) Economic Analysis of Waiting Lines
Total Hourly Cost = (Total salary cost per hour) + (Total hourly cost for orders in the system) = ($20 per trader per hour) x (Number of traders) + ($30 waiting cost per hour) x (Average number of orders in the system) = 20k + 30L. Thus, L must be determined for k = 2 traders and for k = 3 traders with = 40/hr. and  = 30/hr. (since the average service time is 2 minutes (1/30 hr.).

Example: SJJT, Inc. (C) Cost of Two Servers
= 1 / [1+(1/1!)(40/30)]+[(1/2!)(40/30)2(60/(60-40))] = 1 / [1 + (4/3) + (8/3)] = 1/5 Thus, µ( /µ)k (40)(30)(40/30)2 Lq = P0 = (1/5) = 16/15 (k-1)!(kµ - ) !(60-40)2 L = Lq + ( /µ) = 16/15 + 4/3 = 12/5 Total Cost = (20)(2) + 30(12/5) = $ per hour

Example: SJJT, Inc. (C) Cost of Three Servers
[(1/3!)(40/30)3(90/(90-40))] ] = 1 / [1 + 4/3 + 8/9 + 32/45] = 15/59 (30)(40)(40/30)3 Hence, Lq = (15/59) = 128/885 = (2!)(3(30)-40)2 Thus, L = 128/ /30 = 1308/885 (= ) Total Cost = (20)(3) + 30(1308/885) = $ per hour

Example: SJJT, Inc. (C) System Cost Comparison Wage Waiting Total
Cost/Hr Cost/Hr Cost/Hr 2 Traders $ $ $112.00 3 Traders Thus, the cost of having 3 traders is less than that of 2 traders.

Chapter 16 Simulation Advantages and Disadvantages of Simulation
Modeling Random Variables and Pseudo-Random Numbers Time Increments Simulation Languages Validation and Statistical Considerations Examples

Simulation Simulation is one of the most frequently employed management science techniques. It is typically used to model random processes that are too complex to be solved by analytical methods.

Advantages of Simulation
Among the advantages of simulation is the ability to gain insights into the model solution which may be impossible to attain through other techniques. Also, once the simulation has been developed, it provides a convenient experimental laboratory to perform "what if" and sensitivity analysis.

Disadvantages of Simulation
A large amount of time may be required to develop the simulation. There is no guarantee that the solution obtained will actually be optimal. Simulation is, in effect, a trial and error method of comparing different policy inputs. It does not determine if some input which was not considered could have provided a better solution for the model.

Simulation Modeling One begins a simulation by developing a mathematical statement of the problem. The model should be realistic yet solvable within the speed and storage constraints of the computer system being used. Input values for the model as well as probability estimates for the random variables must then be determined.

Random Variables Random variable values are utilized in the model through a technique known as Monte Carlo simulation. Each random variable is mapped to a set of numbers so that each time one number in that set is generated, the corresponding value of the random variable is given as an input to the model. The mapping is done in such a way that the likelihood that a particular number is chosen is the same as the probability that the corresponding value of the random variable occurs.

Pseudo-Random Numbers
Because a computer program generates random numbers for the mapping according to some formula, the numbers are not truly generated in a random fashion. However, using standard statistical tests, the numbers can be shown to appear to be drawn from a random process. These numbers are called pseudo-random numbers.

Time Increments In a fixed-time simulation model, time periods are incremented by a fixed amount. For each time period a different set of data from the input sequence is used to calculate the effects on the model. In a next-event simulation model, time periods are not fixed but are determined by the data values from the input sequence.

Simulation Programs The computer program that performs the simulation is called a simulator. Flowcharts can be useful in writing such a program. While this program can be written in any general purpose language (e.g. BASIC, FORTRAN, C++, etc.) special languages which reduce the amount of code which must be written to perform the simulation have been developed. Special simulation languages include SIMSCRIPT, SPSS, DYNAMO, and SLAM.

Model Verification/Validation
Verification/validation of both the model and the method used by the computer to carry out the calculations is extremely important. Models which do not reflect real world behavior cannot be expected to generate meaningful results. Likewise, errors in programming can result in nonsensical results. Validation is generally done by having an expert review the model and the computer code for errors. Ideally, the simulation should be run using actual past data. Predictions from the simulation model should be compared with historical results.

Experimental Design Experimental design is an important consideration in the simulation process. Issues such as the length of time of the simulation and the treatment of initial data outputs from the model must be addressed prior to collecting and analyzing output data. Normally one is interested in results for the steady state (long run) operation of the system being modeled. The initial data inputs to the simulation generally represent a start-up period for the process and it may be important that the data outputs for this start-up period be neglected for predicting this long run behavior.

Experimental Design For each policy under consideration by the decision maker, the simulation is run by considering a long sequence of input data values (given by a pseudo-random number generator). Whenever possible, different policies should be compared by using the same sequence of input data.

Example: Probablistics, Inc.
The price change of shares of Probablistics, Inc. has been observed over the past 50 trades. The frequency distribution is as follows: Price Change Number of Trades -3/ -1/ -1/ +1/ +1/ +3/ +1/ Total = 50

Relative Frequency Distribution and Random Number Mapping Price Change Relative Frequency Random Numbers -3/ -1/ -1/ +1/ +1/ +3/ +1/ TOTAL

If the current price per share of Probablistics is 23, use random numbers to simulate the price per share over the next 10 trades. Use the following stream of random numbers: 21, 84, 07, 30, 94, 57, 57, 19, 84, 84

Simulation Worksheet Trade Random Price Stock Number Number Change Price / /8 / / /8 /8 / / /8 / / /8

Spreadsheet for Stock Price Simulation

Theoretical Results and Observed Results Based on the probability distribution, the expected price change per trade can be calculated by: (.08)(-3/8) + (.04)(-1/4) + (.16)(-1/8) + (.40)(0) + (.20)(1/8) + (.06)(1/4) + (.04)(3/8) + (.02)(1/2) = +.005 The expected price change for 10 trades is (10)(.005) = Hence, the expected stock price after 10 trades is = Compare this ending price with the spreadsheet simulation and “manual” simulation results on the previous slides.

Example: Mark Off’s Process
Mark Off is a specialist at repairing large metal-cutting machines that use laser technology. His repair territory consists of the cities of Austin, San Antonio, and Houston. His day-to-day repair assignment locations can be modeled as a Markov process. The transition matrix is: Next Day's Location Austin San Antonio Houston Austin This Day's San Antonio Location Houston

Random Number Mappings Currently in Currently in Currently in Austin San Antonio Houston Next-Day Random Next-Day Random Next-Day Random Location Numbers Location Numbers Location Numbers Austin Austin Austin San Ant San Ant San Ant Houston Houston Houston

Assume Mark is currently in Houston. Simulate where Mark will be over the next 16 days. What percentage of time will Mark be in each of the three cities? Use the following random numbers: 93, 63, 26, 16, 21, 26, 70, 55, 72, 89, 49, 64, 91, 02, 52, 69

Simulation Worksheet Starting in Houston Random Day's Random Day's Day Number Location Day Number Location Houston San Ant. Houston San Ant. Houston San Ant. San Ant San Ant. San Ant San Ant. San Ant Austin San Ant Austin San Ant San Ant.

Repeat the simulation with Mark currently in Austin. Use the following random numbers: 13, 08, 60, 13, 68, 40, 40, 27, 23, 64, 36, 56, 25, 88, 18, 74 Compare the percentages with those found with Mark starting in Houston.

Simulation Worksheet Starting in Austin Random Day's Random Day's Day Number Location Day Number Location Austin San Ant. Austin San Ant. San Ant San Ant. Austin San Ant. San Ant San Ant. San Ant San Ant. San Ant Austin San Ant San Ant.

Simulation Summary Starting in Houston Austin = 2/16 = % San Antonio = 11/16 = % Houston = 3/16 = % Starting in Austin Austin = 4/16 = 25% San Antonio = 12/16 = 75% Houston = 0/16 = 0%

Partial Spreadsheet with Variable Look-up Table LRN = Lower Random Number URN = Upper Random Number NDL = Next-Day Location

Partial Spreadsheet with Simulation Table =IF(A13=$A$2,VLOOKUP(C13,$A$6:$C$8,3), IF(A13=$D$2,VLOOKUP(C13,$D$6:$F$8,3), VLOOKUP(C13,$G$6:$I$8,3)))

Example: Wayne International Aiport
Wayne International Airport primarily serves domestic air traffic. Occasionally, however, a chartered plane from abroad will arrive with passengers bound for Wayne's two great amusement parks, Algorithmland and Giffith's Cherry Preserve. Whenever an international plane arrives at the airport the two customs inspectors on duty set up operations to process the passengers. Incoming passengers must first have their passports and visas checked. This is handled by one inspector. The time required to check a passenger's passports and visas can be described by the probability distribution on the next slide.

Time Required to Check a Passenger's Passport and Visa Probability 20 seconds 40 seconds 60 seconds 80 seconds

After having their passports and visas checked, the passengers next proceed to the second customs official who does baggage inspections. Passengers form a single waiting line with the official inspecting baggage on a first come, first served basis. The time required for baggage inspection has the following probability distribution: Time Required For Baggage Inspection Probability No Time 1 minute 2 minutes 3 minutes

Random Number Mapping Time Required to Check a Passenger's Random Passport and Visa Probability Numbers 20 seconds 40 seconds 60 seconds 80 seconds

Random Number Mapping Time Required For Random Baggage Inspection Probability Numbers No Time 1 minute 2 minutes 3 minutes

Next-Event Simulation Records For each passenger the following information must be recorded: When his service begins at the passport control inspection The length of time for this service When his service begins at the baggage inspection

Time Relationships Time a passenger begins service by the passport inspector = (Time the previous passenger started passport service) + (Time of previous passenger's passport service)

Time Relationships Time a passenger begins service by the baggage inspector ( If passenger does not wait in line for baggage inspection) = (Time passenger completes service with the passport control inspector) (If the passenger does wait in line for baggage inspection) = (Time previous passenger completes service with the baggage inspector)

Time Relationships Time a customer completes service at the baggage inspector = (Time customer begins service with baggage inspector) (Time required for baggage inspection)

A chartered plane from abroad lands at Wayne Airport with 80 passengers. Simulate the processing of the first 10 passengers through customs. Use the following random numbers: For passport control: 93, 63, 26, 16, 21, 26, 70, 55, 72, 89 For baggage inspection: 13, 08, 60, 13, 68, 40, 40, 27, 23, 64

Simulation Worksheet (partial) Passport Control Baggage Inspections Pass. Time Rand. Service Time Time Rand. Service Time Num Begin Num. Time End Begin Num. Time End : : : : : :20 : : : : : :20 : : : : : :00 : : : : : :00 : : : : : :00

Simulation Worksheet (continued) Passport Control Baggage Inspections Pass. Time Rand. Service Time Time Rand. Service Time Num Begin Num. Time End Begin Num. Time End : : : : : :00 : : : : : :00 : : : : : :00 : : : : : :00 : : : : : :20

Explanation For example, passenger 1 begins being served by the passport ontrol inspector immediately. His service time is 1:20 (80 seconds) at which time he goes immediately to the baggage inspector who waves him through without inspection. Passenger 2 begins service with passport inspector 1:20 minutes (80 seconds) after arriving there (as this is when passenger 1 is finished) and requires 1:00 minute (60 seconds) for passport inspection. He is waved through baggage inspection as well. This process continues in this manner.

Question How long will it take for the first 10 passengers to clear customs? Answer Passenger 10 clears customs after 9 minutes and 20 seconds.

Question What is the average length of time a customer waits before having his bags inspected after he clears passport control? How is this estimate biased? Answer For each passenger calculate his waiting time: (Baggage Inspection Begins) - (Passport Control Ends) = = 120 seconds. 120/10 = 12 seconds per passenger. This is a biased estimate because we assume that the simulation began with the system empty. Thus, the results tend to underestimate the average waiting time.

Chapter 17 Markov Process
Transition Probabilities Steady-State Probabilities Absorbing States Transition Matrix with Submatrices Fundamental Matrix

Markov Processes Markov process models are useful in studying the evolution of systems over repeated trials or sequential time periods or stages.

Transition Probabilities
Transition probabilities govern the manner in which the state of the system changes from one stage to the next. These are often represented in a transition matrix. A system has a finite Markov chain with stationary transition probabilities if: there are a finite number of states, the transition probabilities remain constant from stage to stage, and the probability of the process being in a particular state at stage n+1 is completely determined by the state of the process at stage n (and not the state at stage n-1). This is referred to as the memory-less property.

Steady-State Probabilities
The state probabilities at any stage of the process can be recursively calculated by multiplying the initial state probabilities by the state of the process at stage n. The probability of the system being in a particular state after a large number of stages is called a steady-state probability. Steady state probabilities can be found by solving the system of equations P =  together with the condition for probabilities that i = 1. Here the matrix P is the transition probability matrix and the vector, , is the vector of steady state probabilities.

Absorbing States An absorbing state is one in which the probability that the process remains in that state once it enters the state is 1. If there is more than one absorbing state, then a steady-state condition independent of initial state conditions does not exist.

Transition Matrix with Submatrices
If a Markov chain has both absorbing and nonabsorbing states, the states may be rearranged so that the transition matrix can be written as the following composition of four submatrices: I, 0, R, and Q: I 0 R Q

Transition Matrix with Submatrices
I = an identity matrix indicating one always remains in an absorbing state once it is reached 0 = a zero matrix representing 0 probability of transitioning from the absorbing states to the nonabsorbing states R = the transition probabilities from the nonabsorbing states to the absorbing states Q = the transition probabilities between the nonabsorbing states

Fundamental and NR Matrices
The fundamental matrix, N, is the inverse of the difference between the identity matrix and the Q matrix: N = (I - Q )-1 The NR matrix, the product of the fundamental matrix and the R matrix, gives the probabilities of eventually moving from each nonabsorbing state to each absorbing state. Multiplying any vector of initial nonabsorbing state probabilities by NR gives the vector of probabilities for the process eventually reaching each of the absorbing states. Such computations enable economic analyses of systems and policies.

Example: North’s Hardware
Henry, a persistent salesman, calls North's Hardware Store once a week hoping to speak with the store's buying agent, Shirley. If Shirley does not accept Henry's call this week, the probability she will do the same next week is On the other hand, if she accepts Henry's call this week, the probability she will not do so next week is .20.

Transition Matrix Next Week's Call Refuses Accepts This Refuses Week's Call Accepts

Steady-State Probabilities Question How many times per year can Henry expect to talk to Shirley? Answer To find the expected number of accepted calls per year, find the long-run proportion (probability) of a call being accepted and multiply it by 52 weeks. . . . continued

Steady-State Probabilities Answer (continued) Let 1 = long run proportion of refused calls 2 = long run proportion of accepted calls Then, [ ] = [ ]

Steady-State Probabilities Answer (continued) Thus,  +  =  (1)  +  =  (2) and,   = (3) Solving using equations (2) and (3), (equation 1 is redundant), substitute  = 1 -  into (2) to give: .65(1 - 2) +  = 2 This gives  = Substituting back into (3) gives  = Thus the expected number of accepted calls per year is (.76471)(52) = or about 40.

State Probability Question What is the probability Shirley will accept Henry's next two calls if she does not accept his call this week? Answer REFUSES P = .35(.35) = .1225 .35 REFUSES ACCEPTS .35 P = .35(.65) = .2275 .65 REFUSES REFUSES .20 P = .65(.20) = .1300 ACCEPTS ACCEPTS .65 .80 P = .65(.80) = .5200

State Probability Question What is the probability of Shirley accepting exactly one of Henry's next two calls if she accepts his call this week? Answer The probability of exactly one of the next two calls being accepted if this week's call is accepted can be week and refuse the following week) and (refuse next week and accept the following week) = = .29

Example: Jetair Aerospace
The vice president of personnel at Jetair Aerospace has noticed that yearly shifts in personnel can be modeled by a Markov process. The transition matrix is: Next Year Same Position Promotion Retire Quit Fired Current Year Same Position Promotion Retire Quit Fired

Transition Matrix Next Year Retire Quit Fired Same Promotion Current Year Retire Quit Fired Same Promotion

Fundamental Matrix -1 N = (I - Q )-1 = =

Fundamental Matrix The determinant, d = aa - aa = (.45)(.80) - (-.70)(-.10) = .29 Thus, .80/ / N = = .70/ /

NR Matrix The probabilities of eventually moving to the absorbing states from the nonabsorbing states are given by: NR = x Retire Quit Fired Same = Promotion

Absorbing States Question What is the probability of someone who was just promoted eventually retiring? quitting? being fired? Answer The answers are given by the bottom row of the NR matrix. The answers are therefore: Eventually Retiring = .12 Eventually Quitting = .64 Eventually Being Fired = .24

Chapter 18 Multicriteria Decision Problems
Goal Programming Goal Programming: Formulation and Graphical Solution Scoring Models The Analytic Hierarchy Process Establishing Priorities Using AHP Using AHP to Develop an Overall Priority Ranking

Goal Programming Goal programming may be used to solve linear programs with multiple objectives, with each objective viewed as a "goal". In goal programming, di+ and di- , deviation variables, are the amounts a targeted goal i is overachieved or underachieved, respectively. The goals themselves are added to the constraint set with di+ and di- acting as the surplus and slack variables. One approach to goal programming is to satisfy goals in a priority sequence. Second-priority goals are pursued without reducing the first-priority goals, etc.

Goal Programming For each priority level, the objective function is to minimize the (weighted) sum of the goal deviations. Previous "optimal" achievements of goals are added to the constraint set so that they are not degraded while trying to achieve lesser priority goals.

Goal Programming Approach
Step 1: Decide the priority level of each goal. Step 2: Decide the weight on each goal. If a priority level has more than one goal, for each goal i decide the weight, wi , to be placed on the deviation(s), di+ and/or di-, from the goal. Step 3: Set up the initial linear program. Min w1d1+ + w2d2- s.t. Functional Constraints, and Goal Constraints Step 4: Solve this linear program. If there is a lower priority level, go to step 5. Otherwise, a final optimal solution has been reached.

Goal Programming Approach
Step 5: Set up the new linear program. Consider the next-lower priority level goals and formulate a new objective function based on these goals. Add a constraint requiring the achievement of the next-higher priority level goals to be maintained. The new linear program might be: Min w3d3+ + w4d4- s.t. Functional Constraints, Goal Constraints, and w1d1+ + w2d2- = k Go to step 4. (Repeat steps 4 and 5 until all priority levels have been examined.)

Example: Conceptual Products
Conceptual Products is a computer company that produces the CP400 and the CP500 computers. The computers use different mother boards produced in abundant supply by the company, but use the same cases and disk drives. The CP400 models use two floppy disk drives and no zip disk drives whereas the CP500 models use one floppy disk drive and one zip disk drive.

The disk drives and cases are bought from vendors. There are 1000 floppy disk drives, 500 zip disk drives, and 600 cases available to Conceptual Products on a weekly basis. It takes one hour to manufacture a CP400 and its profit is $200 and it takes one and one-half hours to manufacture a CP500 and its profit is $500.

The company has four goals which are given below: Priority 1: Meet a state contract of 200 CP machines weekly. (Goal 1) Priority 2: Make at least 500 total computers weekly. (Goal 2) Priority 3: Make at least $250,000 weekly. (Goal 3) Priority 4: Use no more than 400 man-hours per week. (Goal 4)

Variables x1 = number of CP400 computers produced weekly x2 = number of CP500 computers produced weekly di- = amount the right hand side of goal i is deficient di+ = amount the right hand side of goal i is exceeded Functional Constraints Availability of floppy disk drives: x1 + x2 < 1000 Availability of zip disk drives: x2 < 500 Availability of cases: x1 + x2 < 600

Goals (1) 200 CP400 computers weekly: x1 + d1- - d1+ = 200 (2) 500 total computers weekly: x1 + x2 + d2- - d2+ = 500 (3) $250(in thousands) profit: .2x1 + .5x2 + d3- - d3+ = 250 (4) 400 total man-hours weekly: x x2 + d4- - d4+ = 400 Non-negativity: x1, x2, di-, di+ > 0 for all i

Objective Functions Priority 1: Minimize the amount the state contract is not met: Min d1- Priority 2: Minimize the number under computers produced weekly: Min d2- Priority 3: Minimize the amount under $250, earned weekly: Min d3- Priority 4: Minimize the man-hours over 400 used weekly: Min d4+

Formulation Summary Min P1(d1-) + P2(d2-) + P3(d3-) + P4(d4+) s.t x x < 1000 +x < 500 x x < 600 x d1- -d = 200 x x d2- -d = 500 .2x1+ .5x d3- -d = 250 x1+1.5x d4- -d4+ = 400 x1, x2, d1-, d1+, d2-, d2+, d3-, d3+, d4-, d4+ > 0

Graphical Solution, Iteration 1 To solve graphically, first graph the functional constraints. Then graph the first goal: x1 = Note on the next slide that there is a set of points that exceed x1 = 200 (where d1- = 0).

Functional Constraints and Goal 1 Graphed x2 1000 800 600 400 200 2x1 + x2 < 1000 Goal 1: x1 > 200 x2 < 500 x1 + x2 < 600 Points Satisfying Goal 1 x1

Graphical Solution, Iteration 2 Now add Goal 1 as x1 > 200 and graph Goal 2: x1 + x2 = Note on the next slide that there is still a set of points satisfying the first goal that also satisfies this second goal (where d2- = 0).

Goal 1 (Constraint) and Goal 2 Graphed x2 1000 800 600 400 200 2x1 + x2 < 1000 Goal 1: x1 > 200 x2 < 500 x1 + x2 < 600 Points Satisfying Both Goals 1 and 2 Goal 2: x1 + x2 > 500 x1

Graphical Solution, Iteration 3 Now add Goal 2 as x1 + x2 > 500 and Goal 3: .2x1 + .5x2 = Note on the next slide that no points satisfy the previous functional constraints and goals and satisfy this constraint. Thus, to Min d3-, this minimum value is achieved when we Max .2x1 + .5x2. Note that this occurs at x1 = 200 and x2 = 400, so that .2x1 + .5x2 = 240 or d3- = 10.

Goal 2 (Constraint) and Goal 3 Graphed x2 1000 800 600 400 200 2x1 + x2 < 1000 Goal 1: x1 > 200 x2 < 500 x1 + x2 < 600 (200,400) Points Satisfying Both Goals 1 and 2 Goal 2: x1 + x2 > 500 Goal 3: .2x1 + .5x2 = 250 x1

A Scoring Model for Job Selection
A graduating college student with a double major in Finance and Accounting has received the following three job offers: financial analyst for an investment firm in Chicago accountant for a manufacturing firm in Denver auditor for a CPA firm in Houston

The student made the following comments: “The financial analyst position provides the best opportunity for my long-run career advancement.” “I would prefer living in Denver rather than in Chicago or Houston.” “I like the management style and philosophy at the Houston CPA firm the best.” Clearly, this is a multicriteria decision problem.

Considering only the long-run career advancement criterion the financial analyst position in Chicago is the best decision alternative. Considering only the location criterion the accountant position in Denver is the best decision alternative. Considering only the style criterion the auditor position in Houston is the best alternative.

Steps Required to Develop a Scoring Model Step 1: List the decision-making criteria. Step 2: Assign a weight to each criterion. Step 3: Rate how well each decision alternative satisfies each criterion. Step 4: Compute the score for each decision alternative. Step 5: Order the decision alternatives from highest score to lowest score. The alternative with the highest score is the recommended alternative.

Mathematical Model Sj = S wi rij i where: rij = rating for criterion i and decision alternative j Sj = score for decision alternative j

Step 1: List the criteria (important factors). Career advancement Location Management Salary Prestige Job Security Enjoyable work

Five-Point Scale Chosen for Step 2 Importance Weight Very unimportant 1 Somewhat unimportant 2 Average importance 3 Somewhat important 4 Very important 5

Step 2: Assign a weight to each criterion. Criterion Importance Weight Career advancement Very important 5 Location Average importance 3 Management Somewhat important 4 Salary Average importance 3 Prestige Somewhat unimportant 2 Job security Somewhat important 4 Enjoyable work Very important 5

Nine-Point Scale Chosen for Step 3 Level of Satisfaction Rating Extremely low 1 Very low 2 Low 3 Slightly low 4 Average 5 Slightly high 6 High 7 Very high 8 Extremely high 9

Step 3: Rate how well each decision alternative satisfies each criterion. Decision Alternative Analyst Accountant Auditor Criterion Chicago Denver Houston Career advancement Location Management Salary Prestige Job security Enjoyable work

Step 4: Compute the score for each decision alternative. Decision Alternative Analyst in Chicago Criterion Weight (wi ) Rating (ri1) wiri1 Career advancement x = 40 Location Management Salary Prestige Job security Enjoyable work Score

Step 4: Compute the score for each decision alternative. Sj = S wi rij i S1 = 5(8)+3(3)+4(5)+3(6)+2(7)+4(4)+5(8) = 157 S2 = 5(6)+3(8)+4(6)+3(7)+2(5)+4(7)+5(6) = 167 S3 = 5(4)+3(7)+4(9)+3(5)+2(4)+4(6)+5(5) = 149

Step 4: Compute the score for each decision alternative. Decision Alternative Analyst Accountant Auditor Criterion Chicago Denver Houston Career advancement Location Management Salary Prestige Job security Enjoyable work Score

Step 5: Order the decision alternatives from highest score to lowest score. The alternative with the highest score is the recommended alternative. The accountant position in Denver has the highest score and is the recommended decision alternative. Note that the analyst position in Chicago ranks first in 4 of 7 criteria compared to only 2 of 7 for the accountant position in Denver. But when the weights of the criteria are considered, the Denver position is superior to the Chicago job.

Partial Spreadsheet Showing Steps 1 - 3

Partial Spreadsheet Showing Formulas for Step 4

Partial Spreadsheet Showing Results of Step 4

Analytic Hierarchy Process
The Analytic Hierarchy Process (AHP), is a procedure designed to quantify managerial judgments of the relative importance of each of several conflicting criteria used in the decision making process.

Step 1: List the Overall Goal, Criteria, and Decision Alternatives Step 2: Develop a Pairwise Comparison Matrix Rate the relative importance between each pair of decision alternatives. The matrix lists the alternatives horizontally and vertically and has the numerical ratings comparing the horizontal (first) alternative with the vertical (second) alternative. Ratings are given as follows: . . . continued For each criterion, perform steps 2 through

Step 2: Pairwise Comparison Matrix (continued) Compared to the second alternative, the first alternative is: Numerical rating extremely preferred very strongly preferred strongly preferred moderately preferred equally preferred

Step 2: Pairwise Comparison Matrix (continued) Intermediate numeric ratings of 8, 6, 4, 2 can be assigned. A reciprocal rating (i.e. 1/9, 1/8, etc.) is assigned when the second alternative is preferred to the first. The value of 1 is always assigned when comparing an alternative with itself.

Step 3: Develop a Normalized Matrix Divide each number in a column of the pairwise comparison matrix by its column sum. Step 4: Develop the Priority Vector Average each row of the normalized matrix. These row averages form the priority vector of alternative preferences with respect to the particular criterion. The values in this vector sum to 1.

Step 5: Calculate a Consistency Ratio The consistency of the subjective input in the pairwise comparison matrix can be measured by calculating a consistency ratio. A consistency ratio of less than .1 is good. For ratios which are greater than .1, the subjective input should be re-evaluated. Step 6: Develop a Priority Matrix After steps 2 through 5 has been performed for all criteria, the results of step 4 are summarized in a priority matrix by listing the decision alternatives horizontally and the criteria vertically. The column entries are the priority vectors for each criterion.

Step 7: Develop a Criteria Pairwise Development Matrix This is done in the same manner as that used to construct alternative pairwise comparison matrices by using subjective ratings (step 2). Similarly, normalize the matrix (step 3) and develop a criteria priority vector (step 4). Step 8: Develop an Overall Priority Vector Multiply the criteria priority vector (from step 7) by the priority matrix (from step 6).

Determining the Consistency Ratio
Step 1: For each row of the pairwise comparison matrix, determine a weighted sum by summing the multiples of the entries by the priority of its corresponding (column) alternative. Step 2: For each row, divide its weighted sum by the priority of its corresponding (row) alternative. Step 3: Determine the average, max, of the results of step 2.

Determining the Consistency Ratio
Step 4: Compute the consistency index, CI, of the n alternatives by: CI = (max - n)/(n - 1). Step 5: Determine the random index, RI, as follows: Number of Random Number of Random Alternative (n) Index (RI) Alternative (n) Index (RI) Step 6: Compute the consistency ratio: CR = CR/RI.

Example: Gill Glass Designer Gill Glass must decide which of three manufacturers will develop his "signature" toothbrushes. Three factors seem important to Gill: (1) his costs; (2) reliability of the product; and, (3) delivery time of the orders. The three manufacturers are Cornell Industries, Brush Pik, and Picobuy. Cornell Industries will sell toothbrushes to Gill Glass for $100 per gross, Brush Pik for $80 per gross, and Picobuy for $144 per gross. Gill has decided that in terms of price, Brush Pik is moderately preferred to Cornell and very strongly preferred to Picobuy. In turn Cornell is strongly to very strongly preferred to Picobuy.

Select the Best Toothbrush Manufacturer
Example: Gill Glass Hierarchy for the Manufacturer Selection Problem Overall Goal Select the Best Toothbrush Manufacturer Criteria Cost Reliability Delivery Time Cornell Brush Pik Picobuy Cornell Brush Pik Picobuy Cornell Brush Pik Picobuy Decision Alternatives

Example: Gill Glass Forming the Pairwise Comparison Matrix For Cost
Since Brush Pik is moderately preferred to Cornell, Cornell's entry in the Brush Pik row is 3 and Brush Pik's entry in the Cornell row is 1/3. Since Brush Pik is very strongly preferred to Picobuy, Picobuy's entry in the Brush Pik row is 7 and Brush Pik's entry in the Picobuy row is 1/7. Since Cornell is strongly to very strongly preferred to Picobuy, Picobuy's entry in the Cornell row is 6 and Cornell's entry in the Picobuy row is 1/6.

Example: Gill Glass Pairwise Comparison Matrix for Cost
Cornell Brush Pik Picobuy Cornell / Brush Pik Picobuy 1/ /

Example: Gill Glass Normalized Matrix for Cost
Divide each entry in the pairwise comparison matrix by its corresponding column sum. For example, for Cornell the column sum = /6 = 25/6. This gives: Cornell Brush Pik Picobuy Cornell 6/ / /14 Brush Pik 18/ / /14 Picobuy 1/ / /14

Example: Gill Glass Priority Vector For Cost
The priority vector is determined by averaging the row entries in the normalized matrix. Converting to decimals we get: Cornell: ( 6/ /31 + 6/14)/3 = Brush Pik: (18/ /31 + 7/14)/3 = Picobuy: ( 1/ /31 + 1/14)/3 =

Example: Gill Glass Checking Consistency
Multiply each column of the pairwise comparison matrix by its priority: / = 1/ / Divide these number by their priorities to get: .923/.298 = 2.009/.632 = .209/.069 =

Example: Gill Glass Checking Consistency
Average the above results to get max. max = ( )/3 = 3.102 Compute the consistence index, CI, for two terms by: CI = (max - n)/(n - 1) = ( )/2 = .051 Compute the consistency ratio, CR, by CI/RI, where RI = .58 for 3 factors: CR = CI/RI = .051/.58 = .088 Since the consistency ratio, CR, is less than .10, this is well within the acceptable range for consistency.

Example: Gill Glass Gill Glass has determined that for reliability, Cornell is very strongly preferable to Brush Pik and equally preferable to Picobuy. Also, Picobuy is strongly preferable to Brush Pik.

Example: Gill Glass Pairwise Comparison Matrix for Reliability
Cornell Brush Pik Picobuy Cornell Brush Pik 1/ Picobuy 1/ /

Example: Gill Glass Normalized Matrix for Reliability
Divide each entry in the pairwise comparison matrix by its corresponding column sum. For example, for Cornell the column sum = 1 + 1/7 + 1/2 = 23/14. This gives: Cornell Brush Pik Picobuy Cornell 14/ / /8 Brush Pik / / /8 Picobuy / / /8

Example: Gill Glass Priority Vector For Reliability
The priority vector is determined by averaging the row entries in the normalized matrix. Converting to decimals we get: Cornell: (14/ /41 + 2/8)/3 = Brush Pik: ( 2/ /41 + 5/8)/3 = Picobuy: ( 7/ /41 + 1/8)/3 = Checking Consistency Gill Glass’ responses to reliability could be checked for consistency in the same manner as was cost.

Example: Gill Glass Gill Glass has determined that for delivery time, Cornell is equally preferable to Picobuy. Both Cornell and Picobuy are very strongly to extremely preferable to Brush Pik.

Example: Gill Glass Pairwise Comparison Matrix for Delivery Time
Cornell Brush Pik Picobuy Cornell Brush Pik 1/ /8 Picobuy

Example: Gill Glass Normalized Matrix for Delivery Time
Divide each entry in the pairwise comparison matrix by its corresponding column sum. Cornell Brush Pik Picobuy Cornell / / /17 Brush Pik / / /17 Picobuy / / /17

Example: Gill Glass Priority Vector For Delivery Time
The priority vector is determined by averaging the row entries in the normalized matrix. Converting to decimals we get: Cornell: (8/17 + 8/17 + 8/17)/3 = Brush Pik: (1/17 + 1/17 + 1/17)/3 = Picobuy: (8/17 + 8/17 + 8/17)/3 = Checking Consistency Gill Glass’ responses to delivery time could be checked for consistency in the same manner as was cost.

Example: Gill Glass The accounting department has determined that in terms of criteria, cost is extremely preferable to delivery time and very strongly preferable to reliability, and that reliability is very strongly preferable to delivery time.

Example: Gill Glass Pairwise Comparison Matrix for Criteria
Cost Reliability Delivery Cost Reliability 1/ Delivery 1/ /

Example: Gill Glass Normalized Matrix for Criteria
Divide each entry in the pairwise comparison matrix by its corresponding column sum. Cost Reliability Delivery Cost / / /17 Reliability 9/ / /17 Delivery 7/ / /17

Example: Gill Glass Priority Vector For Criteria
The priority vector is determined by averaging the row entries in the normalized matrix. Converting to decimals we get: Cost: (63/ /57 + 9/17)/3 = Reliability: ( 9/ /57 + 7/17)/3 = Delivery: ( 7/ /57 + 1/17)/3 =

Example: Gill Glass Overall Priority Vector
The overall priorities are determined by multiplying the priority vector of the criteria by the priorities for each decision alternative for each objective. Priority Vector for Criteria [ ] Cost Reliability Delivery Cornell Brush Pik Picobuy

Example: Gill Glass Overall Priority Vector (continued)
Thus, the overall priority vector is: Cornell: (.729)(.298) + (.216)(.571) + (.055)(.471) = .366 Brush Pik: (.729)(.632) + (.216)(.278) + (.055)(.059) = .524 Picobuy: (.729)(.069) + (.216)(.151) + (.055)(.471) = .109 Brush Pik appears to be the overall recommendation.

QUANTITATIVE METHODS FOR BUSINESS 8e

Similar presentations

Presentation on theme: "QUANTITATIVE METHODS FOR BUSINESS 8e"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

QUANTITATIVE METHODS FOR BUSINESS 8e

Similar presentations

Presentation on theme: "QUANTITATIVE METHODS FOR BUSINESS 8e"— Presentation transcript:

Similar presentations

About project

Feedback