LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS

LINEAR PROGRAMMING MODELS: GRAPHICAL AND COMPUTER METHODS
Introduction Most of management decisions involve trying to make the most effective use of an organization’s resources, (machinery, labor, time, warehouse space and raw materials. These resources may be used to make products or services. LP is widely used mathematical modeling technique to help in planning and decision making. “Linear programming is a technique that helps in resource allocation decisions”.

REQUIREMENTS OF A LINEAR PROGRAMING PROBLEM
First LP property: Problems seek to maximize or minimize an objective. Second LP Property: Constraints limit the degree to which the objective can be obtained. Third LP Property: There must be alternatives available. Fourth LP Property: Mathematical relationships are linear.

LP Properties and Assumptions
Properties of Linear Programs One objective function One or more constraints Alternative courses of action Objective function and constraints are linear These are above four basic technical requirements of Linear Programs (LP) Assumptions of Linear programs Certainty Proportionality Additively Divisibility Non Negative Variables These are above five basic technical requirements of Linear Programs (LP)

Basic Assumptions of LP
All of first, we make all equation in the equality form i.e. A+B=C First, Let’s certainty exists i.e. nos. in objective and constraints are known with certainty. Second, lets Proportionality exists in objective and constraints. Third, lets Additivity exists i.e. the total no. of activities = sum of individual activities. Fourth, Make divisibility assumed for that solutions need to be a fractional value. Finally, assume that all answers or variables are nonnegative because quantity of the physical good or person etc may not be negative.

History of Linear Programming
It was developed before World War II by the out standing Soviet mathematician A.N. Kolmogorov. Another Russian,Leonid Kantorovich, won the Nobel Prize in Economics for advancing the concepts of optional planning.

History of Linear Programming
An early application of LP was in the area called “diet problems”. Major progress was in 1947and after George D. Dantzig developed thed simplex algorithm. He was an Air Force Mathematician, work on logistic problems. In 1984, N.Karmarkar developed algorithm is superior to the simplex method. Make the divisibility assumption that all answers or variables are nonnegative. Physical quantities are impossible to produce negative numbers of chairs, shirts, lamps, or computers.

Formulating LP Problems
Formulating a linear program developing a mathematical model to solve the managerial problem, if you formulate the LP problems, it is required to understand the managerial problem being faced. Once understand, we develop the mathematical statement of the problem. The steps in formulating a linear program follow: 1- Completely understand the managerial problem being faced. 2- Identify the objective and the constraints. 3- Define the decision variables 4- Use the decision variables to write mathematical expressions for the objective function and the constraints.

“Product mix problems use LP to decide how much of each product to make, given a series of resource restrictions”. we maximize total contribution margin between unit selling price and minimize total costs in production to the quantity of the item produced. A problem of this type is formulated in the following example

Flair Furniture Company
The Flair Company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labor hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hrs in painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hours in painting and varnishing. The current production period, 240 hrs of carpentry time is available and 100 hrs in painting and varnishing time are available. Each table sold yields a profit of $70; each chair produced is sold for a $50 profit. F Flair Furniture’s problem is to determine the best possible combination of tables and chairs to manufacturer in order to reach the maximum profit. The firm would like this production mix situation formulated as an LP Problem.

Flair Furniture Company Data
Hours Required to Production Department Tables Chairs Available Hrs this week Carpentry Painting & Varnishing Profit per unit $70 $50 Maximize profit The constraints are Hrs of carpentry time use cannot exceed 240 hrs per week Hrs of painting and varnishing time use cannot exceed 100 hrs per week T= No. of tables produced per week C= No. of chairs to be produced per week Create the LP objective function in terms of T&C the objective function is Maximize profit = $70T+$50C Mathematical relationship to describe the two constraints. One common relationship is that resource use is to be less than or equal to (≤) amount of resource available

Flair Furniture Company Data
In case of carpentry department, the total time is used (4 hrs/table) (No. of tables produced)+(3hrs/chair) (No. of chair produced) Constraint may be follows. Carpentry time used ≤ carpentry time available 4T+3C≤240 (hrs of carpentry time) second constraint is follows. Painting and varnishing time used ≤ Painting and varnishing time available 2T+IC ≤ 100 (hours of painting & varnishing time) it means each table produced takes 2 hrs of painting & varnishing resource.) hence these constraint represent production capacity restrictions, affect the total profit. e.g. flair furniture cannot produce 80 tables during the period if T=80 constraint will be violated. It cannot make T=50 tables and C= 10 chairs. Because second constraint no more than 100 hours of painting and varnishing time be allocated. To obtain meaningful solutions, value for T and C ne nonnegative numbers, represent real tables and real chairs. Mathematically expressions are- T ≥ 0 (No. of tables produced is greater than or equal to 0) C ≥ 0 ( No. of chairs produced is greater than or equal to 0) the problem restated mathematically as Maximize profit = $70T+$50C

This is a complete mathematical statement of the LP problem
Related to the constraints 4T + 3C ≤ 240 (carpentry constraint) 2T + 1C ≤ 100 (painting & varnishing) T ≥ 0 (first nonnegativity constraint) C ≥ 0 (second nonnegativity constraint) Nonnegative constraints are technically separate constraints, they are written on a single line variables separated by commas, this would be written as T,C ≥ 0

GRAPHICAL SOLUTION TO AN LP PROBLEM
The graphical method works only when there are two decision variables, but it provides valuable insight into how larger problems are structured. Nonnegativity constraints mean we are always in the graphical area when T ≥ 0 and C ≥ 0 Plotting the first constraint involves finding points at which the line intersects the T and C axes.

Isoprofit Line Solution Method
This methods find the optimal solution to the problem. To maximize profit. The isoprofit method is the first method to introduce for finding the optimal solution.

Implemented in the real world Setting Crew Schedules at American Airlines
Defining the Problem: American’s crews exceed $1.4 billion/year, scheduling crews is one of AA’ biggest and most complex problems. Developing a Model: American Airlines Decision Technologies developing an model TRIP (trip reevaluation and improvement program). This models builds crew schedule that meet the maximum extent possible. Acquiring Input Data: Data derived from salary information and unions and FAA rules specify maximum duty lengths, overnight costs, airlines schedules, and plane sizes. Developing a Solution: 500 hrs of mainframe computer time/month to develop crew schedules-these are prepared 40days prior to the targeted time. Testing the Solution: TRIP results compared with manually since 1971 model has improved with new LP techniques, TRIP’s ability to reach more accurate and optimal solutions. Analyzing the Results: LP model improves AA’s efficiency and improving in scheduling of the airlines. Implementing the Results: after implementation, this model generates annual saving more

Solving Flair Furniture’s LP Problem using QM for Windows and Excel
In every organization access to computer usage. Computers may solve the problem with easily and accuracy in minimum time period. Using Excel’s Solver Command to Solve LP Problem, excel uses a tool named solver to find the solutions to linear programming related problems.

Solving Minimization Problems
LP Problems minimizing cost also instead of maximizing profit function. Its objective to meet the several demands at least cost. e.g. Holiday Meal This meal buying two brands feed and blending them to provide good, low-cost diet. Essential nutritional ingredients for fattening. Each Kg of brand 1 purchased, e.g. contains 5 grams of ingredients A, 4 grams of ingredient B, and 0.5 grams of ingredient C, each Kg of brand 2 contains costs the ranch 2 paisa a Kg, brand 2 feed costs 3 paisa a Kg. owner of the ranch use LP to performs the lowest-cost diet that meets the minimum requirement for each nutritional ingredient.

4x1 + 3x2 ≥ 48 Gms (ingredient B constraint)
Solving Minimization Problems Let, x1 = No. of Kg of brand 1 feed purchased x2 = No. of Kg of brand 2 feed purchased formulate linear programming problem-: minimize cost in (Rs) = 2x1+ 3x2 5x1 + 10x2 ≥ 90 Gms (ingredient A constraint) 4x1 + 3x2 ≥ 48 Gms (ingredient B constraint) 0.5x1 ≥ 1.5 Gms (ingredient C constraint) X1 ≥ 0 (nonnegatively constraint) X2 ≥ 0 (nonnegatively constraint)

Composition of Each Rupee of Feed (Gm)
Ingredient Brand 1 Feed Brand 2 Minimum monthly requirement per turkey (Gm) A B C Cost per gm 2 paisa 3 paisa Using the corner Point Method on a Minimization Problem To solve the Holiday Turkey Ranch problem, construct the feasible solution region, by plotting three constraint equations 0.5x1 ≥ 1.5, plotted as X1 ≥ 3. There are three corner points: a, b, and c for point a we find the coordinates at the intersection of the ingredient C and B constraints, i.e. the line X1 = 3 crosses the line 4X1+3X2 =48. Substitute X1 =3 into the B constraint equation, 4(3)+3X2 =48 Or X2 = 12 point a has coordinates (X1 =3, X2 =12).

Find the coordinates of point b
The equation is 4x1+ 3x2 = 48 and 5x1 + 10x2= 90 simultaneously. This yields (X1 = 8.4, X2 = 4.8). Point c are seen by inspection to be (X1 = 18, X2 = 0) Evaluate objection function at each corner point, Cost = 2X1 + 3X2 Cost at point a = 2(3) + 3(12) =42 Cost at point b= 2(8.4) + 3(4.8) = 31.2 Cost at point c= 2(18) + 3(0) = 36 The minimum cost solution is to purchase 8.4 Kg of brand 1 feed and 4.8 kg of brand 2 feed per turkey per month. cost of 31.2 grams per turkey. We plot the three constraints to develop a feasible solution region for the minimization Problem.

NBC uses Linear, Integer, and Goal Programming in Selling Advertising Slots.
The National Broadcasting Company (NBC) its work is to sold television advertising to provide clients program mix, weekly weighting, unit length and a provide negotiated cost per viewers. NBC develop details sales plans, 1996, a project begun more accurately meet customer’s requirements, more quickly, make the most profitable use of its limited inventory of advertising time slots, and reduce rework. Full scale optimization system based on linear, integer, and goal programming. Sales revenue between the 1996 and 2000 increased by over $200 million. By use of this systems

Four Special Cases & difficulties arise at times when using the graphical approach to solving LP problems Infeasibility: Lack of feasible solution region can occur if constraints conflict with one another. Unboundedness: when the profit in a maximization problem can be infinitely large, the problem is unbounded and is missing one or more constraints. Redundancy: a redundant constraint is one that does not affect the feasible solution region. Alternate Optimal Solutions: Multiple optimal solutions are possible in LP problems. Isoprofit or isocost line runs parallel they have the same slope, more than one optimal solution when they formulated the LP problems.

Sensitivity Analysis Sensitivity: Optimal solution is also called deterministic assumption. But it is dependent on the resources, and other constraints, inputs parameters. Sensitive is the optimal solution to changes in profits, resources, or other input parameters, some analyses are used to examine the effects of changes, (1) contribution rates for each variable, (2) technological coefficients and (3) available resources this task is known as parametric programming. An important function of sensitivity analysis is to allow managers to experiment with values of the input parameters. Sensitivity analysis can be used to estimating input parameters to he LP model but also used to management’s experiments with future changes affect profits. Sensitive optimal solution is to changes by using trial-and-error approach its uses by input data in computer each time input data items to change its parameters at possible way.

Sensitivity Analysis Post optimality analysis means examining changes after the optimal solution has been reached. This is done resolving the whole problem. Judge the sensitivity analysis by implemented in production base problem, represent graphically, tableau used to make LP concepts more clear.

Changes in the Objective Function Coefficient
Changes in contribution rates are examined first. A new corner point becomes optimal if an objective function coefficient is decreased or increased too much. The above solution remains optimal unless an objective function coefficient is increased to a value above the upper bound or decreased to a value below the lower bound. The upper and lower bounds relate to changing only one coefficient at a time.

Excel Solver and Changes in Objective Function Coefficients
Excel solver gives allowable increase and decrease rather than upper and lower bounds. Changes in the Technological Coefficients Changes in technological coefficients affect the shape of the feasible solution region, said changes no effect on the objective function of LP problem.

Swift & Company Uses LP to Schedule Production
Greely, Colorada, Swift & Company annual sales more than $8 billion, making beef related products, each head of beef is cut into two sides, some cuts are more demand than others, customer services representatives try to meet the demand for providing discounts to clear some cuts which is excess in supply. March 2001 Swift started a project develop a mathematical programming model it would be optimize the supply chain, dynamically schedule its operations as orders are received. company is able to deliver products customer specification, model cost $6 million to generated benefit of $12.7 million.

Simplex Method A company which produces 2 different types of glass products has 3 factories, and the data for the same is as given Capacity per unit production rate Capacity available Product 1 2 Plant Unit profit (USD) 3 5

Simplex Method The model formulation is as Max Z=3x1+5x2
s.t: x1 <= 4 2x2<=12 3x1+2x2<=18 x1, x2>=0

Simplex Method Introduce slack variables (red coloured) such that equality can be brought into the picture. Thus we have Max Z=3x1+5x2+0x3+0x4+0x5 s.t: x1+x3= 4 2x2+x4=12 3x1+2x2+x5=18 x1, x2, x3, x4, x5,>=0

Initial tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z x x x You main task is to replace the basic variables with the actual real variables or decision variables which for your case is x1 and x2. As this is an maximization problem then make a decision which one will be the entering variables such that it has the largest non-negative coefficient Name the column corresponding to this entering variable as pivot variable, hence we will choose x2. WHY?

Changing tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z x x x Now if a variable enters, one has to leave. We now have to decide which variable leaves. Calculate the ratios of 4/0 (non defined, hence left out), 12/2=6, 18/2=9. Choose the one for which the ratio is the smallest. WHY? The column marked is the pivot column corresponding to x2.

Changing tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z x x x Choose the row corresponding to this as the pivot row and the element common to pivot column and pivot row is the pivot number. Hence x4 leaves and x2 enters and 2 is the pivot number.

Changing tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z /2 0 30 x x /2 0 6 x How are the new rows formed: (New row) = (old row) – (pivot column coefficient) X (new pivot row) Eq # 0 has changed as per the following calculation to ( )-(-5)X(0 1 0 ½ 0 6)=( /2 0 30) Eq # 1 has changed as per the following calculation to ( )-(0)X(0 1 0 ½ 0 6)=( ), i.e., it does not change as we have 0 Eq # 2 has changed as per the following calculation to ½( )=(0 1 0 ½ 0 6) Eq # 3 has changed as per the following calculation to ( )-(2)X(0 1 0 ½ 0 6)=( )

Changing tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z / x x /2 0 6 x

Changing tableau Basic Eq # Z x1 x2 x3 x4 x5 RHS Variable Z /2 1 36 x /3 -1/3 2 x /2 0 6 x /3 1/3 2 How are the new rows formed: (New row) = (old row) – (pivot column coefficient) X (new pivot row) Eq # 0 has changed as per the following calculation to ( /2 0 30)-(-3)X( /3 1/3 2)=( /2 1 36) Eq # 1 has changed as per the following calculation to ( )-(1)X( /3 1/3 2)=( /3 -1/3 2) Eq # 2 has changed as per the following calculation to (0 1 0 ½ 0 6)-(0)X( /3 1/3 2)=(0 1 0 ½ 0 6) Eq # 3 has changed as per the following calculation to 1/3( )=( /3 1/3 2)

Final solution x=(x1, x2, x3, x4, x5)=(2,6,2,0,0) and maximum Z is 36
What are the constraints now X1+x3=4, hence slack is x2=2 2x2+x4=12, hence slack is x3=0 3x1+2x2+x5=18, hence slack is x5=0 Which would be better? To have all slacks as zeros or optimize

Outline The chapter introduces mathematical modeling i.e. Linear programming, use to reach the optimal solution of the problem. By using corner point method isoprofit/isocost for graphically solving problems, explored sensitivity analysis graphically and with computer output by changing some parameters. Chapter is also provide represent the problem by solving with the help of graphical is most accurate and optimal solution of the problems. Input in computer by different parameters.

Glossary Alternate Optimal Solution Constraint
Corner Point, or Extreme Point Corner Point Method Dual Price (value) Feasible Region Inequality Infeasible solution Isocost Line Isoprofit Line Linear Programming Mathematical Programming Nonnegative Programming Objective Function Product Mix Problem Redundancy Sensitivity Analysis Shadow Price Simultaneous Equation Method Technological Coefficients Unboundedness

Problem 7.1 Personal Mini Warehouses is Planning to expand its successful Orlando business into Tampa. In doing so, the company must determine how many storage rooms of each size to build. Its objective and constraints follow: Maximize monthly earnings = 50X1 + 20X2 Subject is 2X1 + 4X2 ≤ 400 (advertising budget available) 100X1 + 50X2 ≤ 8000 (square footage required) X1 ≤ 60 (rental limit expected) X1, X2 ≥ 0 Where X1 = No. of large spaces developed X2 = No. of Small spaces developed Sol: An evaluation of the five corner points of the accompanying graph indicates that corner point C produces the greatest earnings. Refer to the graph and table.

Table Corner point Values of X1, X2 Objective Function value ($)
D (40,80) 36,00 E (0,100) 2,000

Graphical solution for Personal Mini warehouse
100x1+ 50x2 ≤ 8000 X1 ≤60 Feasible Area (A,B,C,D,E,A) X2 100 (E) D 40,80 (D) C 2x1+ 4x2 ≤ 400 X1 A B 60

Solve the LP formulation graphically, using the isocost line approach: Minimize costs = 24X1+ 28 X2 subject 5X1+ 4X2 ≤ 2,000 X1≥ 80 X1 + X2 ≥ 300 X2 ≥ 100 X1, X2 ≥ 0 Sol: A graph of the four constraints follows. The arrows indicate the direction of feasibility for each constraint. The next graph illustrates the feasible solution region and plots of two possible objective function costs lines. The first, $10,000, was selected arbitrarily as a starting point. To find the optimal corner point, we need to move the cost line in the direction of lower cost, that, is down and to the left. The last point where a cost line touches the feasible region as it moves towards the origin is corner point D. thus D, which represents X1 = 200, X2 = 100, and a cost of $7,600, is optimal.

5x1+ 4x2 ≤ 2000 X1 ≥ 80 X2 X1+ X2 ≥ 300 400 300 X2 ≥ 100 100 X1 80 400 300

Optimal Cost Line X2 B 400 Feasible Region Optimal Cost Line 300 A C
$10,000 = 24X1+28X2 B 400 Feasible Region Optimal Cost Line $7,600= 24X1+28X2 300 A C Optimal Solution D X1 300 400

Solving the problem using corner point method:
Maximize profit = 30X1 + 40X2 Subject to 4X1 + 2X2 ≤ 16 2X1-X2 ≥ 2 X2 ≤ 2 X1, X2 ≥ 0 Sol: The graph appears next with the feasible region shaded. Corner Point Coordinates Profits (Rs.) A X1 =1, X2 =0 30 B X1 =4, X2 =0 120 C X1 =3, X2 =2 170 D X1 =2, X2 =2 140

The optimal profit of Rs. 170 is at corner point C
X2 4X1+ 2X2 ≤ 16 8 2X1 -X2 ≥ 2 C D X2 ≤ 2 2 Feasible Region X1 A B -2

Problem-1 The mathematical relationships that follows were formulated by an operations research analyst at the Smith-Lawton Chemical Company. Which ones are invalid for use in an LP Problem, and why? Maximize profit = 4X1+3X1X2+8X2+5X3 Refer to 2X1 + X2 + 2X3 ≤ 50 X1 – 4X2 ≥ 6 1.5X21+ 6X2 +3X3 ≥ 21 19X2 – 0.35X3 ≤ 17 5X1 + 4X2+3X31/2 ≥ 80 -X1 – X2 + X3 = 5

Problem-2 The Electrocomp Corporation manufactures two electrical products: air conditioners and large fans. The assembly process for each is similar in that both require a certain amount of wiring and drilling. Each air conditioner takes 3 hours to wiring and 2 hours of drilling. Each fan must go through 2 hours of wiring and 1 hours of drilling. During the next production period, 240 hours of wiring time are available and up to 140 hours of drilling time may be used. Each air conditioner sold yields a profit of $25. each fan assembled may be sold for a $15 profit. Formulate and solve this LP production mix situation to find the best combination of air conditioners and fans that yields the highest profit. Use the corner point graphical approach.

Problem-3 The outdoor Furniture Corporation manufactures two products, benches and picnic tables, for use in yards and parks. The firm has two main resources: its carpenters (labor force) and a supply of redwood for use in the furniture. During the next production cycle, 1,200 hours of labor are available under a union agreement. The firm also has a stock of 3,500 feet of good-quality redwood. Each bench that Outdoor Furniture produces requires 4 labor hours and 10 feet of redwood; each picnic table takes 6 labor hours and 35 feet of redwood. Completed benches will yield a profit of $9 each, and tables will result in a profit of $20 each. How many benches and tables should Outdoor Furniture produce to obtain the largest possible profit? Use the graphical LP approach.

Problem-4 The dean of the Western College of Business must plan the school’s course offerings for the fall semester. Student demands make it necessary to offer at least 30 undergraduate and 20 graduate courses in the term. Faculty contracts also dictate that at least 60 courses be offered in total. Each undergraduate course taught costs the college an average of $2,500 in faculty wages, and each graduate course costs $3,000. how many undergraduate and graduate courses should be taught in the fall so that total faculty salaries are kept to a minimum?

Problem-5 A winner of the Texas Lotto has decided to invest $50,000 per year in the stock market. Under consideration are stocks for a petrochemical firm and a public utility. Although a long-range goal is to get the highest possible return, some consideration is given to the risk involved with the stocks. A risk index on a scale of 1-10 (with 10 being the most risky) is assigned to each of the two stocks. The total risk of the portfolio is found by multiplying the risk of each stock by the dollars invested in that stock. The following table provides a summary of the return and risk : Stock Estimated Return Risk Index Petrochemical 12% 9 Utility 6% 4 The investor would like to maximize the return on the investment, but the average risk index of the investment should not be higher than 6. How much should be invested in each stock? What is the estimated return for this investment?

Problem-6 A wealthy farmer in Sathyamangalam, Tamil Nadu, owns fifty acres of land. He is going to plant each acre with either sugarcane or corn. Each acre planted with sugarcane yields a profit of Rs.16,000; each with corn yields a profit of Rs8,000. the labor and fertilizer used for each acre are given in the table below. One hundred and fifty workers and two hundred tons of fertilizer are available. Sugarcane Corn Labor (workers) 5 3 Fertilizer (Tons) 6 2 (a) Formulate an LP model that will enable the farmer to determine the number of acres where sugarcane and/or corn should be planted in order to maximize profit from his land. (Assume that the farmer wants to utilize all fifty acres; however, he does not insist on using (all the) workers or fertilizers.) (b) Solve the problem using a graph paper and determine the maximum profit. Show the isoprofit lines or objective function lines and clearly identify the corner points of the feasible region.

Problem -7 The stock brokerage firm of Blank, Leibowitz, and Weinberger has analyzed and recommended two stocks to an investors’ club of college professors. The professors were interested in factors such as short-term growth, intermediate growth, and dividend rates. These data on each stock are as follows: Stock ($) Short-term growth Potential, per dollar invested Intermediate growth Potential (over next 3 years) Per dollar invested Dividend rate potential 4% 8% Each member of the club has an investment goal of (1) an appreciation of no less than $720 in the short term, (2) an appreciation of at least $5,000 in the next three years, and (3) a dividend income of at least $200 per year. What is the smallest investment that a professor can make to meet these three goals?

Problem-8 Graph the following LP problem and indicate the optimal solution point: Maximize profit = $3X+ $2Y subject to 2X + Y ≤ 150 2X + 3Y ≤ 300 (a) Does the optimal solution change if the profit per unit of X changes to $4.50? (b) What happens if the profit function should have been $3X + $3Y?

Problem-9 Graphically analyze the following problem:
Maximize profit = $4X+$6Y Subject to X+2Y ≤ 8 hours 6X + 4Y ≤ 24 hours What is the optimal solutions? If the first constraint is altered to X + 3Y ≤ 8, does the feasible region or optimal solution change?

Problem-10 Consider the following LP problem:
Maximize profit = 5X + 6Y Subject to 2X + Y ≤ 120 2X + 3Y ≤ 240 X,Y ≥ 0 What is the optimal solution to this problem? Solve it graphically. If a technical breakthrough occurred that raised the profit per unit of X to $8, would this affect the optimal solution? Instead of an increase in the profit coefficient X to $8, suppose that profit was overestimated and should only have been $3. Does this change the optimal solution?

Linear Programming Modeling Applications with Computer Analyses in Excel and QM Windows
Outline of the Chapter 8.1 Introduction 8.2 Marketing Applications 8.3 Manufacturing Applications 8.4 Employee Scheduling Applications 8.5 Financial Application 8.6 Transportation Applications 8.7 Transshipment Applications 8.8 Ingredient Blending Applications

8.1 Introduction The graphical method of linear programming (LP) discussed in previous chapter is useful for understanding how to formulate and solve small LP Problem. The motto of this chapter to perform how a large number of real-life problems can be modeled using LP. By presenting e.g. of models in the areas of production mix, labor scheduling and job assignment, production scheduling, marketing research media selection, shipping and transportation, transshipment ingredient mix, and financial portfolio selection. Many of these LP problems solved by using Excel’s Solver and QM for windows. These models are relatively small numerically, the principles developed are applicable to larger problems. The practice in “paraphrasing” LP model formulations should help develop skills in applying the technique to other, less common applications

Marketing Applications
LP programming models have been used in the advertising fields as a decision aid in selecting an effective media mix. One perspectives the objective can be to maximize audience exposure. Secondly minimize advertising costs.

An example follows Win Big Gambling Club promotes gambling junkets from a large Midwestern city to casinos in the Bahamas. The club has budgeted up to $8,000/week for local advertising. The money is to be allocated among four promotional media: TV spots, newspaper ads, and two types of radio advertisements. Win Big’s goal is to reach the largest possible high-potential audience through the various media. The following table presents the number of potential gamblers per advertisement placed and the maximum number of ads that can be purchased per week.

The problem can now be stated mathematically as follows.
Medium Audience Reached/AD Cost/AD MaximumADS/Week TV Spot/1 Min , Daily newspaper , (Full-page ad) Radio spot , (30 seconds, prime time) Radio spot , (1munute, afternoon) Win Big’s contractual arrangements require that at least five radio spots be placed each week. To ensure a broad-scoped promotional campaign, management also insists that no more than $1,800 be spent on radio advertising every week. The problem can now be stated mathematically as follows. X1 = No. of 1-minute TV spots taken each week X2 = No. of full-page daily newspaper ads taken each week X3 = No. of 30-second prime-time radio spots taken each week X4 = No. of 1-minute afternoon radio spots taken each week

Maximize Audience Objective: Maximize audience coverage = 5,000X1 + 8,500X2 + 2,400X ,800X4 Subject to X1 ≤ 12 (maximum TV spots/week) X2 ≤ 5 (maximum newspaper ads/week) X3 ≤ 25 (maximum 30-seconds radio spots/weeks) X4≤ 20 (maximum 1-minute radio spots/week) 800X1+925X2+290X3+380X4 ≤ $8,000 (weekly advertising budget) X X4 ≥ 5 (minimum radio spots contracted) 290X3+380X4 ≤ $1,800 (maximum dollars spent on radio) X1,X2,X3,X4 ≥ 0

The solution to this LP formulation, using Excel’s Solver found to be
X1 = 1.97 spots X2 = 5 newspaper ads X3 = second radio spots X4 = 0 one-minute radio spots This produces an audience exposure of 67,240 contacts. Because X1 and X2 are fractional, Win Big would probably round them to 2 and 6, respectively.

Marketing Research LP has also been applied to marketing research problems and the area of consumer behaviors. Below example illustrates how statistical pollsters can reach strategy decisions with LP. Management Sciences Associates (MSA) is a marketing and computer research firm based in Washington, D.C., that handles consumer surveys. One of its clients is a national press service that periodically conducts political polls on issues of widespread interest. In a survey for the press service, MSA determines that it must fulfill several requirements in order to draw statistically valid conclusions on the sensitive issue of new U.S. immigration laws.

Survey at least 2,300 U.S. households in total.
Survey at least 1,000 households whose heads are 30 years of age or younger. Survey at least 600 households whose heads are between 31 and 50 years of age. Ensure that at least 15% of those surveyed live in a state that borders on Mexico. Ensure that no more than 20% of those surveyed who are 51 years of age or over live in a state that borders on Mexico.

Cost Per Person Survived ($)
MSA decides that all surveys should be conducted in person. It estimates that the costs of reaching people in each region category are as follows: Cost Per Person Survived ($) Region Age ≤ 30 Age Age ≥ 51 State bordering $ 7.50 $ $5.50 Mexico State not bordering $6.90 $ $6.10

MSA’s goal is to meet the five sampling requirements at the least possible cost.
X1 = no. surveyed who are 30 or younger and live in a border state. X2 = No. surveyed who are and live in a border state X3 = No. surveyed who are 51 or older and live in a border state X4 = No. surveyed who are 30 or younger and do not live in a border state X5 = No. surveyed who are and do not live in a border state X6 = No. surveyed who are 51 or older and do not live in a border state Objective function: Minimize total interview costs = $7.50X1 + $6.80X2 + $5.50X2 +$6.90X4 + $7.25X5 + $6.10X6 Subject to X1+X2+X3+X4+X5+X6≥2,300 (total households) X X ≥1,000 (households 30 or younger) X X5 ≥ 600 (households in age) X1+X2+X3≥0.15(X1+X2+X3+X4+X5+X6) (border states) X3≥0.2(X3+X6) (limit on age group 51+who live in border) X1,X2,X3,X4,X5,X6 ≥0

Scheduling of Airlines
Airlines seat is the vital delicate service in the world. Every time many seats are vacant there fore loss of revenue We have to scheduled such that operating cost of flights should be minimize and save the cost of passenger using the LP Model and Computer Hardware having with so many constraints to solve the optimization problem. Balancing arrivals and departures at airports, aircraft maintenance needs, and so on. Objective is to minimize a combination of operating costs and lost passenger’s revenue.

Manufacturing Application
Production Mix: Linear Programming planning for the optimal mix of product to manufacturer. A company must meet a myriad of constraints, ranging from financial concerns to sales demand to material contracts to union labor demands. Its goal is to generate largest profit possible. e.g. Fifth Avenue Industries, a nationally known manufacturer of menswear, produces four varieties of ties. One is an expensive, all-silk tie, one is an all-polyester tie, and two are blends of polyester and cotton. The following table illustrates the cost and availability (per monthly production planning period) of the three materials used in the production process:

Material Cost/yards ($) Material Available/Monthly (yards)
Silk Polyester 6 3,000 Cotton ,600 The firm fixed contracts with several major department store chains to supply ties. The contracts require that Fifth Avenue Industries supply a minimum quality of each tie but allow for a larger demand if Fifth Avenue chooses to meet that demand. (most of the ties are not shipped with the name Fifth Avenue on their label, incidentally, but with “private stock” labels supplied by the stores.) following table summarizes the contract demand for each of the four styles of ties, the selling price per tie, and the fabric requirements of each variety.

Fifth Avenue’s goal is to maximize its monthly profit
Fifth Avenue’s goal is to maximize its monthly profit. It must decide upon a policy for product mix. X1= number of all-silk ties produced per month X2= number of polyester ties X3= number of blend 1 poly-cotton ties X4= number of blend 2 poly-cotton ties First the firm must establish the profit per tie: 1- For all-silk ties (X1), each requires yard of silk, at a cost of $21/yard. Therefore, the cost per tie is $2.62. the selling price per silk tie is $6.70, leaving a net profit of ($6.70-$2.62=) $4.08 per unit of X1. 2- For all polyester ties (X2), each requires 0.08 yard of polyester at a cost of $6 per yard. The cost per tie is, therefore, $0.48. the net profit per unit of X2 is ($3.55-$0.48=) $3.07.

3- For poly-cotton blend 1 (X3), each tie requires 0
3- For poly-cotton blend 1 (X3), each tie requires 0.05 yard of polyester at $6 per yard and 0.05 of cotton at $6 per yard, for a cost of $ $0.45=$0.75 per tie. The profit is $3.56. 4- Try to compute the net profit for blend 2. calculate a cost of $0.81 per tie and a net profit of $4. Table: Variety of Tie S.P/Tie Monthly contract mim. Monthly demand Material required/Tie Material requirements All Silk , , %Silk All polyester , , %Poly. Poly-cotton Blend , , %poly.-50%cotton Blend , , %poly.-70%cotton

The objective function may now be stated as
Maximize profit=$4.08X1+$3.07X2+$3.56X3+$4.00X4 Subject to X1 ≤ 800 (yards of silk) 0.08X2+0.05X3+0.03X4 ≤ 3,000 (yards of polyester) 0.05X3+0.07X4 ≤ 1,600 (yards of cotton) X1 ≥ 6,000 (contract minimum for all silk) X1 ≤ 7,000 (contract maximum) X2 ≥ 10,000 (contract minimum for all polyester) X2 ≤ 14,000 (contract maximum) X3 ≥ 13,000 (contract minimum for blend 1) X3 ≤ 16,000 (contract maximum) X4 ≥ 6,000 (contract minimum for blend 2) X4 ≤ 8,500 (contract maximum) X1,X2,X3,X4 ≤ 0 Using Excel and its Solver commands, the computer-generated solution is to produce 6,400 all-silks ties each month; 14,000 all-polyester ties; 16,000 poly-cotton blend 1 ties; and 8,500 poly-cotton blend 2 ties. This produces a profit of $160,020 per production period.

Production Scheduling
Setting a low-cost production schedule over a period of weeks or months is a difficult and important management problem in most plants. Production manager has to consider many factor: labor capacity, inventory and storage costs, space limitations, product demand, and labor relations. Most companies produce more than one product, the scheduling process quite complex. Objective is either to maximize profit or to minimize the total cost of carrying out the task. Production scheduling is amenable to solution by LP it is a problem that must be solved on a regular basis. The objective function and constraints for a firm are established, the inputs can easily be changed each month to provide an updated schedule.

An example of production scheduling: Greenberg Motors
Greenberg Motors, Inc., manufactures two different electrical motors for sale under contract to Drexel Corp., a well-known producer of small kitchen appliances. Its model GM3A is found in many Drexel food processors, and its model GM3B is used in the assembly of blenders. Three times each year, the procurement officer at Drexel contracts Irwin Greenberg, the founder of Greenberg Motors, to place a monthly order for each of the coming four months. Drexel’s demand for motors varies each month on its own sales forecasts, production capacity, and financial position. Greenberg has just received the January-April order and must begin his own four-month production plan. The demand of motors is shown.

Production planning at Greenberg Motors Must consider four factors:
The desirability of producing the same number of each motor each month. This simplifies planning and the scheduling of workers and machines. The necessity to keep down inventory carrying, or holding, cost. This suggests producing in each month only what is needed in that month. Warehouse limitations that cannot be exceeded without great additional storage costs. The company’s no-layoff policy, which has been effective in preventing a unionization of the shop. This suggests a minimum production capacity that should be used each month.

These four factors often conflict, Greenberg has found that LP is an effective tool in setting up a production schedule that will minimize his total costs of per unit production and monthly holding. Double-subscribed variables can be used here to develop the LP model. XA,I = number of model GM3A motor produced in month i (i=1, 2, 3, 4 for January-April) XB,I = number of model GM3B motors produced in month I Production costs are currently $10 per GM3A motor produced and $6 per GM3B unit. A labor agreement going into effect on March 1 will raise each figure by 10%, however. We can write the part of the objective function that deals with production cost as Cost of production = $10XA1+$10XA2+$11XA3+$11XA4 +$6XB1+$6XB2+$6.60XB3+$6.60XB4

To include the inventory carrying costs in the model, we can introduce a second variable.
IA,i = level of on-hand inventory for GM3A motors at end of month i (i=1, 2, 3, 4) IB,I = level of on-hand inventory for GM3A motors at end of month i Each GM3A motor held in stock costs $0.18/month, and each GM3B has a carrying cost of $0.13/month. Greenberg’s accountants allow monthly ending inventories as an acceptable approximation to the average inventory levels during the month. So the carrying cost part of the LP objective function is Model January Feb Mar April GM3A ,000 1,100 GM3B 1,000 1,200 1,400 1,400

Inventory Carrying Cost
Cost of carrying inventory=$0.18IA1+0.18IA2+0.18IA3+0.18IA4 +0.13IB1+0.13IB2+0.13IB3+0.13IB4 The total objective function becomes Minimize total costs = 10XA1+10XA2+11XA3+11XA4+6XB1+6XB2 +6.6XB3+6.6XB4+0.18IA1+0.18IA2+0.18IA3+0.18IA4 + 0.13IB1+0.13IB2+0.13IB3+0.13IB4

Inventory constraints set the relationship between closing inventory this month, closing inventory last month, this month’s production, and sales this month. [Inventory at the end of this month]=[Inventory at the end of this month] + [current month’s production]-[sales to Drexel this month] Suppose that Greenberg is starting the new four-month production cycle with a change in design specifications that left no old motors in stock on January 1. then, recalling that January’s demand for GM3As is 800 and for GM3Bs is 1,000, we can write IA1=0+XA1-800 IB1=0+XB1-1,000 Transposing all unknown variables to the left of the equal sign and multiplying all terms by-1, these January constraints can be rewritten as XA1-IA1=800 XB1-IB1=1,000

The constraints on demand in February, March, and April follows:
XA2+IA1-IA2=700 February GM3A demand XB2+IB1-IB2=1,200 February GM3B demand XA3+IA2-IA3=1,000 March GM3A demand XB3+IB2-IB3=1,400 March GM3B demand XA4+IA3-IA4=1,100 April GM3A demand XB4+IB3-IB4=1,400 April GM3B demand Greenberg also want to have on hand an additional 450 GM3As and 300 GM3Bs at the end of April, we add the constraints IA4=450 IB4=300

IA1+IB1≤3,300 IA2+IB2≤3,300 IA3+IB3≤3,300 IA4+IB4≤3,300
The constraints discussed address demand; they do not, however, consider warehouse space of labor requirements. First, we note that the storage area of Greenberg Motors can hold a maximum of 3,300 motors of either type (they are similar in size) at any one time then IA1+IB1≤3,300 IA2+IB2≤3,300 IA3+IB3≤3,300 IA4+IB4≤3,300 We return to the issue of employment. So that no worker is ever laid off, Greenberg has a base employment level of labor hrs/month. In a busy period, though, the company can bring two skilled former employees on board (They are now retired) to increase capacity to 2,560 hrs/month. Each GM3A motor produced requires 1.3 hrs of labor, and each GM3B takes a worker 0.9 hrs to assemble:

Employment constraints are set for each month.
1.3XA1+0.9XB1≥2,240 (January minimum worker hrs/month 1.3XA1+0.9XB1≤2,560 (January maximum labor available/month) 1.3XA2+0.9XB2≥2,240 (February labor minimum) 1.3XA2+0.9XB2≤2,560 (February labor maximum) 1.3XA3+0.9XB3≥2,240 (March labor minimum) 1.3XA3+0.9XB3≤2,560 (March labor maximum) 1.3XA4+0.9XB4≥2,240 (April labor minimum) 1.3XA4+0.9XB4≥2,560 (April labor maximum) All variables ≥ Nonnegativity constraints

The solution to the Greenberg Motors problem was found by computer and is shown in the four-month total cost is $76,301.61 Production Schedule Jan Feb Mar Apr Units of GM3A 1,277 1, Units of GM3B 1,000 1,200 1,400 1,700 Inventory of GM3A carried Inventory of GM3B carried Labor Hrs. required 2,560 2,560 2,355 2,560 This example illustrates a relatively simple production planning problem in that there were only two products being considered. The 16 variables and 22 constraints may not seem trivial, but the technique can also be applied successfully with dozens of products and hundreds of constraints.

Employee Scheduling Application Assignment Problems
Assignment problems involve determining the most efficient assignment of people to jobs, machines to tasks, police cars to city sectors, salespeople to territories. Objective is to be minimize travel times or costs or to maximize assignment effectiveness. We can assign people to jobs using LP or use a special assignment algorithm. Scheduling problems can be handled with their own special solution procedures.

The use of LP in solving assignment problems for the law firm of Ivan
Ivan’s Effectiveness Ratings Clients Case Lawyer Divorce Corporate Merger Embezzlement Exhibitionism Adams Brooks Carter Darwin To solve using LP, we again employ double-subscribed variables. Xij ={1= if attorney i is assigned to case j, 0= otherwise} Where i= 1, 2, 3, 4 stands for Adams, Brooks, Carter, and Darwin, respectively j= 1, 2, 3, 4 stands for divorce, merger, embezzlement, and exhibitionism, respectively The LP formulation follows: Maximize effectiveness = 6X11+2X12+8X13+5X14+9X21+3X22 +5X23+8X24+4X31+8X32+3X33+4X34 +6X41+7X42+6X43+4X44

Subject to X11+X21+X31+X41=1 (divorce case) X12+X22+X32+X42=1 (merger)
X13+X23+X33+X43=1 (embezzlement) X14+X24+X34+X44=1 (exhibitionism) X11+X12+X13+X14=1 (Adams) X21+X22+X23+X24=1 (Brooks) X31+X32+X33+X34=1 (Carter) X41+X42+X43+X44=1 (Darwin)

Labor Planning Labor planning problems address staffing needs over a specific time period. They are especially useful when managers have some flexibility in assigning workers to jobs that require overlapping or interchangeable talents. Commercial (Large) banks frequently use LP to tackle their labor scheduling. Example: Hong Kong Bank of Commerce and Industry is a busy bank that has requirements for between 10 and 18 tellers, depending on the time of day. The lunch time, from noon to 2 p.m., is usually heaviest. Table 8.4 indicates the workers needed at various hours that the bank is open.

The bank now employs 12 full-time tellers, but many people are on its roster of available part-time employees. A part-time employee must put in exactly four hours per day but can start anytime between 9 a.m. and 1P.M. Part-Timers are a fairly inexpensive labor pool, since not retirement or lunch benefits are provided for them. Full-timers, on the other hand, work from 9 A.M. to 5 P.M. but are allowed 1 hour for lunch. (Half of the full-timers eat at 11 A.M., the other half at noon.) full-timers thus provide 35 hours per week of productive labor time. Time Period No. of Tellers Required 9 A.M.-10A.M. 10 10 A.M.-11 A.M. 12 11A.M. –Noon 14 Noon -1 P.M. 16 1 P.M.-2P.M 18 2 P.M.-3P.M. 17 3 P.M.-4 P.M. 15 4 P.M.-5 P.M. 10

Let, F= full-time tellers
By corporate policy, the bank limits part-time hours to a maximum of 50% of the day’s total requirement. Part-timers earn $8/hrs (or $32/day) on average, and full-timers earn $100/day in salary and benefits, on average. The bank would like to seta schedule that would minimize its total personnel costs. It will release one or more of its full-time tellers if it is profitable to do so. Let, F= full-time tellers P1 = Part timers starting at 9 A.M. (leaving at 1 P.M.) P2 = part-timers starting at 10 A.M. (leaving at 2 P.M.) P3 = part-timers starting at 11 A.M. (leaving at 3P.M.) P4 = part-timers starting at noon (leaving at 4 P.M.) P5 = part-timers starting at 1 P.M.) leaving at 5 P.M.) Objective function: Minimize total daily personnel cost = $100F32(P1+P2+P3+P4+P5) Constraints: For each hrs, the variable labor hrs must be at least equal to the required labor hrs. F+P1 ≥ 10 (9 A.M.-10 A.M. needs) F+P1+P ≥ 12 (10 A.M.-11 A.M. needs) 0.5F+P1+P2+P ≥ 14 (11 A.M.-noon needs) 0.5F+P1+P2+P3+P4 ≥ 16 (Noon-1 P.M. needs) F P2+P3+P4 +P5 ≥ 18 (1 P.M.-2 P.M. needs) F P3 +P4 +P5 ≥ 17 (2 P.M.-3 P.M. needs) F P4 +P5 ≥ 15 (3 P.M.-4 P.M. needs) F P5 ≥ 10 (4 P.M.-5 P.M. needs) Only 12 full-time tellers are available, so F ≤ 12

Part-time workers hours cannot exceed 50% of total hours required each day, which is the sum of the tellers needed each hrs: 4(P1+P2+P3+P4+P5) ≤ 0.50( ) Or 4P1+4P2+4P3+4P4+4P5 ≤ 0.50(112) F,P1,P2,P3,P4,P5 ≥ 0 There are several alternate optimal schedules that Hong Kong Bank can follow. The first is to employ only 10 full-time tellers (F=10) and to start two part-timers at 10 A.M. (P2=2), 7 part-timers at 11 A.M. (P3 = 7), and 5 part-timers at noon (P4=5). No part-timers would begin at 9 A.M. or 1 p.m. The second solution also employs 10 full-time tellers, but starts 6 part-timers at 9 A.M. (P1=6), 1 part-timer at 10 A.M. (P2=1), 2 part-timers at 11 a.m. and 5 at noon (P3 =2 and P4 =5), and part-timers at 1 p.m. (P5=0). The cost of either of these two policies is $1,448 per day.

Financial Applications
Portfolio Selection A managers of any banks or finance collection companies i.e. insurances, mutual funds etc. have often a problem of specific investment from among a wide range of alternatives. They want to maximize their return on investment, given a set of legal, policy, or risk restraints.

E.g. The ICT invests in short-term trade credits, corporate bonds, gold stocks, and construction loans. To encourage a diversified potfolio, the board of directors has placed limits on the amount that can be committed to any one type of investment. ICT has Rs. 5 lac available for immediate investment and wishes to do two things: 1. Max the interest earned on the investments made over the nest six months and 2. satisfy the diversification requirements as set by the board of directors.

The specifics of the investment possibilities are as given below: Investment Interest Maximum Earned(%) Investment(Rs.) Trade credit corporate bonds Gold stocks Construction loans And the board specifies that at least 55% of the funds ivested must be in gold stocks and construction loans, and no less than 15% be invested in trade credit. After formulation, we get as: ICT’s investment decision as an LP problem, we assume

X1=Rs. Invested in trade credit X2=Rs. Invested in corporate bonds X3=Rs. Invested in gold stocks X4=Rs. Invested in construction loans Goal:Maximize Rs. of interest earned= 0.07X1+0.11X2+0.19X3+0.15X4 Subject to X1 ≤ 10,00,000 X2 ≤ 25,00, X3 ≤ 15,00, X4 ≤ 18,00,000 X3 + X4 ≥ 0.55(X1+X2+X3+X4) X1 ≥ 0.15(X1+X2+X3+X4)

X1+X2+X3+X4 ≤ 50,00,000 X1, X2, X3, X4,≥ 0 ICT maximizes its interest earned by making the following investment: X1≤ Rs. 7,50,000, X2 ≤ Rs.9,50,000, X3 ≤ Rs. 15,00,000,and X4 ≤ Rs. 18,00,000, and the total interest earned is Rs. 7,12,000

TRANSPORTATION APPLICATIONS
Question: What is transportation problem? Transporting goods from several origins to several destinations efficiently is called the “transportation problem”. It can be solved with LP, as we see here, or with a special algorithm. This type of problem involves to determining the amount of goods of items to be transported from a number of sources to a number of destinations. The objective is to minimize to total transportation cost/distances.

New York Chicago Los Angeles From
e.g. The Top Speed Bicycle Co. manufactures and markets a line of 10-speed bicycles nationwide. The firm has final assembly plants in two cities in which labor costs are low, New Orleans and Omaha. Its three major warehouses are located near the large market areas of New York, Chicago, and Los Angeles. The sales requirements for the next year at the New York warehouse are 10,000 bicycles, at the Chicago warehouse 8,000 bicycles, and at the Los Angeles warehouse 15,000 bicycles. The factory capacity at each location is limited. New Orleans can assemble and ship 20,000 bicycles; the Omaha plant can produce 15,000 bicycles per year. The cost of shipping one bicycle from each factory to each warehouse differs, and these unit shipping costs are as follows: TO New York Chicago Los Angeles From New Orleans $2 $3 $5 Omaha

Sol: The company wishes to develop a shipping schedule that will minimize its total annual transportation costs. To formulate this by using LP, we employ the concept of double subscripted variables. We let the first subscript represent the origin (factory) and the second subscript the destination (warehouse). Thus, in general, Xij refers to the number of bicycles shipped from origin i to destination j. we could instead denote X6 as the variable for origin 2 to destination 3, but generally most find the double subscripts more descriptive and easier to use. Let X11=No. of bicycles shipped from New Orleans to New York X12=No. of bicycles shipped from New Orleans to Chicago X13=No. of bicycles shipped from New Orleans to Los Angeles X21=No. of bicycles shipped from New Omaha to New York X22=No. of bicycles shipped from New Omaha to Chicago X23=No. of bicycles shipped from New Omaha to Los Angeles

This problem can be formulated as follows:
Minimize total shipping costs=2X11+3X12+5X13+3X21+1X22+4X23 Subject to: X11+X21 = 10,000 (New York demand) X12+X22 = 8,000 (Chicago demand) X13+X23 = 15,000 (Los Angeles demand) X11+X12 +X13 ≥ 20,000 (New Orleans factory supply) X21+X22 +X23 ≤ 15,000 (Omaha factory supply) All variables ≥ 0 Question: Why are transportation problems a special class of LP problems? Answer: Every coefficient in from of a variable in the constraint equations is equal to 1. This special trait is also seen in another special category of LP problems.

Using Excel and its Solver command, the computer generated solution to Top speed’s problem is shown in the table that follows and programs From\To New York Chicago Los Angeles New Orleans 10, ,000 Omaha 0 8,000 7,000 Truck Loading Problem Truck loading problem involves deciding which items to load on a trucks so as to maximize the value of a load shipped. an example of Goodman Shipping, an Orlando firm owned by Steven Goodman. One of his trucks, with a capacity of 10,000 pounds, is about to be loaded. Awaiting shipment are the following items:

Item Value ($.) Weight (Pounds) 22,500 7,500 24,000 7,500 8,000 3,000
22, ,500 24, ,500 8, ,000 9, ,500 11, ,000 9, ,500 Each of these six items, we see, has an associated dollar value and weight. The objective is to maximize the total value of the items loaded onto the truck without exceeding the truck’s weight capacity. We let X1 be the proportion of each item I loaded on the truck: Maximize load value= $22,500X1+$24,000X2+$8,000X3+$9,500X4+$11,500X5+$9,750X6 Subject to 7,500X1+7,500X2+3,000X3+3,500X4+4,000X5

+3,500X6 ≤ 10,000 lb capacity X1 ≤ 1 X2 ≤ 1 X3 ≤ 1 X4 ≤ 1 X5 ≤ 1 X6 ≤ 1 X1,X2,X3,X4,X5X6 ≤ 0 These final six constraints reflect the fact that at most one “unit” of an item can be loaded onto the truck. In effect, if Goodman can load a portion of an item (say, item 1 is a batch of 1,000 folding chairs, not all of which need be shipped together), the Xi s will all be proportions ranging from 0 (nothing) to 1 (all of that item loaded). to solve this LP problem, we turn to Excel’s Solver. Formulation and input data and program which yields a total load value of $31,500.

What does Goodman do if fractional values of items cannot be loaded
What does Goodman do if fractional values of items cannot be loaded? For example, if luxury cars were the items being loaded, we clearly cannot ship one-third of a Maserati. If the proportion of item 1 was rounded up to 1.00, the weight of the load would increase to 15,000 pounds. This would violate the 10,000 pounds maximum weight constraint. Therefore, the fraction of item 1 must be rounded down to zero. This would drop the value of the load to 7,500 pounds, leaving 2,500 pounds of the load capacity unused. Because no other item weights less than 2,500, the truck cannot be filled up further. we see that by using regular LP and rounding the fractional weights, the truck QM for Windows and spreadsheet optimizers such as Excel’s Solver are capable of dealing with integer programming problems as well; that is, LP problems requiring integer solutions. Using Excel, the integer solution to Goodman’s problem is to load items 3,4, and 6, for a total weight of 10,000 pounds and load value of $27,250.

Using an Excel Spreadsheet and solver to structure Goodman’s LP Problem

Goodman’s Problem, Using Data Input from Program

TRANSSHIPMENT APPLICATIONS
The transportation problem is actually a special case of the transshipment problem. If items are being transported from the source through an intermediate point (called a transshipment point) before reaching a final destination, then the problem is called a transshipment problem, e.g. a company might be manufacturing a product at several factories to be shipped to a set of regional distribution centers. From these centers the items are shipped to retail outlets that are the final destinations. For e.g. Distribution Centers: Frosty Machines manufactures snowblowers in factories located in Toronto and Detroit. These are shipped to regional distribution centers in Chicago and Buffalo, from where they are delivered to the supply houses in New York, Philadelphia, and St. Louis are also seen in this table, as are the available supplies of snowblowers at the two factories. Notice that snowblowers may not be shipped directly from Toronto or Detroit to any of the final destinations. This is why Chicago and buffalo are listed not only as destinations but also as sources.

The number of units shipped from Toronto is not more than 800
FROM CHICAGO BUFFALO NEW YORK CITY PHILADELPHIA ST. LOUIS SUPPLY Toronto $4 $ Detroit $ $ Chicago $6 $ $ Buffalo $2 $ $ Demand Frosty; would like to minimize the transportation costs associated with shipping sufficient snowblowers to meet the demands at the three destinations while not exceeding the supply and demand constraints similar to the transportation problem. Since there are no units being produced in Chicago or Buffalo, anything's shipped from these transshipment points must have arrived from either Toronto or Detroit. Therefore, Chicago and Buffalo will each have a constraint indicating this. The verbal statement of this problem would be as follows: Minimize cost Subject to The number of units shipped from Toronto is not more than 800 The number of units shipped from Detroit is not more than 700

3. The number of units shipped to New York City is 450.
4. The number of units shipped to Philadelphia is 350. 5. The number of units shipped to St. Louis is 300. 6. The number of units shipped out of Chicago is equal to the number of units shipped into Chicago. 7. The number of units shipped out of Buffalo is equal to the number of units shipped into Buffalo. The decision variables should represent the number of units shipped from each source to each transshipment point and the number of units shipped from each transshipment point to each final destination, as these are the decisions management must make. The decision variables are T1= the number of units shipped from Toronto to Chicago T2= the number of units shipped from Toronto to Buffalo D1= the number of units shipped from Detroit to Chicago D2= the number of units shipped from Detroit to Buffalo C1= the number of units shipped from Chicago to New York

C2= the number of units shipped from Chicago to Philadelphia
C3= the number of units shipped from Chicago to St. Louis B1= the number of units shipped from Buffalo to Philadelphia B2= the number of units shipped from Buffalo to St. Louis The linear program is Minimize cost = 4T1+7T2+5D1+7D2+6C1+4C2 +5C3+2B1+3B2+4B3 Subject to T1+T2≤800 (Supply at Toronto) D1+D2 ≤700 (Supply at Detroit) C1+B1 ≤450 (Demand at New York) C2+B2 ≤350 (Demand at Philadelphia) C3+B3 ≤300 (Demand at St. Louis) T1 +D1=C1+C2 +C3 (Shipping through Chicago) T2+D2 =B1+B2+B3 (Shipping through Buffalo) T1,T2,D1,D2,C1,C2,C3,B1,B2,B3 ≥0 (non negativity constraints)

To solve this on the computer, we arrange as below
T1 +D1=C1+C2 +C3 Becomes T1 +D1-C1-C2 -C3=0 And T2+D2 =B1+B2+B3 T2+D2 -B1-B2-B3 =0 Solving this with QM for windows yields the output in Program given below. From this we see that we should ship 650 units from Toronto to Chicago, 150 units from Toronto to Buffalo, and 300 units from Detroit to Buffalo. A total of 350 units will be shipped from Chicago to Philadelphia, 300 from Chicago to St. Louis, and 450 from buffalo to New York. The total cost will be $9,550.

Solving the Frosty Machines Transshipment Problem Using QM for Windows
Linear Programming Results Frosty Machines Transshipments Problem Solution T1 T2 D1 D2 C1 C2 C3 B1 B2 B3 RHS Dual Minimize 4 7 5 6 2 3 Toronto supply 1 800 Detroit supply 700 NV demand 450 -9 Philadelphia demand 350 -8 St. Louis deman 300 Chicago -1 -4 Buffalo -7 Solution 650 150 9550

Ingredient Blending Applications
Diet Problems One of the earliest applications of LP, was originally used by hospitals to determine the most economical diet for patients. The diet problem involves specifying a food or feed ingredient combination that satisfies stated nutritional requirements at a minimum cost level. Whole Food’s Natural Cereal Requirements Grain Cost/Pounds protein (Units/LB) Riboflavin (Units/LB) Phosphorus (Units/LB) Magnesium (Units/LB) A B C The minimum adult daily requirement (called the U.S. Recommended Daily Allowance, or USRDA) for protein is 3 units; for riboflavin, 2 units; for phosphorus, 1 units; and for magnesium, unit. Whole food wants to select the blend of grain that will meet the USRDA at a minimum cost. XA= pounds of grain A in one 2-ounce serving of cereal XB= pounds of grain B in one 2-ounce serving of cereal XC= pounds of grain C in one 2-ounce serving of cereal

Objective function: Minimize total cost of mixing a 2-ounce serving = $0.33XA+$0.47XB+$0.38XC Subject to 22XA+28XB+21XC ≥ 3 (protein units) 16XA+14XB+25XC ≥ 2 (riboflavin units) 8XA+ 7XB+9XC ≥ 1 (phosphorus units) 5XA+0XB +6XC ≥ (magnesium units) XA+XB XC =0.125 (total mix is 2 ounces or pounds) XA,XB,XC ≥ 0 The solution to this problem requires mixing together lb of grain A, lb of grain B, and lb of grain C. another way of stating the solution is in terms of the proportion of the 2-ounce serving of each grain, namely, 0.4 ounce of grain A, 0.8 ounce of grain B, and 0.8 ounce of grain C in each serving. The cost per serving is $0.05. program 8.8 illustrates this solution using the QM for windows software package.

Solving Whole Food’s LP Problem with QM for Windows
Linear Programming Results Whole Food Nutrition Center Solution X1 X2 X3 RHS Dual Minimize Constraint >= Constraint >= Constraint >= Constraint >= Constraint = Solution=>

Ingredient Mix and Blending Problems
Diet and feed mix problems are actually special cases of a more general class of LP problems known as ingredient or blending problems. Blending problems arise when two or more resources to produce one or more products. One or more essential ingredients that must be blended so that each final product contains specific percentages of each ingredient.

Major oil refineries all use LP for blending crude oils to produce gasoline grades.
The Low Knock Oil Company produces two grades of cut-rate gasoline for industrial distribution. The grades, regular and economy, are produced by refining a blend of two types of crude oil, type X100 and type X220. each crude oil differs not only in cost per barrel, but in composition as well. The following table indicates the percentage of crucial ingredients found in each of the crude oils and cost per barrel for each: Crude Oil Type Ingredient A(%) Ingredient B(%) Cost/Barrel ($) X X Weekly demand for the regular grade of Low Knock gasoline is at least 25,000 barrels, and demand for the economy is at least 32,000 barrels/weeks. At least 45% of each barrel of regular must be ingredient A. at most 50% of each barrel of economy should contain ingredient B.

The Low Knock management must decide how many barrels of each type of crude oil to buy each week for blending to satisfy demand at minimum cost. To solve this as an LP problem, the firm lets. X1 = barrels of crude X100 blended to produce the refined regular X2 = barrels of crude X100 blended to produce the refined economy X3 = barrels of crude X220 blended to produce the refined regular X4 = barrels of crude X220 blended to produce the refined economy problem can be formulated as follows: Objective: Minimize cost= $30X1+$30X2+$3480X3+$3480X4 subject to ] X1+X3 ≥ 25,000 (demand for regular) X2+X4 ≥ 32,000 (demand for economy)

Ingredient At least 45% of each barrel of regular must be ingredient A: (X1+X3) = total amount o crude blended to produce the refined regular gasoline demand. Thus, 0.45(X1+X3) = minimum amount of ingredient A required But 0.35X1+0.60X3 = amount of ingredient A in refined regular gas So 0.35X1+0.60X3 ≥ 0.45X1+0.45X3 Or -0.10X1+0.15X3 ≥ 0 (ingredient A in regular constraint) Similarly, at most 50% of each barrel of economy should be ingredient B: X2+X4 = total amount of crude blended to produce the refined economy gasoline demanded 0.50(X2+X4) = maximum amount of ingredient B allowed

0.55X2+0.25X4= amount of ingredient B in refined economy gas So
But 0.55X2+0.25X4= amount of ingredient B in refined economy gas So 0.55X2+0.25X4≤ 0.50X2+0.50X4 Or 0.05X2-0.25X4 ≤ 0 (ingredient B in economy constraint) LP formulation: Minimize cost = 30X1+30X X X4 Subject X X3 ≥ 25,000 X X4 ≥ 32,000 -0.10 X X3 ≥ 0 0.05X X4 ≤ 0 X1, X2, X3, X4 ≥ 0

The cost of this mix is $1,783,600. The details are given below
Using QM for Windows, the solution to Low Knock Oil’s formulation was found to be X1 = 15,000 barrels of X100 into regular X2 = 26, barrels of X100 into economy X3 = 10,000 barrels of X220 into regular X4 = 26, barrels of X100 into economy The cost of this mix is $1,783,600. The details are given below

SUMMARY We learned applications from marketing, production, scheduling, finance, transportation, transshipment, and ingredient blending. We also solved of these problem with the two computer programs for Linear Programming: QM for Windows and Excel’s solver.

Linear Programming The Simplex Method

Introduction LP problem you have to seen earlier chapter that a LP problem having two decision variable can be solved with using graphical method be traced the feasible region and then search for the optimal corner point. and corresponding profit or costs. Thus we can understand of basic concepts of LP. Most real life have more than two variables, which can be simplified by graphical method.

Simplex Method In this method has a large number of variables.
We can not be able to plot the graph the feasible region, but the optimal solution will still lie at a corner point of the many-sided, n-dimensional polyhedron that represents the area of feasible solution. This method determines the corner points in a systematic fashion, using basic algebraic concepts. It does so in an iterative manner, i.e., representing the same set of procedures time after time until an optimal solution is reached. Each iteration brings a higher value for the objective function so that we are always moving closer to the optimal solution. It is very important to understand the ideas used to produce solutions. Its approach yields not only the optimal solution to the decision variables and the maximum profit (or minimum cost), but valuable economic information as well. To be able to use computers successfully and to interpret LP computer printouts. We begin by solving a maximization problem using the simplex method. We then tackle a minimization problem and look at a few technical issues that are faced when employing the simplex procedure. We can also observe how to conduct sensitivity analysis using the simplex table.

HOW TO SET UP THE INITIAL SIMPLEX SOLUTION
The Flair Company produces inexpensive tables and chairs. The production process for each is similar in that both require a certain number of hours of carpentry work and a certain number of labor hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hrs in painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hours in painting and varnishing. The current production period, 240 hrs of carpentry time is available and 100 hrs in painting and varnishing time are available. Each table sold yields a profit of $70; each chair produced is sold for a $50 profit. F Flair Furniture’s problem is to determine the best possible combination of tables and chairs to manufacturer in order to reach the maximum profit. The firm would like this production mix situation formulated as an LP Problem.

Hours Required to Production
Department Tables Chairs Available Hrs this week Carpentry Painting & Varnishing Profit per unit $70 $50 Maximize profit The constraints are Hrs of carpentry time use cannot exceed 240 hrs per week Hrs of painting and varnishing time use cannot exceed 100 hrs per week T= No. of tables produced per week C= No. of chairs to be produced per week Create the LP objective function in terms of T&C the objective function is Maximize profit = $70T+$50C Mathematical relationship to describe the two constraints. One common relationship is that resource use is to be less than or equal to (≤) amount of resource available

We let, T=number of tables produced C= number of chairs produced and that the problem was formulated as Maximize profit = $70T+$50C (objective function) Subject to 2T C ≤100 (painting hours constraint) 4T C ≤100 (carpentry hours constraint) T, C ≥0 (non negativity constraint)

Converting the Constraints to Equations
This method requires that convert each inequality constraint in an LP formulation into an equation. Simplex is a matrix algebra method that requires all mathematical relationships to be equations, with each equation containing all of the variables. Slack variables are added to each less-than-or-equal-to constraint. Each slack variable represents an unused resources.

Flair Problem are converted to equations by adding a slack-variables to each constraint.
Slack variables represent unused resources; these may be in the form of time on a machine, labor hours, money, warehouse space, or any number of such resources in various business problems. Case at hand, S1= slack variable representing unused hours in the painting department S2= slack variable representing unused hours in the carpentry department the constraints to the problem may now be written as 2T+1C+S1= 100 And 4T+3C+S2= 240 If the production of tables (T) and chairs (C) uses less than 100 hours of painting time available, the unused time is the value of the slack variable, S1. for example, if T=0 and C=0 (in other words, if nothing is produced), we have S1=100 hours of slack time in the painting department. If flair produces T=40 tables and C=10 chairs, then 2T+1C+S1=100 2(40)+1(10)+S1=100 S1=10 there will be 10 hours of slack, or unused, painting time available.

Include all variables in each equation, which is a requirement of the next simplex step, slack variables not appearing in an equation are added with a coefficient of 0. this means, in effect, that they have no influence on the equations in which they are inserted; but it does allow us to keep tabs on all variables at all times. The equations now appear as follows: 2T+1C+S1+0S2=100 4T+3C+0S1+1S2=240 T,C,S1,S2=0 because slack variables yield no profit, they are added to the original objective function with 0 profit coefficients. The objective function becomes Maximize profit = $70T+$50C+$0S1+$0S2

Corner Points of the Flair Furniture Company Problem
Figure of Corner Points of the Flair Furniture Company Problem C Feasible area is AECB B= (0,80) 2T+1C≤100 Number of Chairs C= (30,40) 4T+3C≤240 D= (50,0) (0,0) A E T Number of Tables

Find and Initial Solution Algebraically
If there are two equations and four variables. Then we need four equations to solve for unique value algebraically. But when there are four unknowns (T,C,S1, and S2 in this case) and only two equations. Let two of the variables equal to zero and then solve for the other two variables. e.g. If T=C=0, then S1= 100 and S2= 240. This solution is called a basic feasible solution. A basic feasible solution to a system of n equations is found by setting all but n variables equal to 0 and solving for the others variables. The simplex method begins with an initial feasible solution in which all real variables are set equal to 0. This trivial solution always produces a profit of $0, as well as slack variables equal to the constant (RHS) terms in the constraint equations. It is not profitable but it is one of the original corner point solutions as shown in above slide.As mentioned, the simplex method will start at this corner pint (A) and then move up or over to the corner point that yields the most improved profit (B&D). Finally, the technique will move to a new corner point (c), which happens to be the optimal solution to the flair Furniture problem. The simplex method considers only feasible solutions and hence will touch no possible combinations other than the corner points of the region AECB in the above slide. Simplex considers only corner points as it seeks the best solution.

The First Simplex tableau
To simplify the equations and objective function in an LP problem, we put all of the coefficients into a tabular form. Constraint Equation: Flair furniture’s two constraint equations can be given below; SOLUTION QUANTITY MIX T G S S (RHS) S S

Flair Furniture’s Initial Simplex Tableau
Profit per unit Column Production mix Column Slack variables Column Profit per unit Column Constraint Column Cj SOLUTION $ $ $ $ QUANTITY MIX T G S S (RHS) $ S $ S Zj $ $ $ $ $0 Cj - Zj $ $ $ $ $0 Profit per unit row Constraint equation rows Gross profit row Net profit row

The initial solution mix begins with real, or decision, variables set equal to zero.
The numbers(2,1,1,0) in the first row represent the coefficients of the first equation, name sly, 2T+1C+1S1+0S2.. The numbers(4,3,0,1) in the second row are the algebraic equivalent of the constraint, 4T+3C+0S1+1S2.. At the first step, where T=0 and C=0 then the value of others, i.e. S1=100 and S2=240. That is the initial solution mix. These are equal to quantity (RHS) because of T & C are not in the solution mix, their initial values are automatically equal to zero. The initial solution is a basic feasible solution and we get in the vector or column form, Variables in the solution mix are called basic in LP terminology, i.e. S1 & S2. Those not in the solution are called monobasic i.e. T & C. The optimal solution of this LP problem resulted to be T=30, C=40, S1= 0 & S2=0 or

Then T&C would be the final basic variables, and S1 and S2 would be the nonbasic variables. Notice that for any corner point, exactly two of the four variables will equal zero. Substitution Rates Many students are unsure as to the actual meaning of the numbers in the columns under each variables. We know that the entries are the coefficients for that variable. Under T are the coefficients (2/4), under C are (1/3), under S1 are (1/0). Substitution rates are numbers in the body of the table. This paragraph explains how to interpret their meaning. what is their interpretation? The numbers in the body of the simplex tableau (see Table 9.1) can be though of as substitution rates. For example, suppose we now wish to make T larger than 0, that is, produce some tables. For every unit of the T product introduced into the current solution, 2 units of S1 and 4 units of S2 must be removed from the solution. This is to because each table requires 2 hours of the currently unused painting department slack time, S1. It also takes 4 hours of carpentry time; hence 4 units of variable S2 must be removed from the solution for every unit of T that enters. Similarly, the substitution rates for each unit of C that enters the current solution are 1 unit of S1 and 3 units of S2. Another point that you are reminded of throughout this chapter is that for any variable ever to appear in the solution mix column, it must have the number 1 someplace in its columns 0s in every other place in that column. We see that column S1 contains , so variable S1 is in the solution. Similarly, the S2 column is so S2 is also in the solution. Adding the Objective Function: Now we add a row to present the objective function values for each variable. These contribution rates, called Cj,appear just

above each respective variable, as shown in the following table:
The unit profit rates are not just found in the top Cj row : in the left most column, Cj indicates the unit profit for each variable currently in the solution mix. If s1 were removed from the solution and replaced, by C, $50 would appear in the Cj column just to the left of the term C. The Zj and Cj – Zj Rows we can complete the initial Flair Furniture simplex tableau by adding two final rows. These last two rows provide us with important economic information, including the total profit and the answer as to whether the current solution is optimal. we compute the Zj value for each column of the initial solution in table 9.1 by multiplying the 0 contribution value of each number in the Cj column by each number in that row and the jth columns, and summing. The Zj value for the quantity column provides the total contribution (gross profit in this case) of the given solution: Cj $ $ $ $ QUANTITY Solution Mix T C S S (RHS)` $ S $ S

Zj (for gross profit) = (Profit per unit of S1) X (Number of units of S1)
= $0X100 units + $0X240 units = $0 profit The Zj values for the other columns (under the variables T, , S1, and S2) represent the gross profit given up by adding one unit of unit of this variable into the current solution. Their calculations are as follows: Zj = (Profit per unit of S1) X (Substitution rate in row 1) +(profit per unit of S2) X (Substitution rate in row 2) thus, ZJ(for column T) = ($0)(2)+($0)(4) = $0 ZJ(for column C) = ($0)(1)+($0)(3) =$0 ZJ(for column S1)=($0)(1)+($0)(0) =$0 ZJ(for column S1)=($0)(0)+($0)(1) =$0 we see that there is no profit lost by adding one unit of either T (tables), C (chairs), S1 or S2.

The Cj- Zj number in each column represents the net profit, that is, the profit gained minus the profit given up, that will result from introducing 1 unit of each product or variable into the solution. It is not calculated for the quantity column. To compute these numbers, simply subtract the Zj total for each column from the Cj value at the very top of that variable’s column. The calculations for the net profit per unit (the Cj-Zj row) in this example follow: Column T C S1 S2 Cj for column $70 $50 $0 $0 Zj for column Cj-Zj for column $70 $50 $0 $0 We reach an optimal solution when the Cj-Zj row has no positive numbers in it.

Simplex Solution Procedures
Five Simplex steps. 1- Variable entering the solution has the largest positive Cj-Zj. 2- Variable leaving the solution is determined by a ratio we must compute. 3- New pivot-row calculations are done next. 4- Compute the new values for each remaining row. (In our Flair Furniture problem there are only two rows in the LP tableau, but most larger problems have many more rows.) All remaining row (s) are calculated as follows: -[(Number above or below pivot No.) X (Corresponding no. in the new row, that is, the row replaced in step 3)] 5- Compute the Zj and Cj-Zj rows, as demonstrated in the initial tableau. If all numbers in the Cj-Zj row are 0 or negative, an optimal solution has been reached. If this is not the case, return to step 1.

The Second Simplex Tableau
Here we apply the five steps to Flair Furniture. Step 1 - Decide which of the variables will enter the solution next we select the one with the largest positive Cj-Zj value. Variable T, tables, has a Cj-Zj value of $70 to the overall profit. Variable C, chairs, has a Cj-Zj value of only $50. the overall two variables, S1 and S2, have 0 values and can add nothing more than to profit. We select T as the variable to enter the solution mix and identify its column as the pivot column. Step 2 – We must decide that T which variable is to be replaced. There can only be as many basic variables as there are constraints in any LP problem, so either S1 or S2 will have to leave to make room for the introduction of T, tables, into the basis. To identify the pivot row, each number in the quantity column is divided by the corresponding number in the T column.

= 50 tables = 60 tables For the S1 row
100(hours of painting time available) 2(hours of required per table) For the S2 row: 240(hours of carpentry time available) 4(hours of 4(hours required per table) S1 leaves the solution mix because the smaller of the two ratios indicates that the next pivot row will be the first row. Step -3 We have decided which variable is to enter the solution mix (T) and which is to leave (S1), we begin to develop the second, improved simplex tableau. Step 3 involve computing a replacement for the pivot row. This is done by dividing every number in the pivot row by the pivot number: The new pivot row is computed by dividing every number in the pivot row by the pivot number. The entire successive pivot row appears in the accompanying table. T is in the solution mix and that 50 unit of T are being produced. The Cj value is listed as a $70 contribution per unit of T in the solution. This will definitely provide Flair Furniture with a more profitable solution than the $0 generated in the initial tableau. = 50 tables = 60 tables Cj Solution Mix T C S S QUANTITY $ T

Graph of the Flair Furniture Company Problem
Number of Chairs C= (30,40) D= (50,0) E=(60,0) (0,0) A T Number of Tables

Fig: Pivot Row and Pivot Number Identified in the Initial Simplex Tableau
Step 4. On the next step, we intended to the S2 row, it is slightly more complex than replacing the pivot row and uses. We can now recomputed the S2 row. The expression on the RHS of the following equation is used to calculate the LHS. Cj $ $ $ $ QUANTITY Solution Mix T C S S (RHS)` $ S Pivot row $ S Pivot number Zj $ $ $ $ $0 Cj –Zj $ $ $ $0 Pivot column 2 = (4) X (1) = (4) X (0.5) = (4) X (0.5) = (4) X (0) = (4) X (50)

This new S2 row will appear in the second tableau in the following format:
Cj Solution Mix T C S1 S2 Quantity $70 T $0 S We note that the T column contains (1/0) and the S2 column contains (0/1). These Os and 1s indicate that T and S2 are in the basis (the solution mix). Step -5 The final step of the second iteration is to introduce the effect of the objective function. This involves computing the Zj and Cj-Zj rows. We find the new profit in the Z row. Zj(for T column) = ($70)(1) +($0)(0) =$70 Zj(for C column) = ($70)(0.5) +($0)(1) =$35 Zj(for S1 column) = ($70)(0.5) +($0)(-2) =$35 Zj(for S2 column) = ($70)(0) +($0)(1) =$0 Zj(for total profit) = ($70)(50) +($0)(40) =$35,500 the current profit is $3,500.

Completed Second Simplex Tableau for Flair Furniture
Cj $ $ $ $0 Solution mix T C S1 S2 Quantity T S $70 $0 Zj $ $ $ $ $3,500 Cj-Zj $ $ $ $0

The Cj-Zj numbers represent the net profit that will result, given our present production mix, if we add one unit of each variable into the solutions: The Cj-Zj row indicates the net profit, given the current solution, of one more unit of each variable. For example, C has a profit of $15 per unit. Colulmn T C S1 S2 Cj for column $70 $50 $0 $0 Zj for column $70 $35 $35 $0 Cj-Zj for column $0 $15 -$35 $0 The Zj and Cj-Zj rows are inserted into the complete second tableau as shown.

Interpreting the Second tableau
Summarizes all of the information for the Flair Furniture Company’s production mix decision as of the second iteration of the simplex method. We can look at the current solution as a corner point in the graphical method. Resource Information slack variable S2, representing the amount of unused time in the carpentry department, is in the basis. slack variable S1 is nonbasic and has a value of 0 hrs. there is no slack time in the painting department. Substitution Rates Net Profit Row The Cj-Zj row tells us (1) whether the current solution is optimal and (2) if it is not, which variable should enter the solution mix next.

Developing the Third Tableau
C (chairs) will be the next solution mix variable because it has the only positive value in the Cj-Zj row. Step 1. Variable C will enter the solution next by virtue of the fact that is Cj-Zj value of 15 is the largest (and only) positive number in the row. This means that for every unit of C (chairs) we start to produce, the objective function will increase in value by $15. the C column is the new pivot column. Step 2. The next step involves identifying the pivot row. The question is, which variable currently in the solution (T or S2) will have to leave to make room for C to enter? Again, each number in the quantity column is divided by its corresponding substitution rate in the C column: For the T row: 50/0.5= 100 chairs for the S2 row: 40/1 = 40 chairs.

Pivot Row, Pivot Column, and Pivot Number Identified in the Second Simplex Tableau
Fig $70 $50 $0 $0 Cj Solution Mix T C S S2 Quantity Pivot row $ T $ S 1 Pivot number Zj $ $ $ $ $3,500 Cj-Zj $ $ $ $ (total profit) Pivot column

the entire new C row looks like this:
Step 3. The pivot row for the third tableau is replaced here the pivot row is replaced by dividing every number in it by the (circled) pivot number. Since every number is divided by 1, there is no change: 0/1=0 1/1=0 -2/1=-2 1/1=1 40/1=40 the entire new C row looks like this: It will be placed in the new simplex tableau in the same row position that S2 Cj Solution Mix T C S1 S2 Quantity $50 C

Step 4. The new values for the T row may now be computed:
(Number in new T row)=(Number in old T row)-[(Number above pivot Number) X (Corresponding No. in new C row)] = (0.5) X (0) = (0.5) X (1) = (0.5) X (-2) = (0.5) X (1) = (0.5) X (40) The new T row will appear in the third tableau in the following position: Cj Solution mix T C S1 S2 Quantity $70 T $50 C

Step 5. The final step is again computing the Zj and Cj values
Zj (for T Column) = ($70)(1)+($50)(0) = $70 Zj (for C Column) = ($70)(0)+($50)(1) = $50 Zj (for S1 Column) = ($70)(1.5)+($50)(-2) = $5 Zj (for S2 Column) = ($70)(-0.5)+($50)(1) = $15 Zj (for total profit) = ($70)(30)+($50)(40) = $4,100 The net profit per unit row appears as follows: Cj COLUMN T C S S2 Cj for column $ $ $ $0 Zj for column $ $ $ $15 Cj –Zj for column $ $ $ $15

Table no. 6 : Final Simplex Tableau for the Flair Furniture Problem
All results for the third iteration of the simplex method are summarized in below table. Since every number in the tableau’s Cj-Zj row is 0 or negative, an optimal solution has been reached. The final solution is to make 30 tables and 40 chairs at a profit of $4,100. this is the same as the graphical solution presented earlier. T=30 Tables, C=40 chairs S1=0 slack hours in the painting department S2= 0 slack hours in the carpentry department. profit = $4,100 for the optimal solution Table no. 6 : Final Simplex Tableau for the Flair Furniture Problem Cj $ $ $ $0 Solution Mix T C S S QUANTITY $ T $ C Zj $ $ $ $ $4,100 Cj –Zj $ $ $ $15

T and C are the final basic variables, and S1 and S2 are no basic (i.e. 0)
This solution corresponds to corner point C in fig. 2 Verifying that the solution does not violate any of the original constraints is a good way to check that no mathematical errors were made. This can be done in part by looking at the Original Flair Furniture Company constraints and objective function. First constraint: T+1C ≤ painting department hours. 2(30)+1(40) ≤ 100 100 ≤ 100 Second constraint: 2T+1C ≤ painting department hours. 4(30)+3(40) ≤ 240 240 ≤ 240 Objective Function : profit = $70T + $50C = $70(30) + $50(40) = $4,100

REVIEW OF PROCEDURES FOR SOLVING LP MAXIMIZATION PROBLEMS
Formulate the LP problem’s objective function and constraints. Add slack variables to each less-than-or-equal-to constraint and to the problem’s objective function. Develop an initial simplex tableau with slack variables in the basis and the decision variables set equal to 0. Compute the Zj and Cj-Zj values for this tableau. Follow these five steps until an optimal solution has been reached: Choose the variable with the greatest positive Cj-Zj to enter the solution. This is the pivot column. Determine the solution mix variable to be replaced and the pivot row by selecting the row with the smallest (non negative ) ratio of the quantity –to-pivot column substitution rate. This row is the pivot row. Calculate the new values for the pivot row. Calculate the new values for the other row(s). Calculate the Zj and Cj-Zj values for this tableau. If there are any Cj-Zj numbers greater than 0, return to step 1. If there are no Cj-Zj numbers that are greater than 0, an optimal solution has been reached

Surplus and Artificial Variables
To handle ≥ and = constraints, the simplex method makes a conversion like it made to ≤ Constraints. Constraint 1: 5X1+10X2+8X3 ≥ 210 Constraint 2: 25X1+30X = 900 Surplus Variables We subtract a surplus variable to form an equality when dealing with a ≥ constraint. Constraint 1 rewritten: 5X1+10X2+8X3-S1=210 If, for example, a solution to an LP problem involving this constraint is X1=20, X2=8, X3=5, the amount of surplus could be computed as follows: 5X1+10X2+8X3-S =210 5(20)+10(8)+8(5)-S1 =210 S1 =210 -S1 = S1= 10 surplus units

Artificial Variables One small problem use the first constraint in setting up an initial simplex solution. All “real” variables such as X1, X2, and X3 are set to 0 in the initial tableau, takes on a negative value: 5(0)+10(0)+8(0)-S1= 210 0-S1= 210 S1= -210 Artificial variables are needed in ≥ and = constrain. All variables in LP problems, be they real , slack, or surplus, must be nonnegative at all times. If S1=-210, this important condition is violated. Constraint 1 completed: 5X1+10X2+8X3-S1+A1=20 Now, not only the X1, X2, and X3 variables may be set to 0 in the initial simplex solution, but the S1 surplus variable as well. This leaves us with A1=210 our attention to constraint 2 for a moment. This constraint is already an equality, so why worry about it? To be included in the initial simplex solution, it turns out, even an equality must have an artificial variable added to it: Constraint 2 rewritten: 25X1+30X2+A2=900

Surplus and Artificial Variables in the Objective Function
Artificial variables have no physical meaning and drop out of the solutions mix before the final tableau. Surplus and Artificial Variables in the Objective Function - To make sure that an artificial variable is forced out before the final solution is reached, it is assigned a very high cost (m). If a problem had an objective function that read Minimize cost = $5X1+$9X2+$7X3 and constraints such as the two mentioned previously, the completed objective function and constraints would appear as follows: Minimize cost = $5X1+$9X2+$7X3+$0S1+$MA1+$MA2 Subject to X1+10X2+8X3- 1S1+1A1+0A2 = 210 25X1+30X2+0X3- 0x3+0S1+0A1+1A2 = 900

Solving Minimization Problems The muddy river Chemical Company Example
The muddy river Chemical Corporation must produce exactly 1,000 pounds of a special mixture of phosphate and potassium for a customer. Phosphate costs $5 per pound and potassium costs $ 6 per pound. No more than 300 pounds of phosphate can be used, and at least 150 pounds of potassium must be used. The problem is to determine the least-cost blend of the two ingredients. Mathematical formulation of the minimization problem for Muddy River Chemical Corp. Minimize cost = $5X1 + $6X2 Subject to X1 + X2 =1,000 lb X1≤300 lb X2≥150 lb X1, X2 ≥ 0 Where X1=Number of pounds of phosphate X2=Number of pounds of potassium

Graphical Analysis Looking at a graphical solution first will help us understand the steps in the simplex method. Muddy River Chemical Corporation’s Feasible Region Graph X2 1,000 800 600 400 200 100 A X1≤ 300 B X1+X2=1,000 X2 ≥150 G F H E D c X1

Don’t need the simplex method to solve the Muddy River Chemical problem, of course. But we can guarantee you that few problems will be this simple. In general, you can expect to see several variables and many constraints. The purpose of this section is to illustrate the straightforward application of the simplex method to minimization problems. When the simplex procedure is used to solve this, it will methodically move from corner point to corner point until the optimal solution is reached. An above figure the simplex method will begin at point E, then move to point f, then to point G, and finally to point B, which is the optimal solution.

Converting the Constraints and Objective Function
First insert slack, surplus, and artificial variables. This makes it easier to set up the initial simplex tableau. First step is to apply what we learn in the preceding section to convert the constraints and objective function into the proper form for the simplex method. The equality constraint, X1+X2=1,000, just involves adding an artificial variable, A1: X1+X2+A1=1,000 The second constraint, X1≤ 300, requires the insertion of a slack variable-let’s call it S1: X1+S1=300 The last constraint is X2≥150, which is converted to an equality by subtracting a surplus variable, S2, and adding an artificial variable, A2 X2-S2+A2 = 150 Finally, the objective function, cost =$5X1+$6X2, is rewritten as Minimize cost = $5x1+$6X2+$0S1+$0S2+$MA1+$MA2

Rules of the Simplex Method for Minimization Problems
The complete set of constraints can now be expressed as follows: 1X1+1X2 +0S1+0S2 +1A1 + 0A2 = 1,000 1X1+0X2 +1S1+0S2 +0A1 + 0A2 = 300 0X1+1X2 +0S1-1S2 +0A1 + 1A2 = 150 X1,X2,S1,S2,A1,A2 ≥ 0 Rules of the Simplex Method for Minimization Problems The minimization simplex rules are slightly different. Now, the new variable to enter the solution mix will be in the column with the negative Cj-Zj value indicating the greatest improvement.

Steps for Simplex Minimization Problems
1- Choose the variable with a negative Cj-Zj that indicates the largest decrease in cost to enter the solution. The corresponding column is the pivot column. 2- Determine the row to be replaced by selecting the one with the smallest (nonnegative) quantity-to-pivot column substitution rate ratio. This is the pivot row. 3- Calculate new values for the pivot row. 4- Calculate new values for the other rows. 5- Calculate the Zj and Cj-Zj values for this tableau. If there are any Cj-Zj numbers less than 0, return to step 1.

First Simplex Tableau for the Muddy River Chemical Corporation Problem
Solve the Muddy River Chemical Corporation’s LP formulation using the simplex method. Earlier example its first three rows are shown in the accompanying table. We note the presence of the $M costs associated with artificial variables A1 and A2 but we treat them as if they were any large number. Effect of forcing the artificial variables out of the solution quickly because of their large costs. Cj Solution Mix X X S S A A Quantity $M A ,000 $ S $M A

The numbers in the Zj row are computed by multiplying the Cj column on the far left of the tableau times the corresponding numbers in each other column. Zj (for X1 column) = $M(1) +$(1) +M(0) =$M Zj (for X2 column) = $M(1) +$0(0) +$M(1) =$2M Zj (for S1 column) = $M(0) +$0(1) +$M(0) =$0 Zj (for S2 column) = $M(0) +$0(0) +$M(-1) = - $M Zj (for A1 column) = $M(1) +$0(0) +$M(0) =$M Zj (for A2 column) = $M(0) +$0(0) +$M(1) =$M Zj (for total cost) = $M(1,000)+$0(300)+$M(150) =$1,150M

The Cj-Zj entries are determined as follows:
Column X1 X2 S1 S2 A1 A2 Cj for Column $5 $6 $0 $0 $M $M Zj for column $M $2M $0 -$M $M $M Cj-Zj for column -$M+$5 -$2M+$6 $0 $M $0 $0

Initial Simplex Tableau for the Muddy River Chemical Corporation Problem
Cj $5 $6 $0 $0 $M $M Solution Mix X1 X2 S1 S2 A1 A2 Quantity $M A ,000 $0 S $M A Zj $M $2M 0 -$M $M $M $1,150M Cj-Zj -$M $2M+6 $0 $M $ (total cost) Pivot column 1 Pivot row Pivot number

The above initial solution was obtained by letting each of the variables X1,X2, and S2 assume a value of 0. the current basic variables are A1=1,000, S1=300, and A2=150. this complete solution could be expressed in vector, or column, for as X X S = 300 S A ,000 A An extremely high cost, $1,150M from associated with this answer. We know that this can be reduced significantly and now move on to the solution procedures.

Developing a Second Tableau
We examine whether the current solution is optimal by looking at the Cj-Zj row. Which variable will leave the solution to make room for the new variable, X2? To find out, we divide the elements of the quantity column by the respective pivot column substitution rates: for the A1 row = =1,000 for the S1 row = (this is an undefined ratio, so we ignore it) for the A2 row = =150 (smallest quotient, indicating pivot row) The pivot row is the A2 row, and pivot number (circled) is at the intersection of the X2 column and the A2, row.

The entering row for the next simplex tableau is found by dividing each element in the pivot row by the pivot number, 1. this old pivot row unchanged, except that it now represents the solution variable X2. The other rows are altered one at a time by again applying the formula. (New row numbers)=(Numbers in old row) -[(Number above or below pivot number) X (Corresponding number in newly replaced row)] A1 Row S1 Row 1=1-(1)(0) 1=1-(0)(0) 0=1-(1)(1) 0=0-(0)(1) 0=0-(1)(0) 1=1-(0)(0) 1=0-(1)(-1) 0=0-(-0)(-1) 1=1-(1)(0) 0=0-(0)(0) -1=0-(1)(1) 0=0-(0)(1) 850=1,000-(1)(150) =300-(0)(150)

Developing a Third Tableau
New pivot column is the X1 column. To determine which variable will leave the basis to make room for X1, we check the quantity column-to-pivot column ratios again. The third tableau is developed in this section. For the A1row =850/1=850 For the S1 row =300/1=300 (smallest ratio) for the X2 row = 150/0= undefined

Third Simplex Tableau for the Muddy River Chemical Corporation Problem.
Cj $ $ $ $ $M $M Solution Mix X1 X2 S1 S2 A1 A QTY $M A $S X $6 X Zj $5 $6 -$M+5 $M-6 $M -$M $550M+2,400 Cj-Zj $0 $0 $M-5 -$M+6 $0 $2M-6 Hence variables S1, will be replaced by X1*5 the pivot number, row, and column are labeled in tabled in above table. To replace the pivot row, we divide each number in the S1 row by 1 (the circled pivot number), leaving the row unchanged. The new X1 row is shown in this table. Pivot row 1 Pivot No. Pivot column

Here are the computations for the third tableau.
A1 Row S1 Row 0=1-(1)(1) 0=0-(0)(1) 0=0-(1)(0) 1=1-(0)(0) -1=0-(1)(1) 0=0-(0)(1) 1=1-(1)(0) -1=-1-(0)(0) 1=1-(1)(0) 0=0-(0)(0) -1=-1-(1)(0) 1=1-(0)(0) 550=850-(1)(300) 150=150-(0)(300)

The Zj and Cj-Zj rows are computed next:
Zj(for X1) = $M(0) +$5(1) +$6(0) = $5 Zj(for X2) = $M(0) +$5(0) +$6(1) = $6 Zj(for S1) = $M(-1) +$5(1) +$6(0) = -$M+5 Zj(for S2) = $M(1) +$5(0) +$6(-1) = -$M-6 Zj(for A1) = $M(1) +$5(0) +$6(0) = -$M Zj(for A2) = $M(-1) +$5(0) +$6(1) = -$M+6 Zj(for total cost) = $M(550)+$5(300)+$6(150) =$550M+2,400 Column X1 X2 S1 S2 A1 A2 Cj for column $5 $6 $0 $0 $M $M Zj for column $5 $6 -$M+5 $M-6 $M -$M+6 Cj_Zj for column $0 $0 $M-5 -$M+6 $0 $2M-6

The third solution is still not optimal
The solution at the end of the three iterations is still not optimal because the S2 column contains a Cj-Zj value that is negative. The total cost is nonetheless lower than at the end of the second tableau, which in turn is lower than the initial solution cost. We are headed in the right direction but have one more tableau to go!

Fourth Tableau for the Muddy River Chemical Corporation Problem
The pivot column is now the S2 column. The ratios that determine the row and variable to be replaced are computed as follows: For the A1 row: (row to be replaced) For the X1 row: 300/0 (undefined) for the X2 row: 150/-1 (not considered because it is negative)

Computation for the Fourth Solution
X1 Row X2 Row 1=1-(0)(0) =0-(-1)(0) 0=0-(0)(0) =1-(-1)(0) 1=1-(0)(-1) =0-(-1)(-1) 0=0-(0)(1) = -1-(-1)(1) 0=0-(0)(1) =0-(-1)(1) 0=0-(0)(-1) =1-(-1)(-1) 300=300-(0)(-1) =150-(-1)(-1)

Computations of Forth Solution
Zj(for X1) =$0(0) +$5(1) +$6(0) =$5 Zj(for X2) =$0(0) +$5(0) +$6(1) =$6 Zj(for S1) =$0(-1)+$5(1) +$6(-1) =-$1 Zj(for S2) =$0(1) +$5(0) +$6(0) =$0 Zj(for A1) =$0(1) +$5(0) +$6(1) =$6 Zj(for A2) =$0(-1)+$5(0) +$6(0) =$0 Zj (for total cost) =$0(550)+$5(300)+$6(700)=$5,700 Column X1 X2 S1 S2 A1 A2 Cj for column $5 $6 $0 $0 $M $M Zj for column $5 $6 -$1 $0 $6 $0 Cj-Zj for column $0 $0 $1 $0 $M-6 $M

Fourth Optimal Solution to the Muddy River Chemical Corporation Problem
The optimal solution has been reached because only positive or zero values appear in the Cj-Zj row. Cj $5 $6 $0 $0 $M $M Solution X1 X2 S1 S2 A1 A2 Quantity Mix $0 S $5 X $6 X Zj $5 $6 -$1 $0 $6 $ $5,700 Cj-Zj $0 $0 $1 $0 $M-6 $M

Procedures for Solving LP Minimization Problems
Summarized the steps for solving LP maximization problems Formulate the LP problem’s objective function and constraints. Include slack variables in each less-than-or-equal-to constraints, artificial variables in each equality constraint, and both surplus and artificial variables in each greater-than-or-equal-to constraint. Then add of these variables to the problem’s objective function. Develop an initial simplex tableau with artificial and slack variables in the basis and the other variables set equal to 0. Compute the Zj and Cj-Zj values for this tableau.

4- these five steps until an optimal solution has been reached:
Choose the variable with the negative Cj-Zj indicating the greatest improvement to enter the solution. This is the pivot column. Determine the row to be replaced by selecting the one with the smallest (nonnegative) quantity-to-pivot column substitution rate ratio. This is the pivot row. Calculate the new values for the pivot row. Calculate the new values for the other row (s). Calcualte the Zj and Cj-Zj values for the tableau. If there are any Cj-Zj numbers less than 0, return to step 1. if there are no Cj-Zj numbers that are less than o, an optimal solution has been reached.

SPECIAL CASES Infeasibility
Infeasibility, when there is no solution that satisfies all of the problem’s constraints. In the simplex method, an feasible solution is indicated by looking at the final tableau. In it, all Cj-Zj row entries will be of the proper sign to imply optimality, but an artificial variable (A1) will still be in the solution mix. Unbounded Solutions Unboundedness describes linear programs that do not have finite solutions. It occurs in maximization problems. No finite solution may exist in problems that are not bounded. This means that a variable can be infinitely large without violating a constraint.

ILLUSTRATION OF INFEASIBILITY
Cj $5 $8 $0 $0 $M $M Solution Mix X1 X2 S1 S2 A1 A2 Qty $ X $ X $M A Zj $ $ $ $31-M $21-M $M $1,800+20M Cj-Zj $ $ $ $M $2M $0

Problem with an Unbounded Solution
$ X $ S Zj -$9 $9 $18 $0 $270 Cj-Zj $15 $0 -$18 $0 Cj $6 $9 $0 $0 Sol. Mix X1 X2 S1 S2 Quantity Pivot column

Degeneracy Degeneracy is situation that can occur for solving an LP problem using the simplex method. Tied ratios in the simplex calculations signal degeneracy. Cycling may result from degeneracy. Degeneracy could lead to a situation known as cycling, in which the simplex algorithm alternates back and forth between the same nonoptimal solutions; that is, it puts a new variable in, then takes it out in the next tableau, puts it back in, and so on. One simple way of dealing with the issue is to select either row (S2 or S3 in this case) arbitrarily. If we are unlucky and cycling does occur, we simply go back and select the other row.

Problem Degeneracy Cj $5 $8 $2 $0 $0 $0 Solution
Mix X1 X2 X3 S1 S2 S3 Qty $8 X $0 S $0 S Zj $2 $8 $8 $16 $0 $0 $80 Cj-Zj $3 $0 -$6 -$16 $0 $0 Pivot column

Sensitivity Analysis with the Simplex Tableau
Sensitivity analysis it applies to LP problems that we have solved graphically. This valuable concept shows how the optimal solution and the value of its objective function change, given changes in various inputs to the problem. Graphical analysis is useful in understanding intuitively and visually how feasible regions and slopes of objective functions can change as model coefficients change. Computer programs handling LP problems of all sizes provide sensitivity analysis as an important output feature.

High Note Sound Company Revisited
High Note Sound Company to illustrate sensitivity analysis graphically. High Note is a firm that makes compact disk (CD) players (called X1) and stereo receivers (called X2). Its LP formulation is repeated here: Maximize profit=$50X1+$120X2 Subject to 2X1+4X2≤ 80 (hours of electricians’ time available) 3X1+1X2 ≤ 60 (hours of audio technicians’ time available)

High Note Sound Company Graphical Solution
(Receivers) X2 60 40 a=(0,20) 20 10 Optimal Solution at point a X1=0CD Players X2=20 Receivers Profits=$2,400 B=(16,12) Isoprofit Line: $2,400+50X1+120X2 X1 c=(20,0) (CD players)

Changes in the Objective Function Coefficients
Second way of illustrating the sensitivity analysis of objective function coefficients is to consider the problem’s final simplex tableau. For the High Note Sound Company. X2= 20 stereo receivers (Basic variables) S2= 40 hours of slack time of audio technicians (Basic variables) X1= 0 CD players (non basic variables) S1= 0 hours of slack time of electricians (non basic variables) Basic variables (those in the solution mix) and nonbasic variables (those set equal to 0) must be handled differently using sensitivity analysis. Let us first consider the case of a nonbasic variable.

Optimal Solution by the Simplex Method
Nonbasic variables are variables that have a value of zero. The solution is optimal as long as all Cj-Zj ≤ 0 Cj-Zj ≤ 0 This is the same as writing Cj ≤ Zj Cj $50 $120 $0 $0 Sol. Mix X1 X2 S1 S2 Quantity $120 X $0 S Zj $60 $120 $30 $0 $2,400 Cj-Zj -$10 $0 -$30 $0

The range over which Cj rates for nonbasic variables can vary without causing a change in the optimal solution mix is called the range of insignificance. ≤ Cj (for X1) ≤ $60 ≤ Cj (for C1) ≤ $30 Testing basic variables involves reworking the final simplex tableau. △ ≤ 0 -10 ≤ 0.5△ -20 ≤ or △ ≥ -20

Basic Objective Function Coefficient
Testing basic variables involves reworking the final simplex tableau. △ ≤ 0 -10 ≤ 0.5△ -20 ≤ or △ ≥ -20 The range of optimality is the range of values over which a basic variable’s coefficient can change without causing a change in the optimal solution mix.

THE DUAL Every LP primal has a dual. The dual provides useful economic information. The dual variables represent the potential value of resources. These are the five steps for formulating a dual. If the primal is a maximization, the dual is a minimization, and vice versa. The RHS values of the primal constraints become the dual’s objective function coefficients. The primal objective function coefficients become the RHS values of the dual constraints. The transpose of the primal constraint coefficients become the dual constraint. Constraint inequality signs are reversed.

Karmarkar’s Algorithm
The biggest change in the field of LP solution techniques in four decades was 1984, alternative to the simplex algorithm. By Narendra Karmarkar, takes significantly less computer time to solve very large-scale LP problems. Algorithm finds a solution by moving from one adjacent corner point to the next, following the outside edges of the feasible region. In contrast, Karmarkar’s method follows a path of point on the inside of the feasible region. Karmarkar’s method is unique in its ability to handle an extremely large number of constraints and variables, thereby giving LP users the capacity to solve previously unsolvable problems. Simplex method to be used for many LP problems, Delta Air Lines became the first commercial airline to use the Karmarkar program, called KORBOX, was developed and sold by AT&T. Increased efficiency in allocating limited resources, Delta saves millions of dollars in crew time and related costs.

Problems 1- The Dreskin Development Company is building two apartment complexes. It must decide how many units to construct in each complex subjects to labor and material constraints. The profit generated for each apartment in the first complex is estimated at $900, for each apartment in the second complex, $1500. A partial initial simplex tableau for Dreskin is given in the following table. (a) Complete the initial tableau. (b) Reconstruct the problem’s original constraints (excluding slack variables). (c) Write the problem’s original objective function. (d) What is the basis for the initial solution? Cj $900 $1,500 $0 $0 Solution X1 X2 S1 s2 QUANTITY Mix ,360 ,600 Zj Cj-Zj

(e) Which variable should enter the solution at the next interaction?
(f) Which variable will leave the solution at the next iteration? (g) How many units of the variable entering the solution next will be in the basis of the second tableau? (h) How much will profit increase in the next solution? Problem-2 A Pharmaceutical firm is about to begin production of three new drugs. An objective function designed to minimize ingredient costs three production constraints are as follows: Minimize cost = 50X1+10X2+75X3 Subject to X1-X2 =1,000 2X2+2X3 =2,000 X1 ≤1,500 X1,X2,X3 ≥ 0 (a) Convert these constraints and objective function to the proper form for use in the simplex tableau. (b) Solve the problem by the simplex method. What is the optimal solution and cost?

(a) Formulate this problem so as to minimize total production costs.
The Roniger Company produces two products: bed mattresses and box springs. A prior contract requires that the firm produce at least 30 mattresses or box springs, in any combination. In addition, union labor agreements demand that stitching machines be kept running at least 40 hrs per week, which is one production period. Each box spring takes 2 hrs of stitching time, and each mattresses takes 1 hrs on the machine. Each mattress produced costs $20: each box spring costs $24. (a) Formulate this problem so as to minimize total production costs. (b) Solve using the simplex method.

Rule for formulating dual problems
unrestricted

Primal and dual formulation
Max: 2x1+1.5x2+0.5x3+9x4+6x5 s.t.: 0.3x1+0.4x2+0.2x3+1x4<=20 0.4x1+0.2x2+0.3x3+1x5<=15 x1,x2,x3>=0, x4,x5<=0 Min: 20v1+15v2 s.t.: 0.3v1+0.4v2>= v1+0.2v2>= v1+0.3v2>=0.5 1v1<=9 1v2<=6 v1, v2>=0

INVENTORY CONTROL MODELS
Inventory is any stored resource that is used to satisfy a current or future need. All organizations have different type of inventory planning and control system. Bank has control of cash, Hospital has control of blood supplies, state and federal governments, school, and virtually and control. Inventory Planning and Control Planning on what Inventory to Stock And How to Acquire it Forecasting Parts/Product Demand Controlling Inventory Levels Feedback Measurements to Revise Plans and Forecasts

Importance of Inventory Control
Inventory control important functions of flexibility operation of the firm: 1- Decoupling Function: Inventory is to decouple manufacturing process, inventory can act as a buffer. 2- Storing Resources: Agricultural and seafood products definite seasons over which they can be harvested or caught, but the demand for these products constant during the year, inventory can be used to store these resources. Resources can be stored in work-in-process.

Important of Inventory Control
3- Irregular Supply an Demand: When supply or demand for an inventory item is irregular, storing certain amounts in inventory can be important. If the greatest demand for Diet-Delight beverage is during the summer, make planning to enough supply to meet this irregular demand. Inventory will be needed in the summer, true for irregular supplies. 4- Quantity Discounts: Another use of inventory is to take advantage of quantity discounts. Suppliers offer discounts of large orders. Purchasing in larger quantities can substantially reduce the cost of products.

Importance of Inventory Control
5- Avoiding Stockouts and Shortages: Important fucntion of inventory is to avoid shortages or stockouts. If the firm has repeatedly out of stock, customers are likely to go elsewhere to satisfy their needs.

How much to order When to order Inventory Decisions
A major objective of all inventory models is to minimize inventory costs. Ordering Cost Factors Carrying Cost Factors Developing and sending orders Cost of capital Processing and inspecting incoming inventory Taxes Bill Paying Insurance Inventory inquiries Spoilage Utilities, phone bills, and so on, for the purchasing Dept. Theft Salaries and wages for purchasing department employees Obsolescence Supplies such as forms and paper for the purchasing dept. Salaries and wages for warehouse employees Utilities and building costs for the warehouse Supplies such as forms and paper for the warehouse

Most significant inventory costs follow:
1- Cost of the items (purchase cost or material cost) 2- Cost of ordering 3- Cost of carrying, or holding, inventory 4- Cost of stockouts

ECONOMIC ORDER QUANTITY: DETERNINING HOW MUCH TO ORDER
The economic order quantity is most commonly inventory control techniques. It is used by large number of organizations. It is easy to use, but it some of the more assumptions follow: 1- Demand is known and constant. 2- The lead time, that is, the time between the placement of the order and the receipt of the order, is known and constant. 3- The receipt of inventory is instantaneous. The inventory from an order arrives in one batch, at one point in time. 4- The purchase cost per unit is constant throughout the year. Quantity discounts are not possible. 5- The only variable costs are the cost of placing an order, ordering cost, and the cost of holding or storing inventory over time, holding or carrying cost. The holding cost per unit per year and the ordering cost per order are constant throughout the year. 6- Orders are placed so that stockouts or shortages are avoided completely.

The inventory usage curve has a sawtooth shape
Inventory Usage over Time Inventory Level Order Quantity=0=Maximum Inventory Level Minimum Inventory Time

Inventory Costs in the EOQ Situation
The objective of the simple EOQ model is to minimize total inventory cost. The relevant costs are the ordering and holding costs The average inventory level is one-half the maximum level. Order quantity is Q, average inventory will be Q/2. Q=number of pieces to order EOQ=Q*=optimal number of pieces to order D=annual demand in units for the inventory item Co=ordering cost of each order Ch=holding or carrying cost per unit per year

Sensitivity Analysis with the EQM Model
EOQ Model all values are fixed and certainty. These values are change over time, and know to the order quantity might change when different values are used. These changes are said to be sensitivity analysis. EOQ formula as follows: EOQ= The changes in the inputs (D,Co,Ch) will relatively minor changes in optimal order quantity.

REORDER POINT: DETERMINING WHEN TO ORDER
The reorder point (RoP) determines when to order inventory. It is found by multiplying the daily demand times the lead time in days. When to order is usually expressed a reorder point, inventory level which order should be placed. The time between placing and receipt of an order called lead time, inventory must be maintain to meet the demand during this time. ROP= (Demand per day) X (Lead time for a new order in days) =dxl

Reorder Point Graphically
Reorder Point Curve Inventory Level Units Q* Slope=Units/Day=d ROP (Units) Lead Time=L Time (Days)

EOQ WITHOUT THE INSTANTANEOUS RECEIPT ASSUMPTION
The production run model eliminates the instantaneous receipt assumptions. Solving the production run model involves setting setup costs equal to holding costs and solving for Q. Inventory Control and the Production Process. Inventory Level Part of Inventory Cycle During Which Production is Taking Place There is No Production During This Part of the Inventory Cycle Maximum Inventory Time

Annual Setup Cost or Annual Ordering
A Product is produced over time, setup cost replaces ordering cost. Both of these are independent at order and size of the production run, this cost is simply the number of order times the ordering cost. Annual setup cost= D/QCs Annual ordering cost = D/QCo Determining the Optimal production Quantity Assumptions of the production run model are met cost are minimized when the setup cost equals the holding cost. Find the optimal quantity by setting these costs equal and solving for Q.

Annual holding cost = Annual setup cost
Q/2 (1-d/p) Ch=D/QCs Formula for the optimal production quantity similar to the basic EOQ Model. Quantity Discount Models Companies offer quantity discounts, if all EOQ assumptions are met find the minimizing total inventory cost. In this methods purchase cost and material cost becomes relevant cost. Relevant cost as follows: Total cost =Material cost+Ordering cost+Carrying cost

Relevant Costs: Total cost=DC+D/QCo+Q/2Ch Where D=annual demand in units Co= ordering cost each order C=cost per unit Ch=holding or carrying cost per unit per year holding cost per unit per year is based on cost of the items, it is express as Ch=IC I= holding cost as a percentage of the unit cost (C)

The overall objective of the quantity discount model is to minimize total inventory cost, which now include actual material costs. Quantity Discount Model 1- For each discount price (C), compute EOQ= 2- If EOQ< Minimum for discount, adjust the quantity to Q= Minimum for discount. 3- for each EOQ or adjusted Q, compute total cost=DC+D/QCo+Q/2Ch. 4- Choose the lowest-cost quantity.

USE OF SAFETY STOCK Safe stock helps in avoiding stockouts. It is stock kept on hand. Necessary to carry additional inventory, to prevent stockout. Safety stock is included in the ROP. demand and lead time are constant, ROP=dXL d= daily demand L= order lead time When demand during the lead time is uncertain and safety stock is necessary, the ROP becomes ROP=dXL+SS SS= Safety stock

Use of Safety Stock Inventory on hand Time Stockout

Use of Safety Stock Inventory on Hand Safety Stock, SS
0 Units Stockout is Avoided Time

Safety Stock with Unknown Stockout Costs
Determining stockout costs may be difficult or impossible. An alternative to determining safety stock is to use service level and the normal distribution. Service level = - Probability of a stockout or Probability of a stockout = 1- Service level

Safety Stock and the Normal Distribution
Safety level can be determined when the probability of demand over the lead time follows a normal curve. Safety Stock and the Normal Distribution 5% Area of Normal Curve SS µ =350 X=?

µ= Mean demand = 350

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