REV 01 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 PROGRAMMING IN OPERATIONAL RESEARCH
DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

CHAPTER 1: INTRODUCTION TO QUANTITATIVE ANALYSIS
REV 01 CHAPTER 1: INTRODUCTION TO QUANTITATIVE ANALYSIS Quantitative analysis uses a scientific approach to decision making Both quantitative and qualitative factors must be considered Quantitative factors: Determine how much our investment will be worth in the future when deposited at a bank at a given interest rate for a certain number of years DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH
REV 01 REV 00 Computing financial ratios whose stock we are considering Analyse cash flow and rates of return for investment property Qualitative factors: Weather State and federal legislation New technological breakthroughs Outcome of an election DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Quantitative analysis approach Defining a Problem Testing the Solution Developing a Model Analysing the Result Acquiring Input Data Implementing the Result Developing a Solution DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 1. Defining the Problem Defining the problem can be the most important step Concentrate on only a few problems 2. Acquiring Input Data The types of models include physical, scale, schematic and mathematical models Most models contain one or more variables and parameters DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 3. Acquiring Input data Obtaining accurate data for a model is essential Sources of data can be company reports or any data sources 4. Developing a Solution Involves manipulating the model to arrive at the best (optimal) solution to the problem A series of steps or procedures that are repeated is called algorithm DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 5. Testing the Solution Includes determining the accuracy and completeness of the data used by the model, done before the solution can be analysed and implemented 6. Analysing the Result Starts with determining the implications of the solution Sensitivity analysis or post optimality analysis determines how the solutions will change with a different model or input data DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 7. Implementing the Result Incorporating the solution in the company DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Models of Cost revenue and Profit Profit = Revenue – Expenses Revenue = Selling Price per Unit (P) × Number of units sold (x) Expenses = Fixed Cost (F) + Variable Cost per unit (V) × Number of Units Sold (x) Breakeven Point = When Total Revenue Equals Total Expenses DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Breakeven Chart DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 1.1: Swatch Time Plc buys, sells and repairs clock and clock parts. It sells rebuilt spring for a price of RM10. The fixed cost of the equipment to build the spring is RM1, 000. The variable cost per unit is RM5. What is the Breakeven Point? What is the profit earned if the company sells 500 units of springs? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

CHAPTER 2: DECISION THEORY
REV 01 CHAPTER 2: DECISION THEORY Decision theory is an analytical and systematic way to tackle problems A good decision is based on logic Six steps in Decision Making: 1. Clearly defined the problem at hand Whether to expand manufacturing line or market new product DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 2. List the possible alternatives Alternative is defined as a course of action or a strategy that the decision maker can choose 3. Identify the possible outcomes or states of nature States of nature are those outcomes which the decision maker has little or no control on it, e.g. favourable market or unfavourable market DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 4. List the payoff or profit of each combination of alternatives and outcomes Example 2.1: Alternative State of Nature Favourable Market (RM) Unfavourable Market (RM) Construct a large plant 200, 000 - 180, 000 Construct a small plant 100, 000 - 20, 000 Do nothing DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 5. Select one of the mathematical decision theory models 6. Apply the model and make the decision Types of Decision Making Environments 1. Decision making under certainty Decision makers know with certainty the consequences of every alternative or decision choice. E.g. You have RM1, 000 to invest. You have two options: Saving account with 6% interest or Government bond with 10% interest. Which is your recommendation? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 2. Decision making under risk Decision maker knows the probability of occurrence of each outcome. Decision theory models normally employ two equivalent criteria: maximisation of expected monetary value and minimisation of expected loss 3. Decision making under uncertainty Here the probability are not known to the decision maker DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Decision Making Under Risk Models
REV 01 Decision Making Under Risk Models (1) Expected Monetary Value (EMV) EMV is the weighted sum of possible payoffs for each alternative EMV (alternative i) = (payoff)1 × (probability)1 (payoff)2 × (probability)2 + ………….. + (payoff)i × (probability)i DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.2: John of 1st Silicon has the options of the following: Which alternative would you recommend to John? Alternative State of Nature Favourable Market (RM) Unfavourable Market (RM) Construct a large plant 200, 000 - 180, 000 Construct a small plant 100, 000 - 20, 000 Do nothing Probabilities 0.5 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 (2) Expected Value of Perfect Information (EVPI) EVPI places an upper bound on what to pay for information EVPI = Expected value with Perfect Information (EVwPI) – Maximum EMV EVwPI = (best payoff)1 × (probability)1 + (best payoff)2 × (probability)2 + ……..+ (best payoff)i × (probability)i DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.3: John has been approached by Xcel Marketing to help John to make decision. It will change his environment from one decision-making under risk to one decision-making under certainty. Xcel Marketing would charge John RM65, 000 for the information. What would you recommend? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 (3) Expected Opportunity Loss (EOL) EOL is the cost of not picking the best solution An alternative approach to maximise EMV is to minimise EOL Opportunity loss, sometimes called regret, refers to the difference between optimal profit for a given state of nature and the actual payoff received Solving steps: (i) Construct the opportunity loss table, (ii) Compute the EOL DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 EOL will always result in the same decision as the maximum EMV Minimum EOL = Maximum EMV Example 2.4: Construct the EOL for John of Example 2.2. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Decision Making Under Uncertainty Models
REV 01 Decision Making Under Uncertainty Models (1) Maximax It finds the alternative that maximises the maximum payoff for every alternative Maximax is an optimistic approach (2) Maximin It finds the alternative that maximises the minimum payoff for every alternative Maximin is a pessimistic approach DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.5: Alternative State of Nature Maximum in Row (RM) Favourable Market (RM) Unfavourable Market (RM) Construct a large plant 200, 000 - 180, 000 Construct a small plant 100, 000 - 20, 000 Do Nothing DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.6: Alternative State of Nature Minimum in Row (RM) Favourable Market (RM) Unfavourable Market (RM) Construct a large plant 200, 000 - 180, 000 Construct a small plant 100, 000 - 20, 000 Do Nothing DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 (3) Criterion of Realism (Hurwicz Criterion) Uses the weighted average approach It is a compromise between an optimistic and a pessimistic decision A coefficient of realism, , is selected Note that when  = 1, this is the same as the optimistic criterion, when  = 0, this is the same as the pessimistic criterion Weighted average =  (Maximum in row) + (1 - )(Minimum in row) DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.7: Assuming  for John is 0.8. What is your recommendation? (4) Equally Likely (Laplace) Computes the highest average outcome Select the alternative with the maximum average payoff Example 2.8: Compute the Equally Likely criterion for John in Example 2.2 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 (5) Minimax Regret Based on Opportunity Loss It finds the alternative that minimises the maximum opportunity loss within each alternative Example 2.9: Develop an opportunity loss table for John in Example 2.2. What is the minimax regret decision? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Decision Tree Any problem that can be presented in a decision table can also be graphically illustrated in a decision tree All decision trees are similar in that they contain decision points or nodes and state of nature or nodes: A decision node from which one of several alternatives may be chosen A state of nature node out of which one state of nature will occur DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Five steps of Decision Tree Analysis: Define the problem Structure or draw the decision tree Assign probabilities to the state of nature Estimate payoffs for each possible combination of alternatives and states of nature Solve the problem by computing expected monetary values (EMV) for each state of nature. This is done by working backward that is, starting at the right of the tree and working back to decision nodes on the left. At each decision node, the alternative with the best EMV is selected DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.10: John of 1st Silicon has the options of the following: Which alternative would you recommend to John? Draw the decision tree. Alternative State of Nature Favourable Market (RM) Unfavourable Market (RM) Construct a large plant 200, 000 - 180, 000 Construct a small plant 100, 000 - 20, 000 Do nothing Probabilities 0.5 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 When a sequence of decisions need to be made, decision trees are much more powerful tools than decision tables: All outcomes and alternatives must be considered Most of the probabilities are conditional probabilities The cost of the survey had to be subtracted from the original payoffs Start by computing the EMV of each branch DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.11: DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Expected Value of Sample Information (EVSI) measures the value of sample information EVSI = (EV with sample information + cost) – (EV without sample information) In John’s case, EVSI = (RM49, RM10,000) – (RM40, 000) = RM19, 200 This means that John could have paid up to RM19, 200 for a market survey and still come out ahead. Since it costs only RM10, 000, the survey is indeed worthwhile. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Utility Theory The overall value of the result of a decision is called utility Rational people make decisions that maximise the expected utility Utility assessment assigns the worst outcome a utility of 0 and the best outcome utility of 1 When you are indifferent, the expected utilities are equal Utility values replace monetary value in utility as a decision making criterion. The objective is to maximise the expected utility DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Expected utility of alternative 2 = expected utility of alternative 1 Utility of other outcome = (p)(utility of best outcome, 1) + (1 – p)(utility of worst outcome, 0) Utility of other outcome = (p)(1) + (1 – p)(0) = p DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 2.12: Jane would like to construct a utility curve revealing her preference for money between RM0 and RM10, 000. A utility curve is the graph that plots utility value versus monetary value. She can either invest her money in a bank savings account or she can invest the same money in a real estate deal. If the money is invested in the bank, in three years Jane would have RM5, 000. Is she invested in real estate, after three years she could either have nothing or RM10, 000. Jane, however, is very conservative. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Unless there is an 80% chance of getting RM10, 000 from the real estate deal Jane would prefer to have money in the bank, where it is safe. Construct the utility assessment. Construct the utility curve for the following information: U (RM0) = 0 U (RM3, 000) = 0.50 U (RM5, 000) = 0.80 U (RM7, 000) = 0.90 U (RM10, 000) = 1.00 Is Jane a risk avoider or a risk taker? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

CHAPTER 3: LINEAR PROGRAMMING MODEL
REV 01 CHAPTER 3: LINEAR PROGRAMMING MODEL Linear programming is a technique for solving problems for profit maximisation or cost minimisation and resource allocation The word programming is simply used to denote a series of events The various aspects of the problem are expressed as linear equations DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Setting up a linear programming model: Step 1: Define the variables What are the quantities that the company can vary? Step 2: Establish constraints Having defined the variables we can now translate the two constraints into inequalities involving the variables DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Step 3: Establish objective function An objective function is the mathematical expression of the aim of a linear programming exercise There are two methods in linear programming: Graphical methods – for problems with two variables only Simplex method – for problems with more than two variables DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Graphical Method In the graphing model, the area on a graph of a linear programming problem within which all of the inequalities are satisfied in known as a feasible area y y y = 6 4x + 3y = 24 x x DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.1: 1.Draw the feasible area for the following inequalities: 2x + 3y < 12 y > 2x 2.Yung Kong Limited manufactures plastic covered steel fencing in two qualities, standard and heavy gauge. Both product pass through the same processes, involving steel-forming and plastic bonding. Standard gauge fencing sells at RM18 a roll and heavy gauge fencing at RM24 a roll. Variable costs per roll are RM16 and RM21 respectively. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 There is an unlimited market for the standard gauge, but demand for the heavy gauge is limited to 1, 300 rolls a year. Factory operations are limited to 2, 400 hours in each of the two production processes. Determine the production mix, which will maximize total contribution. Calculate the total contribution. Processing hours per roll Gauge Steel-forming Plastic-bonding Standard 0.6 0.4 Heavy 0.8 1.2 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 3. Dell Computer has undertaken a contract to supply a customer with at least 260 units in total of two products, X and Y, during the next month. At least 50% of the total output must be in units of X. The products are each made by two grades of labour, as follows. X Y Hours Grade A labour 4 6 Grade B labour 2 Total 8 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Although additional labour can be made available at short notice, the company wishes to make use of 1, 200 hours of Grade A labour and 800 hours of Grade B labour which has already assigned to working on the contract next month. The total variable cost per unit is RM120 for X and RM100 for Y. Dell Computer Ltd. Wishes to minimise expenditure on the contract next month. Determine the production mix. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Simplex Method Recall that the theory of LP states that the optimal solution will lie at a corner point of the feasible region In large LP problems, the feasible region cannot be graphed because it has many dimensions, but the concept is the same The simplex method systematically examines corner points, using algebraic steps, until an optimal solution is found DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 In setting up the initial simplex solution: Slack variables are added to each less-than-or-equal-to constraint Each slack variable represents an unused resource Slack considers only corner points as it seeks the best solution Then construct the initial simplex table See attached notes for steps of Simplex Method for solving Maximisation Problems and Minimisation Problems DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.2: The Flair Furniture 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 labour hours in the painting and varnishing department. Each table takes 4 hours of carpentry and 2 hours in the painting and varnishing shop. Each chair requires 3 hours in carpentry and 1 hour in painting and varnishing. During the current production period, 240 hours of carpentry time are available and 100 hours in painting and varnishing time are available. Each table sold yields a profit of RM7; each chair produced is sold for a RM5 profit. Formulate and solve the problem using simplex method. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 To handle > and = constraints, the simples method makes a conversion like it made to < constraints When dealing with a > constraint, we subtract a surplus variable to form an equality The surplus variable tells how much the solution exceeds the constraints resource Because of its analogy to a slack variable, surplus is sometimes simply called negative slack DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.3: Constraint 1: 5X1 + 10X2 + 8X3 > 210 Rewritten as: 5X1 + 10X2 + 8X3 – S1 = 210 In setting up an initial simplex table, since all real variables such as X1, X2 and X3 are set to “0”, S1 takes a negative value All variables in LP problems must be nonnegative at all times DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Hence, we need to introduce another type of variable called artificial variable Artificial variables are needed in > and = constraints Constraint 1 completed: 5X1 + 10X2 + 8X3 – S1 + A1 = 210 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.4: Constraint 2: 25X1 + 30X2 = 900 Rewritten as: 25X1 + 30X2 + A2 = 900 Artificial variables have no physical meaning and drop out of the solution mix before the final table To make sure that an artificial variable is forced out before the final solution is reached, it is assigned a very high cost DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.5: Minimise cost: Z = $5X1 +$9X2 + $7X3 Constraints: 5X1 +10X2 + 8X3 > 210 25X1 +30X2 = 900 What would the completed objective function and constraints be? DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 3.6: The Tetra Chemical Corporation must produce exactly 1, 000kg of a special mixture of phosphate and potassium for a customer. Phosphate costs $5 per kg and potassium costs $6 per kg. No more than 300kg of phosphate can be used, and at least 150kg of potassium must be used. The problem is to determine the least-cost blend of the two ingredients. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Four special cases in LP: Infeasibility Unboundedness Redundancy Alternate optimal solutions DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

CHAPTER 4: TRANSPORTATION AND ASSIGNMENT MODELS
REV 01 CHAPTER 4: TRANSPORTATION AND ASSIGNMENT MODELS Transportation problem: Distribution of items from several sources to several destinations Supply capacities and destination requirements are known Assignment problem: One-to-one assignment of people to jobs DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Importance of special purpose algorithms: Fewer, less complicated, computations than with simplex Less computer memory required Produce integer solutions Methods for solving transportation problem: Northwest corner rule Stepping-stone method MODI method Vogel’s approximation DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Northwest Corner Rule: Start in the upper left-hand cell and allocate units to shipping routes as follows: Exhaust the supply (factory capacity) of each row before moving down to the next row Exhaust the demand (warehouse) requirements of each column before moving to the next column to the right Check that all supply and demand requirements are met DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 4.1: Transportation costs, factory capacity and warehouse requirement are as follows. What is the optimal distribution method? From (Sources) To (destinations) Factory Capacity Everise Upwell Choice Daily Jacobs $5 $4 $3 100 Danone $8 300 Julie’s $9 $7 Warehouse Requirement 200 700 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Stepping-Stone Method: Select any unused square to evaluate Begin at this square. Trace a closed path back to the original square via squares that are already being used (only horizontal or vertical moves allowed) Place + in unused square: alternate – and + on each corner square of the closed path Calculate improvement index: add together the unit cost figures found in each square containing a +; subtract the unit cost figure in each containing a – Repeat steps 1 ~ 4 for each unused square DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 MODI Method: Compute the values for each row and column: set Ri + Kj = Cij for those squares currently used or occupied After writing all equations, set R1 = 0 Solve the system of equations for Ri and Kj values Compute the improvement index for each unused square by the formula improvement index: Cij – Ri + Kj Select the largest negative index and proceed to solve the problem as you did using the stepping stone method DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Vogel’s Approximation Method: For each row/column of table, find difference between two lowest cost (opportunity cost) Find the greatest opportunity cost Assign as many units as possible to lowest cost square in row/column with greatest opportunity cost Eliminate row or column, which has been completely satisfied Begin again, omitting eliminated rows/columns DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The Hungarian Method (Assignment Model): Subtract the smallest number in each row from every number in that row Subtract the smallest number in each column from every number in that column Draw the minimum number of vertical and horizontal straight lines necessary to cover zeros in the table If the number of lines equals the number of rows or columns, than one can make an optimal assignment DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 If the number of lines does not equal the number of rows and columns: Subtract the smallest number not covered by a line from every other uncovered number Add the same number to any number lying at the intersection of any two lines Return to Step 2 Make optimal assignments at locations of zeros within the table DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 4.2: What is the optimal assignment? Person Project No. 1 2 3 Adams 11 14 6 Brown 8 10 Cooper 9 12 7 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

CHAPTER 5: NETWORK MODELS
REV 01 CHAPTER 5: NETWORK MODELS The presentation will cover three network models that can be used to solve a variety of problems: the minimal-spanning tree technique, the maximal-flow technique, the shortest-route technique, and PERT / CPM DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Minimal-Spanning Tree Technique
REV 01 Minimal-Spanning Tree Technique Definition: The minimal-spanning tree technique determines the path through the network that connects all the points while minimizing total distance. For example: If the points represent houses in a subdivision, the minimal spanning tree technique can be used to determine the best way to connect all of the houses to electrical power, water systems, etc. in a way that minimizes the total distance or length of power lines or water pipes. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

The Maximum Flow Technique
REV 01 The Maximum Flow Technique Definition: The maximal-flow technique finds the maximum flow of any quantity or substance through a network. For example: This technique can determine the maximum number of vehicles (cars, trucks, etc.) that can go through a network of roads from one location to another. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Shortest Route Technique
REV 01 Shortest Route Technique Definition: Shortest route technique can find the shortest path through a network. For example: This technique can find the shortest route from one city to another through a network of roads. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Minimal-Spanning Tree Steps
REV 01 Selecting any node in the network. Connecting this node to the nearest node minimizing the total distance. Finding and connecting the nearest unconnected node. If there is a tie for the nearest node, one can be selected arbitrarily. A tie suggests that there may be more than one optimal solution. Repeating the third step until all nodes are connected. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Example 5.1: Network for Lauderdale Construction
REV 01 Example 5.1: Network for Lauderdale Construction DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Solving the network for Melvin Lauderdale construction Start by arbitrarily selecting node 1. Since the nearest node is the third node at a distance of 2 (200 feet), connect node 1 to node 3. Considering nodes 1 and 3, look for the next-nearest node. This is node 4, which is the closest to node 3 with a distance of 2 (200 feet). Once again, connect these nodes DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Maximal-Flow Technique
REV 01 Maximal-Flow Technique The maximal-flow technique allows the maximum amount of a material that can flow through a network to be determined. For example: It has been used to find the maximum number of automobiles that can flow through a state highway system. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 5.2: Waukesha is in the process of developing a road system for downtown. City planners would like to determine the maximum number of cars that can flow through the town from west to east. The road network is shown in next slide DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Road Network for Waukesha
REV 01 Road Network for Waukesha Traffic can flow in both directions DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Maximal-Flow Technique (continued)
REV 01 Maximal-Flow Technique (continued) The Four Maximal-Flow Technique Steps: Pick any path from the start (source) to the finish (sink) with some flow. If no path with flow exists, then the optimal solution has been found. Find the arc on this path with the smallest flow capacity available. Call this capacity C. This represents the maximum additional capacity that can be allocated to this route. For each node on this path, decrease the flow capacity in the direction of flow by the amount C. For each node on this path, increase the flow capacity in the reverse direction by the amount C. Repeat these steps until an increase in flow is no longer possible. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Solving the Waukesha Start by arbitrarily picking the path 1–2–6, at the top of the network What is the maximum flow from west to east? It is 2 because only 2 units (200 cars) can flow from node 2 to node 6 Now we adjust the flow capacities. As you can see, we subtracted the maximum flow of 2 along the path 1–2–6 in the direction of the flow (west to east) and added 2 to the path in the direction against the flow (east to west) The result is the new path in next slide DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Capacity Adjustment 1 2 6 3 East Point West Add 2 Subtract 2 Iteration 1 4 New path DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Final Network Flow The maximum flow of 500 cars per hour is summarized in the following table: (Cars per Hour) PATH FLOW Total =500 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

The Shortest-Route Technique
REV 01 The Shortest-Route Technique The shortest-route technique minimizes the distance through a network. The shortest-route technique finds how a person or item can travel from one location to another while minimizing the total distance traveled. The shortest-route technique finds the shortest route to a series of destinations. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

Example 5.3: From Ray’s Plant to Warehouse
REV 01 Example 5.3: From Ray’s Plant to Warehouse For example, Every day, Ray Design, Inc., must transport beds, chairs, and other furniture items from the factory to the warehouse. This involves going through several cities. Ray would like to find the route with the shortest distance. The road network is shown on the next slide. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Shortest-Route Technique (continued) Roads from Ray’s Plant to Warehouse: 1 2 3 4 5 6 100 150 200 50 40 Warehouse Plant DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Steps of the Shortest-Route Technique Find the nearest node to the origin (plant). Put the distance in a box by the node. Find the next-nearest node to the origin (plant), and put the distance in a box by the node. In some cases, several paths will have to be checked to find the nearest node. Repeat this process until you have gone through the entire network. The last distance at the ending node will be the distance of the shortest route. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Shortest-Route Technique (continued) Ray Design: 1st Iteration 100 1 2 3 4 5 6 100 150 200 50 40 Warehouse Plant The nearest node to the plant is node 2, with a distance of 100 miles. Thus, connect these two nodes. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

PROJECT MANAGEMENT: PERT / CPM
REV 01 PROJECT MANAGEMENT: PERT / CPM Project management can be used to manage complex projects. The first step in planning and scheduling a project is to develop the work breakdown structure. This involves identifying the activities that must be performed in the project. Each detail and each activity may be broken into its most basic components. The time, cost, resource requirements, predecessors, and person(s) responsible are identified. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 One of the earliest techniques was the Gantt chart (Used by US Navy). This type of chart shows the start and finish times of one or more activities, as shown below: DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The program evaluation and review technique (PERT) and the critical path method (CPM) are two popular quantitative analysis techniques that help managers plan, schedule, monitor, and control large and complex projects. They were developed because there was a critical need for a better way to manage. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 There are six steps common to both PERT and CPM. Define the project and all of its significant activities or tasks. Develop the relationships among the activities. Decide which activities must precede others. Draw the network connecting all of the activities. Assign time and/or cost estimates to each activity. Compute the longest time path through the network; this is called the critical path. Use the network to help plan, schedule, monitor, and control the project. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The critical path is important because activities on the critical path can delay the entire project. PERT is probabilistic, whereas CPM is deterministic. Almost any large project can be subdivided into a series of smaller activities or tasks that can be analyzed with PERT. Projects can have thousands of specific activities and it is important to be able to answer many associated questions. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Example 5.4: Background: General Foundry, Inc., a metal works plant has long been trying to avoid the expense of installing air pollution control equipment. The local environmental protection group has recently given the foundry 16 weeks to install a complex air filter system on its main smokestack. General Foundry was warned that it will be forced to close unless the device is installed in the allotted period. They want to make ensure that installation of the filtering system progresses smoothly and on time. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Define the project and all project activities. Immediate predecessors are determined in the second step. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Activities and events are drawn and connected. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The fourth step is to assign activity times. Time is usually given in units of weeks. Without solid historical data, managers are often uncertain as to activity times. The developers of PERT thus employed a probability distribution based on three time estimates for each activity: Optimistic time Pessimistic time Most likely time DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Optimistic time (a) = time an activity will take if everything goes as well as possible. There should be only a small probability (say, 1/1000) of this occurring. Pessimistic time (b) = time an activity would take assuming very unfavorable conditions. There should also be only a small probability that the activity will really take this long. Most likely time (m) = most realistic time estimate to complete the activity. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Time Estimates (weeks) for General Foundry, Inc. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The fifth step is to compute the longest path through the network— the critical path. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 To find the critical path, need to determine the following quantities for each activity in the network: Earliest start time (ES): the earliest time an activity can begin without violation of immediate predecessor requirements. Earliest finish time (EF): the earliest time at which an activity can end. Latest start time (LS): the latest time an activity can begin without delaying the entire project. Latest finish time (LF): the latest time an activity can end without delaying the entire project. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The earliest times are found by beginning at the start of the project and making a forward pass through the network The latest times are found by beginning at the finish of the project and making a backward pass through the network Slack is the length of time an activity can be delayed without delaying the whole project. Mathematically, Slack = LS-ES or LF-EF DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 The Network Diagram: DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Slack Time: DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 CHAPTER 6: FORECASTING Forecasting techniques: Qualitative techniques Time series methods: moving average, exponential smoothing, trend projections Causal methods: regression analysis, multiple regression DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Time series can be decomposed into: Trend (T): gradual up or down movement over time Seasonality (S): pattern of fluctuations above or below trend line that occurs every year Cycles (C): patterns in data that occur every several years Random variations (R): ‘blips’ in the data caused by change and unusual situations DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Components of Decomposition DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Scatter Diagram DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Multiplicative model: Demand = T × S × C × R Additive model: Demand = T + S + C + R Moving Average = Demand in previous n periods n Exponential Smoothing: New forecast = Previous forecast +  (Previous actual – Previous forecast) DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Forecast errors allow one to see how well the forecast model works and compare that model with other forecast models. Forecast error = actual value – forecast value DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Measures of forecast accuracy include: Mean Absolute Deviation (MAD) Mean Squared Error (MSE) Mean Absolute Percent Error (MAPE) = å |forecast errors| n = å (errors) n 2 error = å actual n × 100% DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Trend Projection The formula for the trend projection is: Y = b + b X where: Y = predicted value b1 = slope of the trend line b0 = intercept X = time period (1,2,3…n) DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Correlation Coefficient The correlation coefficient (r) measures the strength of the linear relationship. = ( ) [ ] å - 2 Y n X XY r Note: -1 < r < 1 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 Tracking signals measure how well predictions fit actual data. DDC 2313 PROGRAMMING IN OPERATIONAL RESEARCH

REV 01 PROGRAMMING IN OPERATIONAL RESEARCH

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