Chapter 3 Formulation of LP Models

Transcription

1 Chapter 3 Formulation of LP Models The problems we have considered in Chapters 1 and 2 have had limited scope and their formulations were very straightforward. However, as the complexity of the problem increases so does the need for a systematic method of formulating the LP model. In this chapter we introduce a technique that has proven useful in setting up and solving complex LP problems. 3.1 Introduction to Model Formulation The approach to modeling we will use is through the identification of different parts of the model, includes a schematic diagram of each activity at its unit level, and includes a diagram of the total (exogenous) flow of resources and products through the system being modeled. The steps in this modeling process are as follows. 1. Identify the objective that is the goal of the system. Decide whether it is to be maximized or minimized. 2. Identify and name each decision or activity variable. These are the things that a decision maker can control to work toward achieving the objective stated in Identify and name the resources that are consumed by the decisions or activities stated in Identify and name the products or items that are produced by the decisions or activities stated in Diagram each activity or decision variable. Resources that are required by the activity are shown as positive inputs entering the diagram from the left. Products that are generated from the activity are shown as negative outputs leaving the diagram on the right. 6. Diagram the total flow of resources and products of the system. The amount of each resource that is available for use is shown as a positive input from the left. The number or quantity of each product that is produced by the activity is shown as a negative output on the right. 7. Convert the diagrams to constraints and build the initial tableau. 8. Look for additional constraints. Formulate them and add them to the initial tableau. 9. Record non-negative constraints and note which variables are not restricted in sign. 10. Construct the formulation tableau, having as column headings the activities or decision variables and having as row headings the resources and products. To illustrate the procedure, consider the following problem from Hillier and Lieberman, Introduction to Operations Research, Sixth Edition, 1995, McGraw-Hill. The SOUTHERN CONFEDERATION OF KIBBUTZIM is a group of 3 kibbutzim (communal farming communities) in Israel. Overall planning for this group is done in its Coordinating Technical Office. The office currently is planning agricultural production for the coming year. The agricultural output of each kibbutz is limited by both the amount of available irrigable land and the quantity of water allocated for irrigation by the Water Commissioner (a national government official). These data are given in Table

2 page 2 Table 3.1: Resource Data for the Southern Confederation of Kibbutzim Kibbutz Usable Land (Acres) Water Allocation (Acre-feet) Crops suited for this region include sugar beets, cotton, and sorghum, and these are the three being considered for the upcoming season. These crops differ primarily in their expected net return per acre and their consumption of water. In addition, the Ministry of Agriculture has set a maximum quota for the total acreage that can be devoted to each of these crops by the Southern Confederation of Kibbutzim, as shown in table 3.2. Table 3.2: Crop Data for the Southern Confederation of Kibbutzim Maximum Water Consumption Net Return Crop Quota (Acres) (Acre-feet / acre) ($ / acre) Sugar Beets ,000 Cotton Sorghum Because of the limited water available for irrigation, the Southern Confederation of Kibbutzim will not be able to use all its irrigable land for planting crops in the upcoming season. To ensure equity between the three kibbutzim, it has been agreed that every kibbutz will plant the same proportion of its available irrigable land. For example, if kibbutz 1 plants 200 of its available 400 acres, then kibbutz 2 must plant 300 of its 600 acres, while kibbutz 3 plants 150 of its 300 acres. However, any combination of the crops may be grown at any of the kibbutzim. The job facing the Coordinating Technical Office is to plan how many acres to devote to each crop at the respective kibbutzim while satisfying the given restrictions. The objective is to maximize the total net return to the Southern Confederation of Kibbutzim as a whole. The respective steps in the formulation of this problem follow. 1. Identify the objective to be obtained and note whether or not it is to be maximized or minimized. In this case, the objective is clearly stated. OBJECTIVE: (max) total net return 2. Identify the decisions that must be made to achieve this objective. DECISIONS: number of acres at each kubbutz (1, 2, or 3) to devote to each crop (sugar beets, cotton, or sorghum). The decisions must be translated into quantitative terms. This means a variable for each decision. DECISION VARIABLES: SB(i) for the number of acres of sugar beets at kubbutz i, i = 1, 2, 3. C(i) for the acres of cotton at kubbutz i. S(i) for the acres of sorghum at kubbutz i.

3 page 3 3. Identify the available resources that are needed in each decision. RESOURCES: Land at kubbutz i (400, 600, 300 acres) Water at kubbutz i (600, 800, 375 acre-feet) Crop quota for sugar beets, cotton and sorghum (600, 500, and 325 acres) 4. Identify any products (including revenue) that result from the decision. PRODUCTS: Net return ($) 5. Construct flow diagrams for each decision variable at the unit level. a. Consider the decision to plant 1 acre of sugar beets at kubbutz i. This requires 1 acre of land at kubbutz i, L(i), 3 acre-feet of water, 3 W(i), and 1 acre of sugar beet quota, Q(sb). Since these are requirements (inputs), arrows are drawn on the left going into the decision box. The only output is revenue or return ($). Its arrow is drawn on the right coming out of the decision box and it is opposite in sign to the inputs. L(i): 1 W(i): 3 Q(sb): 1 SB(i) 0 Plant 1 acre of sugar beets at kibbutz i return: -$1,000 b. The decision boxes for cotton and sorgham are done similarily. L(i): 1 W(i): 2 Q(c): 1 C(i) 0 Plant 1 acre of cotton at kibbutz i return: -$750 L(i): 1 W(i): 1 Q(s): 1 S(i) 0 Plant 1 acre of sorghm at kibbutz i return: -$250

4 page 4 6. Diagram the total flow through the system. L(1): 400 L(2): 600 L(3): 300 W(1): 600 W(2): 800 W(3): 375 Q(sb): 600 Q(c): 500 Q(s): 325 Total System return: -Z (max) 7. Each of the corresponding arrows in the flow diagrams will generate simple constraints or the objective function. For example, consider the arrows for the land required for each decision. The land at kubbutz 1 is limited to 400 acres. This results in the constraint SB ( 1) + C(1) + S (1) 400. The land constraints for the other kubbutzim are found similarly as SB ( 2) + C(2) + S (2) 600 and SB ( 3) + C(3) + S (3) 300. The water arrows translate into the constraints 3* SB (1) + 2 * C(1) + S(1) 600, 3* SB (2) + 2 * C(2) + S(2) 800, and 3* SB (3) + 2 * C(3) + S(3) 375. The quota arrows give SB ( 1) + SB(2) + SB(3) 600, C ( 1) + C(2) + C(3) 500, and S ( 1) + S (2) + S (3) 325. The return ($) arrows generate the objective function, 1000* SB (1) + SB(2) + SB(3) + 750* C(1) + C(2) + C(3) + 250* S(1) + S(2) + S(3) Z= ( ) ( ) ( ) 8. There may be additional constraints that were not generated by the flow diagrams. In this case this includes the proportion constraint that the proportion of land used will be the same at each kubbutz. One way to express this is via two equations, SB ( 1) + C(1) + S(1) SB(2) + C(2) + S(2) SB ( 1) + C(1) + S(1) SB(3) + C(3) + S (3) = and = Notice that these are also linear equations. They reduce to the following constraint equations. These will be referred to as the balance equations and given the names bal1 and bal SB(1) C(1) S(1) 400 SB(2) 400 C(2) 400 S(2) = 0, and 300 SB(1) C(1) S(1) 400 SB(3) 400 C(3) 400 S(3) = 0 9. Non-negative constraints should be recorded, SB(i) 0, C(i) 0, and S(i) 0 for i = 1, 2, The formulation tableau with activities in the columns and resources in the rows. This is shown in Table 3.3.

5 page 5 Table 3.3 Formulation Matrix for Kibbutzim Problem SB(1) C(1) S(1) SB(2) C(2) S(2) SB(3) C(3) S(3) relation rhs L(1) <= 400 L(2) <= 600 L(3) <= 300 W(1) <= 600 W(2) <= 800 W(3) <= 375 Q(sb) <= 600 Q(s) <= 500 Q(c) <= 325 bal = 0 bal = 0 return max -Z In 3.2, we will discuss how to use the Excel solver to find a solution to this LP. Exercise: 1. (Farmer Paul s Annual Planning Problem) Farmer Paul has 1010 acres of land and must develop a plan for next year s operations. He can grow corn, soybeans, raise steers, or get a part time job in a chicken packing plant. His land has an expected yield of 140 bushels of corn per acre, and 38 bushels of soybeans per acre. The expected profit of corn is $0.90 per bushel and for soybeans is $2.45 per bushel. Corn uses 3 hours of labor per acre and soybeans uses 2 hours of labor per acre. Corn requires $70 per acre in capital cost and soybeans requires $60 per acre in capital costs. The profit per steer is $90 and each animal requires $280 in capital costs, 4 hours of labor, 0.05 acres of land, and 45 bushels of corn. The job at the chicken packing plant pays $7.50 per hour and he can work there up to 2,000 hours per year. Farmer Paul can hire part-time farm hands for $8.70 per hour out of pocket costs. Hired hands, however, are not as efficient as Paul and, on the average, require 1.25 hours to do what he can in one hour. Farmer Paul has 2500 hours of available labor time, and a credit line of $90,000 to cover capital costs. Also he is able to rent his land out for $80 per acre per year. Formulate this problem as an LP model. 3.2 Using the Microsoft Excel Solver The organization of the spreadsheet used to solve an LP may be customized to suit the user. The way I prefer to do it is as follows. See Figure Identify the problem. This may be done using a name and a short description of the problem. See rows 1 and 2 in Figure List the activity names or decision variables in the top row of the formulation table, row 4 in Figure The second row is where the current values of the decision variables are stored. These are in row 5, columns B through J, in Figure 3.1. Initially all are set to zero. 4. The third row is another list of the decision variables, row 6. This is just to separate the values of the variables from the objective function and functional constraints that follow. 5. The fourth row, row 7 of Figure 3.1, consists of the objective function coefficients, columns B through J, and the value of the objective function at the current solution,

6 page 6 column K. This value is calculated using the SUMPRODUCT() function as shown in the command line of Figure 3.1. Notice that the coefficients in the objective row are positive, not negative as was the case in the simplex method tableaus in Chapter The remainder of the table (rows 9 18) is made up of the coefficients of the functional constraints (columns B J), their value for the current solution (column K), the relationship they must satisfy (column L), and the right hand side bound on their value (column M). If the reference to the cells containing the current values of the decision variables is fixed, as shown in the command line of Figure 3.1, column K may be filled in by copying the formula in row 7 column K and pasting in the remaining cells. Figure 3. 1: Kibbutzim Problem Initial Spreadsheet Once the spreadsheet has been built, then the formulation of the LP problem in the Excel solver can begin. The solver is accessed under the Tools menu. If it does not appear on the menu when it is fully opened, then you must install it. Selecting the Add-Inns menu item, see Figure 3.2, should do this but may require the Microsoft Office installation disk. When the Solver menu item on the Tools menu is selected, a dialog box is opened that is used to define the LP model in Excel, see Figure 3.3. The target cell is set to $K$7, the current value of the objective function, since that cell was selected when the menu item was chosen. The default objective for the target cell is to maximize, which is what we want for this problem. Entering the reference to the cells that are to be changed to achieve this objective is the next step in the formulation process. This is done by clicking on the icon next to the text box, selecting the cells containing the current solution, $B$5:$J$5, and pressing the Enter key.

7 page 7 Figure 3. 2: Installing the Solver Add-in Figure 3. 3: Initial Solver Dialog Box Step 3 in the formulation process is defining the constraints. This is done using the lower left portion of the dialog box. First select the Add button. This opens another dialog box as shown in Figure 3.4. The cell reference, $K$8 on the left, is to the current value of the first constraint. The relationship is <=, which is what is desired. The constraint reference, =$M$8, is to the rhs value for the constraint. Selecting the Add button on the dialog box completes the definition of the constraint. This results in the constraint being added to the formulation of the LP and keeps the Add Constraint dialog box open for the definition of additional constraints. Select the OK button for the last constraint to close the dialog box. The constraint definitions should now appear in the text box at the lower left of the solver dialog box, see Figure 3.5.

8 page 8 Figure 3. 4: Add Constraint to Solver Formulation Figure 3. 5: Solver Dialog Box after Constraint Definitions There is one last step before starting the solve process. This is to set the options so that the solver knows this is a linear problem with non-negative constraints. Selecting the Options button at the right on the solver dialog box begins this process. This causes another dialog box to be opened which presents the user with a number of options, see Figure 3.6. We want the Assume Linear Model and Assume Non-Negative check boxes selected. Clicking the OK button returns us to the solver dialog box. We are now ready to have Excel try to solve the LP.

9 page 9 Figure 3. 6: Solver Options The Excel solution algorithm is initiated by selecting the Solve button at the top left of the solver dialog box, as shown in Figure 3.5. If a solution is found, then the dialog box shown in Figure 3.7 is shown. Select the Answer and Sensitivity Reports to obtain much useful information about the solution. The user has the option of keeping the solver solution in the changed cells or returning to the original values. Figure 3. 7: Solver Results Dialog Box

10 page 10 Notice that the values of cells B5:J5 have been changed. These new values are the solution found by the Excel solver. The user has the option of keeping the solver solution in the changed cells or returning to the original values. A more detailed listing of the solution is found on the Answer Report 1 sheet. It contains the information given in Table 3.4. In particular, the maximum return the Southern Confederation of Kibbutzim can achieve is $633, This results from the following planting plan for the individual Kibbutzim. 1. Kibbutz 1 plants / 3 acres of sugar beets, 100 acres of cotton and 0 acres of sorghum. 2. Kibbutz 2 plants 100 acres of sugar beets, 250 acres of cotton and 0 acres of sorghum. 3. Kibbutz 3 plants 25 acres of sugar beets, 150 acres of cotton and 0 acres of sorghum. Microsoft Excel 9.0 Answer Report Worksheet: [kibbutzim.xls]sheet1 Report Created: 2/19/01 10:51:32 AM Table 3.4 Microsoft Excel Answer Report Target Cell (Max) Cell Name Original Value Final Value $K$7 profit value Adjustable Cells Cell Name Original Value Final Value $B$5 solution SB(1) $C$5 solution C(1) $D$5 solution S(1) 0 0 $E$5 solution SB(2) $F$5 solution C(2) $G$5 solution S(2) 0 0 $H$5 solution SB(3) 0 25 $I$5 solution C(3) $J$5 solution S(3) 0 0 Constraints Cell Name Cell Value Formula Status Slack $K$10 L(3) value 175 $K$10<=$M$10 Not Binding 125 $K$11 W(1) value 600 $K$11<=$M$11 Binding 0 $K$12 W(2) value 800 $K$12<=$M$12 Binding 0 $K$13 W(3) value 375 $K$13<=$M$13 Binding 0 $K$14 Q(sb) value $K$14<=$M$14 Not Binding $K$15 Q(c) value 500 $K$15<=$M$15 Binding 0 $K$16 Q(s) value 0 $K$16<=$M$16 Not Binding 325 $K$17 balance1 value 0 $K$17=$M$17 Binding 0 $K$18 balance2 value 0 $K$18=$M$18 Binding 0 $K$8 L(1) value $K$8<=$M$8 Not Binding $K$9 L(2) value 350 $K$9<=$M$9 Not Binding 250 Information about the constraints is also provided. In particular, notice that the binding constraints are water at each of the Kibbutzim and the quota on cotton. We may conclude that cotton has the highest return for unit of water. Since there is indication that some sugar beets are

11 page 11 to be grown, they have the next highest return per unit of water. Sorghum has the least return per unit of water, and the solution indicates that none of it is to be grown. The sensitivity report also contains useful information that will be discussed later in the notes. Exercise: 1. Use the Excel Solver to find a solution to Farmer Paul s Annual Planning Problem. 2. Notice that the optimal solution does not have Farmer Paul planting soybeans. How much must the price of soybeans increase to cause him to grow soybeans? Find that profit for soybeans to the nearest nickel. Case 2 3. Farmer Paul can no longer find hired hands, but he has been offered a job as a truck mechanic at $14 per hour. Should he take it? What is the lowest offer he would accept to take a full time position (2000 hours per year)? Consider increments of $0.50. Case 3 4. Working off Case 1 develop Farmer Paul s supply curve for soybeans. 5. Working off Case 1 develop Farmer Paul s supply curve for corn. 3.3 Personnel Scheduling Example To further illustrate the formulation and solution of more complex LP problems, consider the Personnel Scheduling problem from Hillier and Lieberman, Introduction to Operations Research, Sixth Edition, 1995, McGraw-Hill. UNION AIRWAYS is adding more flights to and from its hub airport, and so it needs to hire additional customer service agents. However, it is not clear just how many more should be hired. Management recognizes the need for cost control while also consistently providing a satisfactory level of service to customers. Therefore, an OR team is studying how to schedule the agents to provide satisfactory service with the smallest personnel cost. Based on the new schedule of flights, an analysis has been made of the minimum number of customer service agents that need to be on duty at different times of the day to provide a satisfactory level of service. The rightmost column of Table 3.4 shows the number of agents needed for the time periods given in the first column. The other entries in this table reflect on of the provisions in the company s current contract with the union that represents the customer service agents. The provision is that each agent work an 8-hour shift 5 days per week, and the authorized shifts are Shift 1: 6:00 A.M. to 2:00 P.M. Shift 2: 8:00 A.M. to 4:00 P.M. Shift 3: 12:00 A.M. (noon) to 8:00 P.M. Shift 4: 4:00 P.M. to 12:00 P.M. (midnight) Shift 5: 10:00 P.M. to 6:00 A.M. Checkmarks in the main body of Table 3.4 show the hours covered by the respective shifts. Because some shifts are less desirable than others, the wages specified in the contract differ by shift. For each shift, the daily compensation (including benefits) for each agent is shown in the bottom row. The problem is to determine how many agents should be assigned to the respective

12 page 12 shifts each day to minimize the total personnel cost for agents, based on this bottom row, while meeting (or surpassing) the service requirements given in the rightmost column. Table 3.4 Data for the Union Airways Personnel Scheduling Problem Time Periods Covered Shift Minimum Number of Time Period Agents Needed 6:00 A.M. to 8:00 A.M. 48 8:00 A.M. to 10:00 A.M :00 A.M. to 12:00 A.M :00 A.M. to 2:00 P.M. 87 2:00 P.M. to 4:00 P.M. 64 4:00 P.M. to 6:00 P.M. 73 6:00 P.M. to 8:00 P.M. 82 8:00 P.M. to 10:00 P.M :00 P.M. to 12:00 P.M :00 P.M. to 6:00 A.M. 15 Daily cost per agent $170 $160 $175 $180 $195 The formulation of the LP model proceeds as follows. 1. OBJECTIVE: (min) total personnel cost 2. DECISIONS: number of agents to assign to each shift. DECISION VARIABLES: x i = number of agents assigned to shift i = 1, 2, 3, 4, RESOURCES: capital ($) 4. PRODUCTS: service required (agents), R j, j = 1, 2,, flow diagrams capital:$170 x 1 0 assign 1 agent to shift 1 R 1 : -1 R 2 : -1 R 3 : -1 R 4 : -1 capital:$160 x 2 0 assign 1 agent to shift 2 R 2 : -1 R 3 : -1 R 4 : -1 R 5 : -1 x 3 0 R 4 : -1 x 4 0 R 6 : -1 capital:$175 assign 1 agent to shift 3 R 5 : -1 R 6 : -1 R 7 : -1 capital:$180 assign 1 agent to shift 4 R 7 : -1 R 8 : -1 R 9 : -1

13 page 13 x 5 0 capital:$195 assign 1 agent to shift 5 R 9 : -1 R 10 : -1 R 1 : -48 R 2 : -79 R 3 : Flow diagram for the total system capital: Z (min) Total System R 4 : -87 R 5 : -64 R 6 : -73 R 7 : -82 R 8 : -43 R 9 : -52 R 10 : simple constraints R 1 : x 1 48; (equivalent to x 1-48), R 2 : x 1 + x 2 79; R 3 : x 1 + x 2 65; (not needed because of previous constraint) R 4 : x 1 + x 2 + x 3 87; R 5 : x 2 + x 3 64; R 6 : x 3 + x 4 73; (not needed because of next constraint) R 7 : x 3 + x 4 82; R 8 : x 4 43; R 9 : x 4 + x 5 52; R 10 : x objective function. Z = 170 x x x x x 5 8. additional constraints none 9. non-negative constraints x i 0 for i = 1, 2, 3, 4, formulation tableau The formulation of the Union Airways Personnel Scheduling problem is shown in Figure 3.8 and 3.9.

14 page 14 Figure 3. 8: Union Airways Personnel Scheduling Problem Initial Tableau The solution is shown in Table 3.5. The minimum cost is $30, when 48 agents are assigned to shift 1, 31 agents to shift 2, 39 agents to shift 3, 43 agents to shift 4, and 15 agents to shift 5. The binding (exactly met) requirements are 1, 2, 7, 8, and 10. The other requirements are non-binding and more agents are assigned to these time intervals than the minimum required. Exercises: Formulate and solve the following LP problems. 1. A farmer is raising pigs for market and wishes to determine the quantities of the available types of feed (corn, tankage, and alfalfa) that should be given to each pig. Since pigs will eat any mix of these feed types, the objective is to determine which mix will meet certain nutritional requirements at a minimum cost. The number of units of each type of basic nutritional ingredient contained within 1 kilogram of each feed type is given in the following table, along with the nutritional requirements and feed costs. Nutritional Ingredient Corn (1 kg) Tankage (1 kg) Alfalfa (1 kg) Carbohydrates Protein Vitamins Cost ( ) Min. Daily Requirement

15 page 15 Figure 3. 9: Formulation of Union Airways Personnel Scheduling Problem 2. A cargo plane has 3 compartments for storing cargo: front, center, and back. These compartments have capacity limits on both weight and space, as summarized in the table below. Compartment Weight Capacity (tons) Front 12 Center 18 Back 10 Space Capacity (cubic feet) 7,000 9,000 5,000 Furthermore, the weight of the cargo in the respective compartments must be the same proportion of that compartment s weight capacity to maintain the balance of the airplane. The following four cargoes have been offered for shipment on an upcoming flight as space is available Cargo Weight (tons) Volume (cubic feet / ton) Profit ($ / ton)

16 page 16 Any portion of these cargos can be accepted. The objective is to determine how much (if any) of each cargo should be accepted and how to distribute each among the compartments to maximize the total profit for the flight. Table 3.5 Solution to Union Airways Personnel Scheduling Problem Microsoft Excel 9.0 Answer Report Target Cell (Min) Cell Name Original Value Final Value $G$7 cost value Adjustable Cells Cell Name Original Value Final Value $B$5 #assgnd shift $C$5 #assgnd shift $D$5 #assgnd shift $E$5 #assgnd shift $F$5 #assgnd shift Constraints Cell Name Cell Value Formula Status Slack $G$10 Rqd 3 value 79 $G$10>=$I$10 Not Binding 14 $G$11 Rqd 4 value 118 $G$11>=$I$11 Not Binding 31 $G$12 Rqd 5 value 70 $G$12>=$I$12 Not Binding 6 $G$13 Rqd 6 value 82 $G$13>=$I$13 Not Binding 9 $G$14 Rqd 7 value 82 $G$14>=$I$14 Binding 0 $G$15 Rqd 8 value 43 $G$15>=$I$15 Binding 0 $G$16 Rqd 9 value 58 $G$16>=$I$16 Not Binding 6 $G$17 Rqd 10 value 15 $G$17>=$I$17 Binding 0 $G$8 Rqd 1 value 48 $G$8>=$I$8 Binding 0 $G$9 Rqd 2 value 79 $G$9>=$I$9 Binding 0

17 page Multi-period Planning Models A class of problems that are frequently solved using linear programming are planning problems that extend over several time periods. For these problems, the output from one period becomes the input to the next period. Constructing flow diagrams does their formulation nicely. A simple example of a multi-period planning problem follows. Consider the problem of stockpiling a product for sale at a later date. The idea is to earn money by buying the item when the price is low and selling when it is high. The location used for storing the item has a finite capacity of 200 items. The storage costs are $1.00 per item per time period. At the beginning of a time period items may either be purchased or sold at the current market price, which is shown in Table 3.6. The initial stock is 50 items. The problem is to determine the numbers of items to be sold and/or stored in each of the next 3 time periods. At the beginning of the 4 th time period, all but 50 items (original stock size) in the inventory will be sold. Table 3.6: Market Price of Item Time period Market Price ($) It is often useful to break the formulation of a multi-period problem into pieces so that the complexity of the problem is not overwhelming. We will do that to this example. The first part will be solving the problem for just the first period, selling all items at the beginning of period 2. Once this is done, we will extend the problem to 2 time periods, selling all items at the beginning of period 3. If these are done successfully, it should be an easy step to then formulate the entire problem. Part 1: First Period Only. In this part, we begin at time 0 with an inventory of 50 items. We must decide the number of these to sell at the beginning of period 1 and the number of them to store during period 1. Complicating the formulation is the option of buying items at the beginning of period 1, storing them, then selling them at the beginning of period 2. Thus items stored in period 1 may come from the current inventory or may be purchased at the beginning of the period. One way to handle this complication is to separately make the decision to store an item currently in inventory (which produces an item at the beginning of period 1) and the decision to store an item on hand at the beginning of period Objective: maximize profits. 2. Decision or activity variables: Store_0 = number of items in inventory to store (i.e., not sell) during period 1. Sell_1 = number of items in inventory to sell at beginning of time period 1. Buy_1 = number of items to buy at beginning of time period 1 Store_1 = number of items to store during time period 1 Sell_2 = number of items to sell at beginning of period Resources:

18 page 18 Item_0 = number of items in initial inventory. Storage_1 = storage space in period 1. Item_1 4. Products: Cost = money gained (negative) or lost (positive). 5. Diagram of activity variables Deciding to keep an item in the initial inventory and store it in period 1 requires one Item_0 unit and produces one Item_1 unit. Store_0 0 Item_0: 1 keep 1 item Item_1: -1 in initial inventory to store in period 1 Selling an item at the beginning of period 1 requires an item from the initial inventory and produces the market price in revenue. Sell_1 0 Item_0: 1 sell 1 item Cost: -15 at beginning of period 1 Buying an item at the beginning of period 1 requires the market price and produces one unit of Item_1. Cost: 15 Buy_1 0 buy 1 item at beginning of period 1 Item_1: -1 Storing an item during period 1 requires one unit of Item_1, $1 cost, 1 unit of storage space in period 1, and produces 1 unit of Item_2. Item_1: 1 Cost: 1 Storage_1: 1 Store_1 0 store 1 item during period 1 Item_2: -1

19 page 19 Selling an item at the beginning of period 2 requires a unit of Item_1 and produces the market price in revenue. Item_1: 1 Sell_2 0 sell 1 item at beginning of period 2 Cost: -18 The formulated matrix of coefficients and the Excel solution is shown in Table 3.7. Notice that the solution is to purchase 150 items at the beginning of period 1 at $15 each, then sell the 200 items at the beginning of period 2 at $18 each. The profit is the return from selling the items minus the cost of buying the items and the storage costs, or 200*18 150* = This solution is rather obvious and does not require linear programming to find. Table 3.7: Multi-period Planning Example, First Period Only Store0 Sell1 Buy1 Store1 Sell Store0 Sell1 Buy1 Store1 Sell2 Value relation rhs profit MAX item = 50 item = 0 storecap <= 200 item = 0 Part 2: First Two Periods. We now extend the model to include the second period, selling all items at the beginning of the third period. To do this, we add activities for buying at the beginning of period 2, for storing items during period 2, and for selling items at the beginning of period 3. In addition, we add items for units of items on hand at the beginning of period 3 and units of storage during period 2. The flow diagrams are similar to those for period 1 and will not be shown here. The formulation matrix and solution to this problem are shown in Table 3.8. Table 3.8: Multi-period Planning Example, First Two Periods Store0 Sell1 Buy1 Store1 Sell2 Buy2 Store2 Sell Store0 Sell1 Buy1 Store1 Sell2 Buy2 Store2 Sell3 Value rel rhs profit MAX item = 50 item = 0 storecap <= 200 item = 0 storecap2 1 0 <= 200 item = 0 Notice that the solution has not changed from the single period part. This makes sense, since the selling price at the beginning of period 3 is lower than that of period 2.

20 page 20 Part 3: Three Periods. If all the items were to be sold at the beginning of period 4, we could simply extend the table as we did in Part 2. However, we want to keep 50 of the items in inventory and this increases the complexity of the model. We can handle this much the same way as what we did for the initial inventory. That is, insert a store activity that doesn t incur the storage costs and doesn t require any storage space. The formulation matrix and its solution are shown in Table 3.9. Note that the solution makes sense because it indicates that we should buy when the price is low and sell when it is high. Table 3.8: Multi-period Planning Example, First Two Periods Store0Sell1Buy1 Store1Sell2 Buy2Store2 Sell3Buy3 Store3Sell4Keep Store0Sell1Buy1 Store1Sell2 Buy2Store2 Sell3Buy3 Store3Sell4Keep Value rel rhs profit MAX item = 50 item = 0 storecap <= 200 item = 0 storecap2 1 0 <= 200 item = 0 storecap <= 200 item = 0 endreq 1 50 = 50 Exercises: 1. (Single Period) The DelMar Furniture Company manufactures tables, chairs, and bookshelves. The scarce resources used in the making of its furniture are lumber and two types of skilled labor. The amounts of these resources needed to make the three types of furniture are given in the following table. Resource Table Chair Bookshelf Lumber (board ft) Finishing (hrs) Carpentry (hrs) Next week the company will take delivery of 1760 board feet of lumber, and have available 200 finishing hours, and 160 carpentry hours. DelMar Furniture has a contract to deliver at least 10 dinning room sets and not more than 15 sets next week. Each set is made up of 1 table and 4 chairs. The contract price results in a contribution towards overhead and profit of $100 per set. A discount furniture dealer will take all the bookshelves that DelMar can produce at a price resulting in a contribution towards overhead and profits of $18 per bookshelf. The same dealer will take up to 200 chairs at a price resulting in a contribution towards overhead and profits of $12 per chair. DelMar wants to maximize the total contribution to profit and overhead.

21 Formulate and solve an LP model to produce an optimal production plan. Provide a list of activities and items included in your model. page (Multi-Period) The DelMar Furniture Company has learned it must plan ahead in scheduling its manufacturing operations. Currently they schedule operations every week using a two-week planning horizon. Their hope is to move to a longer planning horizon after building an LP model to aid in this planning. For each of the next two weeks the company will take delivery of 1760 board feet of lumber, and have available 200 finishing hours, and 160 carpentry hours. DelMar Furniture has a contract to deliver 10 dinning room sets next week and at least 15 sets and not more than 20 sets the following week. Each set is made up of 1 table and 4 chairs. The contract price results in a contribution towards overhead and profit of $100 per set. A discount furniture dealer will take all the bookshelves that DelMar can produce at a price resulting in a contribution towards overhead and profits of $18 per bookshelf. The same dealer will take up to 200 chairs at a price resulting in a contribution towards overhead and profits of $12 per chair. Tables produced next week for delivery the following week can be stored at a cost of $1.50 per table. Lumber can be store for use the following week at a cost of $4.00 per 100 board feet. DelMar wants to maximize the total contribution to profit and overhead. Formulate and solve an LP model to produce an optimal production plan. Provide a list of activities and items included in your model. Present the optimal production schedule in a table for use by the DelMar s production manager. 3. (Product Mix) DelMar s President has decided that it is time to implement some changes in the company and to grow the company. A new piece of furniture is to be introduced, a chest of drawers. New employees are to be hired and he directed the sales department to negotiate better, more profitable agreements with the furniture dealers. The new chest of drawers will require 42 board feet of lumber, 4 hours of finishing time, and 4 hours of carpentry time per chest. Production will begin two Mondays from today. Each chest sold will earn the company $150 contribution to profit and overhead. The estimate market for the chest is between 15 and 25 per week. Chests can be stored for a week at the cost of $2.00 per chest. As result of the sales department s efforts the company s dealers will require at least 12 and not more than 22 dinning room sets per week. A higher price for chairs and bookshelves has been negotiate with the dealers. The new prices result in a contribution to profits and overhead from the sale of a chair of $17 per chair and $22 per bookshelf. However at the new, higher price the dealers will only accept no more than 50 chairs per week and 30 bookshelves per week. These prices are effective starting next week. One new man has been hired to work in the finishing department and two new men have been hired for the carpentry work. The new employees will begin productive work two Mondays from today. DelMar s workers work a 40-hour week. The factory will start next week with 135 board feet and 5 tables in storage. The limit on the delivery of lumber has been increased to 1950 board feet per week for the week starting two Mondays from today. The rest of the data needed is the same as in Version 2 of the DelMar problem.

22 page 22 Create the LP model for this new situation by modifying the old LP model for use in planning the production schedule for the two-week period starting next Monday. Also make recommendations about further changes in the size of DelMar s workforce and about the size of the weekly lumber delivery.