Deterministic Operations Research, ME 366Q and ORI 391 Chapter 2: Homework #2 Solutions 11. Consider the following linear program. Maximize z = 6x 1 + 3x 2 subject to x 1 + 2x 2 2x 1 + x 2 20 x 1 x 2 x 1 + x 2 3 Sketch the feasible region and several isovalue contours for the objective function in the (x 1,x 2 )-space. Show the optimal solution on the graph. x 2 Z = 36 Z = 60 Note that the objective function line is parallel to the second constraint. All feasible points on that constraint line are alternative optima. 0 0 x 1 12. Ten jobs are to be completed by three men during the next week. Each man works a 40- hour week. The times for the men to complete the jobs are shown in the table below. The values in the cells assume that each job is completed by a single person; however, jobs can be shared with completion times being determined proportionally. If no entry exists in a particular cell, it means that the corresponding job cannot be performed by the corresponding person. Set up and solve a linear programming model that will determine the optimal assignment of men to jobs. The goal is to minimize the total time required to complete all the jobs. - 1 -
Man \ Task 1 2 3 4 5 6 7 8 9 A 7 3 18 13 6 9 B 12 5 12 4 22 17 13 C 18 6 8 19 8 15 VARIABLE DEFINITIONS x ij : Proportion of job j performed by man i, for i = A, B, C, and j = 1,...,. Variables should be defined for only the man-job combinations not eliminated with a dash in the table. Thus there are 20 variables in the problem. CONSTRAINTS Each man can only work 40 hours. Man A: + 7x A2 + 3x A3 +18x A6 +13x A7 + 6x A8 + 9x A 40 Man B: +12x B1 + 5x B2 +12x B4 + 4x B5 +22x B6 +17x B8 +13x B9 40 Man C: +18x C1 + 6x C3 + 8x C4 +x C5 +19x C7 + 8x C9 +15x C 40 Each job must be done. TASK1: x B1 + x C1 = 1 TASK: x A + x C = 1 NONNEGATIVITY AND SIMPLE UPPER BOUNDS OBJECTIVE FUNCTION 0 x ij 1 for all i and j for which variables are defined. MINIMIZE: Total time required for the jobs. Min z = 7x A2 + 3x A3 + 18x A6 + 13x A7 + 6x A8 + 9x A 12x B1 + 5x B2 + 12x B4 + 4x B5 + 22x B6 + 17x B8 + 13x B9 18x C1 + 6x C3 + 8x C4 + x C5 + 19x C7 + 8x C9 +1 5x C Assignment Value: z = 88-2 -
1 2 3 4 5 6 7 8 9 Man A 0 1 0.5 1 1 1 Man B 1 1 0 1.5 0 0 Man C 0 0 0 1 0 0 1 0 Hours Dual Man A 40 0.22 Man B 32 0 Man C 16 0 13. A company has two manufacturing plants (A and B) and three sales outlets (I, II, and III). Shipping costs from the plants to the outlets are as follows. Outlet Plant I II III A 4 6 8 B 7 4 3 The company wants to plan production, shipping and sales for the next two periods. The data for the periods is shown below. The plants can store products produced in one period for sale in the next. The maximum storage at each plant is 50 and the inventory cost is $1 per unit. Find the solution that maximizes profit. Plant A Manufacturing data Plant B Period Unit cost ($) Capacity Unit cost ($) Capacity 1 8 175 7 200 2 150 8 170 Selling price ($) Demand Data Maximum sales Period I II III I II III 1 15 20 14 0 200 150 2 18 17 21 150 300 150-3 -
Variables P1A, P1B, P2A, P2B: production at A and B in the two periods S11, S21, S31: sales in period 1 at three locations S12, S22, S32: sales in period 2 at three locations x1a1, x1a2, x1a3, x1b1, x1b2, x1b3: shipments in period 1 x2a1, x2a2, x2a3, x2b1, x2b2, x2b3: shipments in period 2 IA, IB: Inventory storage at A and B Max profit: Z = 15S11 + 20S21 + sales revenue 8P1A 7P1B production costs 4x1A1 6x1A2 8x1A3 transportation costs 1IA 1IB Conservation constraints and upper limits on variables (see Excel spreadsheet) Solution Profit: Z = $4450-4 -
1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 A B C D E F G H I J K L M N O P Q R S T U Linear Model Name: E_13 Solver: Jensen LP/IP Ph. 1 Iter. TRUE Type: LP1 Type: Linear Total Iter. 26 FALSE Change Goal: Max Sens.: No Comp. Time 00:01 TRUE Profit: 4450 Side: No Status Optimal FALSE Solve FALSE Variables 1 2 3 4 5 6 7 8 9 11 12 13 14 0 Change Relation Name: P1A P1B S11 S12 S13 X1A1 X1A2 X1A3 X1B1 X1B2 X1B3 I1A I1B P2A 0 Values: 175 200 0 200 0 0 75 0 0 125 0 0 75 150 0 Lower Bounds: 0 0 0 0 60 Upper Bounds: 175 200 0 200 150 000 000 000 000 000 000 0 0 150 Linear Obj. Coef.: -8-7 15 20 14-4 -6-8 -7-4 -3-1 -1 - Constraints Num. Name Value Rel. RHS Linear Constraint Coefficients 1 A1 0 = 0 1 0 0 0 0-1 -1-1 0 0 0-1 0 0 2 B1 0 = 0 0 1 0 0 0 0 0 0-1 -1-1 0-1 0 3 S11 0 = 0 0 0-1 0 0 1 0 0 1 0 0 0 0 0 4 S12 0 = 0 0 0 0-1 0 0 1 0 0 1 0 0 0 0 5 S13 0 = 0 0 0 0 0-1 0 0 1 0 0 1 0 0 0 6 A2 0 = 0 0 1 0 1 7 B2 0 = 0 0 0 1 0 8 S21 0 = 0 0 0 0 0 9 S22 0 = 0 0 0 0 0 S23 0 = 0 0 0 0 0 PA PB S1 S2 S3 XA1 XA2 XA3 XB1 XB2 XB3 IA IB Period 1 175 200 0 200 0 0 75 0 0 125 0 0 75 Period 2 150 170 150 95 150 150 0 0 0 95 150 The two rows give a table with the numeric solution to the problem. Period 1 shows the decisions in period 1 and Period 2 shows the decisions in period 2. For example, the 175 in the upper left corner (cell 1,1) means that 175 units of A are produced in period 1. - 5 -
1 2 3 4 5 6 7 8 9 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 V W X Y Z AA AB AC AD AE AF 15 16 17 18 19 20 21 22 23 24 P2B S21 S22 S23 X2A1 X2A2 X2A3 X2B1 X2B2 X2B3 170 150 95 150 150 0 0 0 95 150 170 150 300 150 000 000 000 000 000 000-8 18 17 21-4 -6-8 -7-4 -3 0 0 0 0-1 -1-1 0 0 0 1 0 0 0 0 0 0-1 -1-1 0-1 0 0 1 0 0 1 0 0 0 0-1 0 0 1 0 0 1 0 0 0 0-1 0 0 1 0 0 1-6 -