Chapter 14 Goal Programming Goal programming is used

Chapter 14: Goal Programming Goal programming is used to solve linear programs with multiple objectives, with each objective viewed as a "goal". In goal programming, di+ and di- , deviation variables, are the amounts a targeted goal i is overachieved or underachieved, respectively. The goals themselves are added to the constraint set with di+ and di- acting as the surplus and slack variables. Slide 1

Goal Programming One approach to goal programming is to satisfy goals in a priority sequence. Second-priority goals are pursued without reducing the first-priority goals, etc. For each priority level, the objective function is to minimize the (weighted) sum of the goal deviations. Previous "optimal" achievements of goals are added to the constraint set so that they are not degraded while trying to achieve lesser priority goals. Slide 2

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

Goal Programming Formulation Step 5: Set up the new linear program. Consider the next-lower priority level goals and formulate a new objective function based on these goals. Add a constraint requiring the achievement of the next-higher priority level goals to be maintained. The new linear program might be: Min w 3 d 3+ + w 4 d 4 s. t. Functional Constraints, Goal Constraints, and w 1 d 1+ + w 2 d 2 - = k Go to step 4. (Repeat steps 4 and 5 until all priority levels have been examined. ) Slide 4

GP Example: Conceptual Products produces CP 400 and CP 500 computers that use memory modules, external hard drives, and cases. The CP 400 model uses two memory modules and no external hard drive, whereas the CP 500 uses one memory module and one external hard drive. Both models use one case. Suppliers can provide Conceptual Products with 1000 memory modules, 500 external hard drives, and 600 cases on a weekly basis. It takes one hour to manufacture a CP 400 and its profit is $200 and it takes one and one-half hours to manufacture a CP 500 and its profit is $500. Slide 5

GP Example: Conceptual Products The company has four goals: Priority 1: Meet a state contract of 200 CP 400 machines weekly. (Goal 1) Priority 2: Make at least 500 total computers weekly. (Goal 2) Priority 3: Make at least $250, 000 weekly. (Goal 3) Priority 4: Use no more than 400 man-hours per week. (Goal 4) Slide 6

GP Example: Formulation Variables x 1 = number of CP 400 computers produced weekly x 2 = number of CP 500 computers produced weekly di- = amount the right hand side of goal i is deficient di+ = amount the right hand side of goal i is exceeded Functional Constraints Availability of memory modules: 2 x 1 + x 2 < 1000 Availability of external hard drives: x 2 < 500 Availability of cases: x 1 + x 2 < 600 Slide 7

GP Example: Formulation Goals (1) 200 CP 400 computers weekly: x 1 + d 1 - - d 1+ = 200 (2) 500 total computers weekly: x 1 + x 2 + d 2 - - d 2+ = 500 (3) $250(in thousands) profit: . 2 x 1 +. 5 x 2 + d 3 - - d 3+ = 250 (4) 400 total man-hours weekly: x 1 + 1. 5 x 2 + d 4 - - d 4+ = 400 Non-negativity: x 1, x 2, di-, di+ > 0 for all i Slide 8

GP Example: Formulation Objective Functions Priority 1: Minimize the amount the state contract is not met: Min d 1 Priority 2: Minimize the number under 500 computers produced weekly: Min d 2 Priority 3: Minimize the amount under $250, 000 earned weekly: Min d 3 Priority 4: Minimize the man-hours over 400 used weekly: Min d 4+ Slide 9

GP Example: Formulation Summary Min P 1 ( d 1 -) + P 2 ( d 2 -) + P 3 ( d 3 -) + P 4 ( d 4 + ) s. t. 2 x 1 < 1000 +x 2 500 < 600 = 200 500 +x 2 x 1 x 1 < + d 1 - - d 1+ +x 2 . 2 x 1+. 5 x 2 + d 2 - - d 2+ = + d 3 - - d 3+ = Slide 10

GP Example: Graphical Solution Iteration 1 To solve graphically, first graph the functional constraints. Then graph the first goal: x 1 = 200. Note on the next slide that there is a set of points that exceed x 1 = 200 (where d 1 - = 0). Slide 11

GP Example: Graphical Solution Functional Constraints and Goal 1 Graphed x 2 1000 2 x 1 + x 2 < 1000 800 Goal 1: x 1 > 200 x 2 < 500 x 1 + x 2 < 600 400 Points Satisfying Goal 1 200 400 600 800 1000 1200 x 1 Slide 12

GP Example: Graphical Solution Iteration 2 Now add Goal 1 as x 1 > 200 and graph Goal 2: x 1 + x 2 = 500. Note on the next slide that there is still a set of points satisfying the first goal that also satisfies this second goal (where d 2 - = 0). Slide 13

GP Example: Graphical Solution Goal 1 (Constraint) and Goal 2 Graphed x 2 1000 2 x 1 + x 2 < 1000 800 Goal 1: x 1 > 200 600 x 1 + x 2 < 600 400 x 2 < 500 Points Satisfying Both Goals 1 and 2 200 Goal 2: x 1 + x 2 > 500 200 400 600 800 1000 1200 x 1 Slide 14

GP Example: Graphical Solution Iteration 3 Now add Goal 2 as x 1 + x 2 > 500 and Goal 3: . 2 x 1 +. 5 x 2 = 250. Note on the next slide that no points satisfy the previous functional constraints and goals and satisfy this constraint. Thus, to Min d 3 -, this minimum value is achieved when we Max. 2 x 1 +. 5 x 2. Note that this occurs at x 1 = 200 and x 2 = 400, so that. 2 x 1 +. 5 x 2 = 240 or d 3 - = 10. Slide 15

GP Example: Graphical Solution Goal 2 (Constraint) and Goal 3 Graphed x 2 2 x 1 + x 2 < 1000 Goal 1: x 1 > 200 800 x 1 + x 2 < 600 x 2 < 500 (200, 400) Points Satisfying Both Goals 1 and 2 400 Goal 2: x 1 + x 2 > 500 200 Goal 3: . 2 x 1 +. 5 x 2 = 250 200 400 600 800 1000 1200 x 1 Slide 16

Goal Programming: More Complex Problems Now we will formulate and solve a GP problem that involves multiple goals within the same priority level. No tradeoffs occur in the achievement of goals with different priorities. However, tradeoffs can occur in the achievement of goals with the same priority and different weights. A GP problem with p preemptive goal priorities can be solved as a series of p linear programming problems. Solving a GP problem as a series of LPs, the objective functions (always to be minimized) will include only relevant deviation variables and not decision variables. Slide 17

GP Example 2: Conceptual Products is a computer company that produces the CP 400, CP 500, and CP 600 computers. Many of the components used in the three computer models are produced in abundant supply by the company. However, the memory modules, external hard drives, and cases are bought from suppliers. The CP 400 model uses two memory modules and no external hard drive, the CP 500 uses one memory module and one external hard drive, and the CP 600 uses two memory modules and one external hard drive. All three models use the same case. Slide 18

GP Example 2: Conceptual Products Suppliers can provide Conceptual Products with 1000 memory modules, 500 external hard drives, and 600 cases on a weekly basis. It takes one hour to manufacture a CP 400 and its profit is $200; it takes one and one-half hours to manufacture a CP 500 and its profit is $500; and it takes two hours to manufacture a CP 600 and its profit is $900. Slide 19

GP Example 2: Conceptual Products The company has four goals: Priority 1: Meet a state contract of 200 CP 400 machines weekly. (Goal 1) Priority 2: Make at least 500 total computers weekly. (Goal 2) Priority 3: Make at least $250, 000 profit weekly. (Goal 3) Priority 3: Use no more than 400 man-hours per week. (Goal 4) Each $1000 underachieved from its profit goal is five times as important as an extra man-hour. Slide 20

GP Example 2: Formulation Variables x 1 = number of CP 400 computers produced weekly x 2 = number of CP 500 computers produced weekly x 3 = number of CP 600 computers produced weekly di- = amount the right hand side of goal i is deficient di+ = amount the right hand side of goal i is exceeded Functional Constraints Availability of memory modules: 2 x 1 + x 2 + x 3 < 1000 Availability of external hard drives: x 2 + x 3 < 500 Availability of cases: x 1 + x 2 + x 3 < 600 Slide 21

GP Example 2: Formulation Goals (1) 200 CP 400 computers weekly: x 1 + d 1 - - d 1+ = 200 (2) 500 total computers weekly: x 1 + x 2 + x 3 + d 2 - - d 2+ = 500 (3) $250(in thousands) profit: . 2 x 1 +. 5 x 2 +. 9 x 3 + d 3 - - d 3+ = 250 (4) 400 total man-hours weekly: x 1 + 1. 5 x 2 + 2 x 3 + d 4 - - d 4+ = 400 Non-negativity: x 1, x 2, x 3, di-, di+ > 0 for all i Slide 22

GP Example 2: Conceptual Products Objective Functions Priority 1: Minimize the number of CP 400 under 200 units: Min d 1 Priority 2: Minimize the number under 500 computers produced weekly: Min d 2 Priority 3: Minimize the amount under $250, 000 earned weekly: Min 5 d 3 Priority 3: Minimize the man-hours over 400 used weekly: Min d 4+ Slide 23

GP Example 2: Conceptual Products Priority 1 Formulation Min d 1 s. t. 2 x 1 +x 2 +x 3 < 1000 < 500 x 1 < 600 x 1 +d 1 - -d 1+ = 200 x 1 +x 2 +x 3 +d 2 - -d 2+ = 500. 2 x 1+. 5 x 2 +. 9 x 3 +d 3 - -d 3+ = 250 x 1+1. 5 x 2 +2 x 3 +d 4 - -d 4+ = 400 x 1, x 2, x 3, d 1 -, d 1+, d 2 -, d 2+, d 3 -, d 3+, d 4 -, d 4+ > 0 Slide 24

GP Example 2: Conceptual Products Computer Solution (First LP) Objective Function Value = 0. 000 Variable Value Reduced Cost x 1 200. 000 x 2 0. 000 x 3 233. 333 0. 000 d 10. 000 1. 000 d 1+ 0. 000 d 266. 667 0. 000 d 2+ 0. 000 d 30. 000 d 3+ 0. 000 d 40. 000 d 4+ 266. 667 0. 000 Slide 25

GP Example 2: Conceptual Products Priority 2 Formulation Min d 2( include the previous 7 constraints, i. e. 3 functional constraints and 4 goal constraints ) ( add the constraint: d 1 - = 0 ) Slide 26

GP Example 2: Conceptual Products Computer Solution (Second LP) Objective Function Value = 0. 000 Variable Value Reduced Cost x 1 285. 714 0. 000 x 2 0. 000 x 3 214. 286 0. 000 d 10. 000 d 1+ 85. 714 0. 000 d 20. 000 1. 000 d 2+ 0. 000 d 30. 000 d 3+ 0. 000 d 40. 000 d 4+ 314. 286 0. 000 Slide 27

GP Example 2: Conceptual Products Priority 3 Formulation Min 5 d 3 - + d 4+ ( include the previous 8 constraints i. e. 3 functional constraints and 4 goal constraints and d 1 - = 0 ) ( add the constraint: d 2 - = 0 ) Slide 28

GP Example 2: Conceptual Products Computer Solution (Third LP) Objective Function Value = 314. 286 Variable Value Reduced Cost x 1 285. 714 0. 000 x 2 0. 000 0. 071 x 3 214. 286 0. 000 d 10. 000 d 1+ 85. 714 0. 000 d 20. 000 d 2+ 0. 000 0. 714 d 30. 000 3. 571 d 3+ 0. 000 1. 429 d 40. 000 1. 000 d 4+ 314. 286 0. 000 Slide 29

GP Example 2: Conceptual Products Final Solution Thus the optimal recommendation is to produce 285. 714 CP 400 computers, 0 CP 500 computers, and weekly and 214. 286 CP 600 computers weekly. All goals will be met except goal 4. 314. 286 extra man -hours or a total of 714. 286 man-hours will be used. Slide 30