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Chapter 14: Goal Programming

Chapter 14: Goal Programming. Goal programming is used to solve linear programs with multiple objectives, with each objective viewed as a "goal".

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Chapter 14: Goal Programming

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  1. 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.

  2. 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.

  3. 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 w1d1+ + w2d2- 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. Otherwise, a final solution has been reached.

  4. 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 w3d3+ + w4d4- s.t. Functional Constraints, Goal Constraints, and w1d1+ + w2d2- = k Go to step 4. (Repeat steps 4 and 5 until all priority levels have been examined.)

  5. GP Example: Conceptual Products Conceptual Products produces CP400 and CP500 computers that use memory modules, external hard drives, and cases. The CP400 model uses two memory modules and no external hard drive, whereas the CP500 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 CP400 and its profit is $200 and it takes one and one-half hours to manufacture a CP500 and its profit is $500.

  6. GP Example: Conceptual Products The company has four goals: Priority 1: Meet a state contract of 200 CP400 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)

  7. GP Example: Formulation Variables x1 = number of CP400 computers produced weekly x2 = number of CP500 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: 2x1 + x2< 1000 Availability of external hard drives: x2< 500 Availability of cases: x1 + x2< 600

  8. GP Example: Formulation Goals (1) 200 CP400 computers weekly: x1 + d1- - d1+ = 200 (2) 500 total computers weekly: x1 + x2 + d2- - d2+ = 500 (3) $250(in thousands) profit: .2x1 + .5x2 + d3- - d3+ = 250 (4) 400 total man-hours weekly: x1 + 1.5x2 + d4- - d4+ = 400 Non-negativity: x1, x2, di-, di+> 0 for all i

  9. GP Example: Formulation Objective Functions Priority 1: Minimize the amount the state contract is not met: Min d1- Priority 2: Minimize the number under 500 computers produced weekly: Min d2- Priority 3: Minimize the amount under $250,000 earned weekly: Min d3- Priority 4: Minimize the man-hours over 400 used weekly: Min d4+

  10. GP Example: Formulation Formulation Summary Min P1(d1-) + P2(d2-) + P3(d3-) + P4(d4+) s.t. 2x1 +x2< 1000 +x2< 500 x1 +x2< 600 x1 +d1- -d1+ = 200 x1 +x2 +d2- -d2+ = 500 .2x1+ .5x2 +d3- -d3+ = 250 x1+1.5x2 +d4- -d4+ = 400 x1, x2, d1-, d1+, d2-, d2+, d3-, d3+, d4-, d4+> 0

  11. GP Example: Graphical Solution Iteration 1 To solve graphically, first graph the functional constraints. Then graph the first goal: x1 = 200. Note on the next slide that there is a set of points that exceed x1 = 200 (where d1- = 0).

  12. GP Example: Graphical Solution Functional Constraints and Goal 1 Graphed x2 1000 800 600 400 200 2x1 + x2< 1000 Goal 1: x1> 200 x2 < 500 x1 + x2< 600 Points Satisfying Goal 1 x1 200 400 600 800 1000 1200

  13. GP Example: Graphical Solution Iteration 2 Now add Goal 1 as x1> 200 and graph Goal 2: x1 + x2 = 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 d2- = 0).

  14. GP Example: Graphical Solution Goal 1 (Constraint) and Goal 2 Graphed x2 1000 800 600 400 200 2x1 + x2< 1000 Goal 1: x1> 200 x2 < 500 x1 + x2< 600 Points Satisfying Both Goals 1 and 2 Goal 2: x1 + x2> 500 x1 200 400 600 800 1000 1200

  15. GP Example: Graphical Solution Iteration 3 Now add Goal 2 as x1 + x2> 500 and Goal 3: .2x1 + .5x2 = 250. Note on the next slide that no points satisfy the previous functional constraints and goals and satisfy this constraint. Thus, to Min d3-, this minimum value is achieved when we Max .2x1 + .5x2. Note that this occurs at x1 = 200 and x2 = 400, so that .2x1 + .5x2 = 240 or d3- = 10.

  16. GP Example: Graphical Solution Goal 2 (Constraint) and Goal 3 Graphed x2 2x1 + x2< 1000 1000 800 600 400 200 Goal 1: x1> 200 x1 + x2< 600 x2 < 500 (200,400) Points Satisfying Both Goals 1 and 2 Goal 2: x1 + x2> 500 Goal 3: .2x1 + .5x2 = 250 x1 200 400 600 800 1000 1200

  17. 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.

  18. GP Example 2: Conceptual Products Conceptual Products is a computer company that produces the CP400, CP500, and CP600 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 CP400 model uses two memory modules and no external hard drive, the CP500 uses one memory module and one external hard drive, and the CP600 uses two memory modules and one external hard drive . All three models use the same case.

  19. 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 CP400 and its profit is $200; it takes one and one-half hours to manufacture a CP500 and its profit is $500; and it takes two hours to manufacture a CP600 and its profit is $900.

  20. GP Example 2: Conceptual Products The company has four goals: Priority 1: Meet a state contract of 200 CP400 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.

  21. GP Example 2: Formulation Variables x1 = number of CP400 computers produced weekly x2 = number of CP500 computers produced weekly x3 = number of CP600 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: 2x1 + x2 + x3< 1000 Availability of external hard drives: x2 + x3< 500 Availability of cases: x1 + x2 + x3< 600

  22. GP Example 2: Formulation Goals (1) 200 CP400 computers weekly: x1 + d1- - d1+ = 200 (2) 500 total computers weekly: x1 + x2 + x3 + d2- - d2+ = 500 (3) $250(in thousands) profit: .2x1 + .5x2 + .9x3 + d3- - d3+ = 250 (4) 400 total man-hours weekly: x1 + 1.5x2 + 2x3 + d4- - d4+ = 400 Non-negativity: x1, x2, x3, di-, di+> 0 for all i

  23. GP Example 2: Conceptual Products Objective Functions Priority 1: Minimize the number of CP400 under 200 units: Min d1- Priority 2: Minimize the number under 500 computers produced weekly: Min d2- Priority 3: Minimize the amount under $250,000 earned weekly: Min 5d3- Priority 3: Minimize the man-hours over 400 used weekly: Min d4+

  24. GP Example 2: Conceptual Products Priority 1 Formulation Min d1- s.t. 2x1 +x2 +x3< 1000 +x2 +x3< 500 x1 +x2 +x3< 600 x1 +d1- -d1+ = 200 x1 +x2 +x3 +d2- -d2+ = 500 .2x1+ .5x2 +.9x3 +d3- -d3+ = 250 x1+1.5x2 +2x3 +d4- -d4+ = 400 x1, x2, x3, d1-, d1+, d2-, d2+, d3-, d3+, d4-, d4+> 0

  25. GP Example 2: Conceptual Products Computer Solution (First LP) Objective Function Value = 0.000 VariableValueReduced Cost x1 200.000 0.000 x2 0.000 0.000 x3 233.333 0.000 d1- 0.000 1.000 d1+ 0.000 0.000 d2- 66.667 0.000 d2+ 0.000 0.000 d3- 0.000 0.000 d3+ 0.000 0.000 d4- 0.000 0.000 d4+ 266.667 0.000

  26. GP Example 2: Conceptual Products Priority 2 Formulation Min d2- ( include the previous 7 constraints, i.e. 3 functional constraints and 4 goal constraints ) ( add the constraint: d1-= 0 )

  27. GP Example 2: Conceptual Products Computer Solution (Second LP) Objective Function Value = 0.000 VariableValueReduced Cost x1 285.714 0.000 x2 0.000 0.000 x3 214.286 0.000 d1- 0.000 0.000 d1+ 85.714 0.000 d2- 0.000 1.000 d2+ 0.000 0.000 d3- 0.000 0.000 d3+ 0.000 0.000 d4- 0.000 0.000 d4+ 314.286 0.000

  28. GP Example 2: Conceptual Products Priority 3 Formulation Min 5d3- + d4+ ( include the previous 8 constraints i.e. 3 functional constraints and 4 goal constraints and d1- = 0 ) ( add the constraint: d2-= 0 )

  29. GP Example 2: Conceptual Products Computer Solution (Third LP) Objective Function Value = 314.286 VariableValueReduced Cost x1 285.714 0.000 x2 0.000 0.071 x3 214.286 0.000 d1- 0.000 0.000 d1+ 85.714 0.000 d2- 0.000 0.000 d2+ 0.000 0.714 d3- 0.000 3.571 d3+ 0.000 1.429 d4- 0.000 1.000 d4+ 314.286 0.000

  30. GP Example 2: Conceptual Products Final Solution Thus the optimal recommendation is to produce 285.714 CP400 computers, 0 CP500 computers, and weekly and 214.286 CP600 computers weekly. All goals will be met except goal 4. 314.286 extra man-hours or a total of 714.286man-hours will be used.

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