1. Lesson 11 Multicriteria Decisions within LP Framework

2. Learning Objectives: By the end of this unit, you should be able to: Understand Goal Programming Formulate model for Goal Programming Find Graphical Solution for Goal Programming

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

4. 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.

5. 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, w i, to be placed on the deviation(s), d i + and/or d i -, from the goal.

6. Goal Programming Formulation 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. Otherwise, a final solution has been reached.

7. 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.)

8. Example: Conceptual Products Conceptual Products is a computer company that produces the CP400 and CP500 computers. The computers use different mother boards produced in abundant supply by the company, but use the same cases and disk drives. The CP400 models use two CD drives and no DVD drives whereas the CP500 models use one CD drive and one DVD disk drive.

9. Example: Conceptual Products The disk drives and cases are bought from vendors. There are 1000 CD drives, 500 DVD disk drives, and 600 cases available to Conceptual Products on a weekly basis. It takes one hour to manufacture a CP400 and its profit is $200 and it takes one and half hours to manufacture a CP500 and its profit is $500.

10. Example: Conceptual Products Formulate the equations and solve the problem assuming that 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)

11. Constrains: Conceptual Products Let’s set : CP400 = X1 and CP500 = X2 We can identify following constrains table: So our constrains are: 1. 2 X1 + X2 ≤ 1000 2. X2 ≤ 500 3. X1 + X2 ≤ 600 Constrains or Goals X1X2Total CD211000 DVD01500 Case11600 Labour11.5 Profit$200$500

12. Goal Equations: Conceptual Products Equation for Goal 1: X1 = 200 + d 1 + - d 1 - or X1 - d 1 + + d 1 - = 200Goal 1: Min d 1 - Equation for Goal 2: X1 + X2 = 500 + d 2 + - d 2 - or X1 + X2 - d 2 + + d 2 - = 500 Goal 2: Min d 2 - Equation for Goal 3: 200X1 + 500X2 = 250000 + d 3 + - d 3 - or 200X1 + 500X2 - d 3 + + d 3 - = 250000 Goal 3: Min d 3 - Equation for Goal 4: X1 + 1.5X2 = 400 + d 4 + - d 4 - or X1 + 1.5X2 – d 4 + + d 4 - = 400 Goal 4: Min d 4 +

13. Final Objective Function, and Goal and Constrain functions Min P 1 (d 1 - ) + P 2 (d 2 - ) + P 3 (d 3 - ) + P 4 (d 4 + ) S.T. 1. 2X1 + X2≤ 1000 2. X2 ≤ 500 3. X1 + X2 ≤ 600 4. X1 - d 1 + + d 1 - = 200 5. X1 + X2 - d 2 + + d 2 - = 500 6. 200X1 + 500X2 - d 3 + + d 3 - = 250000 7. X1 + 1.5X2 – d 4 + + d 4 - = 400 We can solve it using graphical method, pay attention!!!

14. QUESTIONS ?

15. Review Problem 1.

16. Review Problem 1: Answer

17. Review Problem: Answer

18. Review Problem 2.