CASO CONCEPTUAL PRODUCTSConceptual Products is a computer company that produces the CP286 and the CP386 computers. The computers use different mother boards produced in abundant supply by the company, but use the same cases and disk drives. The CP286 models use two floppy disk drives and no hard disks whereas the CP386 models use one floppy disk drive and one hard disk drive. The disk drives and cases are bought from vendors. There are 1000 floppy disk drives, 500 hard disk drives, and 600 cases available to Conceptual Products on a weekly basis. It takes one hour to manufacture a CP286 and its profit is $200 and it takes one and one-half hours to manufacture a CP386 and its profit is $500. The company has four goals which are given below: Priority 1: Meet a state contract of 200 CP286 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) Variables x 1 = number of CP286 computers produced weekly x 2 = number of CP386 computers produced weekly du i = amount the right hand side of goal i is deficient do i = amount the right hand side of goal i is exceeded Functional Constraints Availability of floppy disk drives: 2x 1 + x 2 < 1000 Availability of hard disk drives: x 2 < 500 Availability of cases: x 1 + x 2 < 600 Goals (1) 200 CP286 computers weekly: x 1 + du 1 - do 1 = 200 (2) 500 total computers weekly: x 1 + x 2 + du 2 - do 2 = 500 (3) $250(in thousands) profit: .2x 1 + .5x 2 + du 3 - do 3 = 250 (4) 400 total man-hours weekly: x 1 + 1.5x 2 + du 4 - do 4 = 400 Non-negativity: x 1 , x 2 , du i , do i > 0 for all i Objective Functions Priority 1: Minimize the amount the state contract is not met: Min du 1 Priority 2: Minimize the number under 500 computers produced weekly: Min du 2 Priority 3: Minimize the amount under $250,000 earned weekly: Min du 3 Priority 4: Minimize the man-hours over 400 used weekly: Min do 4 Formulation Summary Min P 1 (du 1 ) + P 2 (du 2 ) + P 3 (du 3 ) + P 4 (do 4 ) s.t. 2x 1 +x 2 < 1000 +x 2 < 500 x 1 +x 2 < 600 x 1 +du 1 -do 1 = 200 x 1 +x 2 +du 2 -do 2 = 500 .2x 1 + .5x 2 +du 3 -do 3 = 250 x 1 +1.5x 2 +du 4 -do 4 = 400 x 1 , x 2 , du 1 , do 1 , du 2 , do 2 , du 3 , do 3 , du 4 , do 4 > 0 The P i in the objective function are not coefficients, they simply indicate us the priority, that is, the order in which the goals should be achieved. The LINDO solution. FIRST ITERATION. In this step the objective function in the formulation contains the variable D1U alone, this is because the only priority one objective refers to D1U. The formulation includes the functional (also called operating or technical) constraints as well as the constraints derived from the goals. ! EXAMPLE: CONCEPTUAL PRODUCTS MIN D1U SUBJECT TO +2 X1 + X2 <= 1000 + X2 <= 500 + X1 + X2 <= 600 + X1 + D1U -D1O = 200 + X1 + X2 + D2U - D2O = 500 +.2 X1 + .5 X2 + D3U - D3O = 250 + X1 + 1.5 X2 + D4U - D4O = 400 END ! LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 0.0000000E+00 Because the objective function is D1U, its optimal value is 0, this means that, given the set of constraints, the priority one goal is met. The contract for 200 CP286 computers is met. VARIABLE VALUE REDUCED COST D1U 0.000000 1.000000 X1 200.000000 0.000000 X2 0.000000 0.000000 D1O 0.000000 0.000000 D2U 300.000000 0.000000 D2O 0.000000 0.000000 D3U 210.000000 0.000000 D3O 0.000000 0.000000 D4U 200.000000 0.000000 D4O 0.000000 0.000000 The optimal solution leads us to produce 200 units of the CP286 computer and no CP386 computers. The second goal is not met, production is 300 units below the goal. Goal 3 is not met, profit is $210000 below the goal. Finally, goal 4 is met, the intended level of manpower used is not exceeded. ROW SLACK OR SURPLUS DUAL PRICES 2) 600.000000 0.000000 3) 500.000000 0.000000 4) 400.000000 0.000000 5) 0.000000 0.000000 6) 0.000000 0.000000 7) 0.000000 0.000000 8) 0.000000 0.000000 NO. ITERATIONS= 1 SECOND ITERATION. Now, let’s solve for the second goal, the priority two goal. Since in the first iteration we already achieve the priority one goal, we do not want to degrade the current solution. We will try to achieve the next goals without degrading the level of the previous goals. The goal in the previous step was to minimize the D1U variable. For the next step we have to fix the optimal value of the D1U variable from the previous iteration. The optimal value is 0, so, we can fix the value in two ways. One way is substituting the zero value instead of the variable D1U and then re-write the constraints, with 0 instead of D1U. The other way is adding a new constraint in which the value of D1U is fixed, D1U=0. This second way is more practical because we do not need to rewrite the previous formulation. Let’s see how this works for the second goal. The second goal involves minimizing D2U, so we use the same formulation as in the previous step, only two things change: the objective function that now becomes MIN D2U and the new added constraint which fixes the value of the variable D1U = 0. In general, in each step we should add a constraint for each variable from the objective function of the previous iteration. ! EXAMPLE: CONCEPTUAL PRODUCTS MIN D2U SUBJECT TO +2 X1 + X2 <= 1000 + X2 <= 500 + X1 + X2 <= 600 + X1 + D1U -D1O = 200 + X1 + X2 + D2U - D2O = 500 +.2 X1 + .5 X2 + D3U - D3O = 250 + X1 + 1.5 X2 + D4U - D4O = 400 + D1U = 0 END ! LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 0.0000000E+00 The result shows us that the optimal value of D2U is zero, this means that the second goal is met as desired. 500 computers of any model are produced. VARIABLE VALUE REDUCED COST D2U 0.000000 1.000000 X1 500.000000 0.000000 X2 0.000000 0.000000 D1U 0.000000 0.000000 D1O 300.000000 0.000000 D2O 0.000000 0.000000 D3U 150.000000 0.000000 D3O 0.000000 0.000000 D4U 0.000000 0.000000 D4O 100.000000 0.000000 We can see that the solution has changed, now the solution suggests us to produce 500 CP286 computers, please notice and explain how other variables also changed. We have solved the problem for the first two goals, which have been met just as desired, since the two objective functions are at their absolute minimum values, both zero. ROW SLACK OR SURPLUS DUAL PRICES 2) 0.000000 0.000000 3) 500.000000 0.000000 4) 100.000000 0.000000 5) 0.000000 0.000000 6) 0.000000 0.000000 7) 0.000000 0.000000 8) 0.000000 0.000000 9) 0.000000 0.000000 NO. ITERATIONS= 1 THIRD ITERATION. Now we go for the third goal. We already have solved for two previous goals and, since these are goals with higher priority than the next ones, we do not want to degrade the current solution. So we need to keep the current values of the previous goals, in order to do so we need to add new constraints. We identify the variables of the previous objective function, in this case the objective function involves D2U only, then we look at the optimal values of these variables in the previous iteration (D2U = 0, in this case), and we add a constraint so the optimal value does not change in the next iteration. We use the same formulation and make two changes, the objective function changes to the priority 3 goal, and we add the new constraint D2U = 0. ! EXAMPLE: CONCEPTUAL PRODUCTS MIN D3U SUBJECT TO +2 X1 + X2 <= 1000 + X2 <= 500 + X1 + X2 <= 600 + X1 + D1U -D1O = 200 + X1 + X2 + D2U - D2O = 500 +.2 X1 + .5 X2 + D3U - D3O = 250 + X1 + 1.5 X2 + D4U - D4O = 400 + D1U = 0 + D2U = 0 END ! LP OPTIMUM FOUND AT STEP 8 OBJECTIVE FUNCTION VALUE 1) 10.00000 The solution shows that the D3U variable does not reach its absolute minimum value, this means that the priority 3 goal is not met as desired. The expected level of profits, $250000, is not achieved, given the demanding higher priority goals, the best thing we can do is to get profits for $240000, $10000 below the desired goal. VARIABLE VALUE REDUCED COST D3U 10.000000 0.000000 X1 200.000000 0.000000 X2 400.000000 0.000000 D1U 0.000000 0.000000 D1O 0.000000 0.300000 D2U 0.000000 0.000000 D2O 100.000000 0.000000 D3O 0.000000 1.000000 D4U 0.000000 0.000000 D4O 400.000000 0.000000 Once we take into account the profits goal, the solution changes again. Now the solution suggests us to produce 200 CP286 computers and 400 CP386 computers. ROW SLACK OR SURPLUS DUAL PRICES 2) 200.000000 0.000000 3) 100.000000 0.000000 4) 0.000000 0.500000 5) 0.000000 -0.300000 6) 0.000000 0.000000 7) 0.000000 -1.000000 8) 0.000000 0.000000 9) 0.000000 0.300000 10) 0.000000 0.000000 NO. ITERATIONS= 8 FOURTH ITERATION. Now we have solved for 3 higher priority goals, we have one last goal. However we do not want to sacrifice any of the goals already achieved, so we have to add new constraints, which fix the values of the variables involved in the objective function of the previous iteration. Remember to add one constraint for each variable of the objective function. In this case we have only one variable in the objective function, D3U, and its optimal value was 10, so we add the constraint D3U = 10, and change the objective function to that of the fourth goal, MIN D4O. ! EXAMPLE: CONCEPTUAL PRODUCTS MIN D4O SUBJECT TO +2 X1 + X2 <= 1000 + X2 <= 500 + X1 + X2 <= 600 + X1 + D1U -D1O = 200 + X1 + X2 + D2U - D2O = 500 +.2 X1 + .5 X2 + D3U - D3O = 250 + X1 + 1.5 X2 + D4U - D4O = 400 + D1U = 0 + D2U = 0 + D3U = 10 END ! LP OPTIMUM FOUND AT STEP 1 OBJECTIVE FUNCTION VALUE 1) 400.0000 The objective was to minimize D4O, we can see that we are not able to achieve this goal, D4O does not reach its absolute minimum value (0), the best thing we can do is to use 400 man-hours beyond the goal. Previous goals force us to produce a product mix that requires 800 man-hours. VARIABLE VALUE REDUCED COST D4O 400.000000 0.000000 X1 200.000000 0.000000 X2 400.000000 0.000000 D1U 0.000000 0.000000 D1O 0.000000 0.400000 D2U 0.000000 0.000000 D2O 100.000000 0.000000 D3U 10.000000 0.000000 D3O 0.000000 3.000000 D4U 0.000000 1.000000 The final solution indicates that: D4O = 400, 400 man-hours are used in excess of the desired level in the priority 4 goal. 800 man-hours are used. X1 = 200, 200 CP286 computers are produced. X2 = 400, 400 CP386 computers are produced. D1U = D1O = 0, the priority 1 goal is achieved exactly as desired. D2O = 100, priority 2 goal has been exceeded, it requires 500 computers being produced, and the solution gives 100 in excess, 600 computers in total. D3U = 10, priority 3 goal is not achieved, profits are $10000 below the desired level. The solution gives profits for $240000 instead of $250000. ROW SLACK OR SURPLUS DUAL PRICES 2) 200.000000 0.000000 3) 100.000000 0.000000 4) 0.000000 0.000000 5) 0.000000 -0.400000 6) 0.000000 0.000000 7) 0.000000 -3.000000 8) 0.000000 1.000000 9) 0.000000 0.400000 10) 0.000000 0.000000 11) 0.000000 3.000000 NO. ITERATIONS= 1