Chapter Two: Linear Programming: Model Formulation and Graphical Solution

Transcription

1 TYLM0_0393.QX //09 :3 M Page hapter Two: Linear Programming: Model Formulation and Graphical Solution PROLEM SUMMRY. Maimization ( continuation), graphical solution. Maimization, graphical solution 3. Minimization, graphical solution. Sensitivity analysis ( 3) 5. Minimization, graphical solution. Maimization, graphical solution 7. Slack analysis ( ). Sensitivity analysis ( ) 9. Maimization, graphical solution. Slack analysis ( 9). Maimization, graphical solution. Minimization, graphical solution 3. Maimization, graphical solution. Sensitivity analysis ( 3) 5. Sensitivity analysis ( 3). Maimization, graphical solution 7. Sensitivity analysis ( ). Maimization, graphical solution 9. Standard form 0. Maimization, graphical solution. Standard form. Maimization, graphical solution 3. onstraint analysis ( ). Minimization, graphical solution 5. Sensitivity analysis ( ). Sensitivity analysis ( ) 7. Sensitivity analysis ( ). Minimization, graphical solution 9. Minimization, graphical solution. Sensitivity analysis ( 9) 3. Minimization, graphical solution 3. Maimization, graphical solution 33. Minimization, graphical solution 3. Maimization, graphical solution 35. Sensitivity analysis ( 3) 3. Minimization, graphical solution 37. Maimization, graphical solution 3. Maimization, graphical solution 39. Sensitivity analysis ( 3) 0. Maimization, graphical solution. Sensitivity analysis ( 0). Maimization, graphical solution 3. Sensitivity analysis ( ). Minimization, graphical solution 5. Sensitivity analysis ( ). Maimization, graphical solution 7. Sensitivity analysis ( ). Maimization, graphical solution 9. Sensitivity analysis ( ). Maimization, graphical solution 5. Maimization, graphical solution 5. Minimization, graphical solution 53. Sensitivity analysis ( 5) 5. Maimization, graphical solution 55. Sensitivity analysis ( 5) 5. Multiple optimal solutions 57. Infeasible problem 5. Unbounded problem PROLEM SOLUTIONS. a) = # cakes = # loaves of bread maimize = $ cups of flour minutes, 0

2 TYLM0_0393.QX //09 :3 M Page 7. a) maimize = + (profit, $) + 0 (line, hr) (line, hr), a) minimize = (cost, $) + (vitamin, mg) + (vitamin, mg), 0 * : = 0 = =. : = /5 = /5 =. : = = 0 =.0 : = 0 =.5 = 5 : = 3. =.33 = 3.9 *: = = 0 = 0 : = 0 = = 0 * : = 3 = 7 = : = = 0 = 3 Point is optimal 0 Point is optimal 0 optimal. The optimal solution point would change from point to point, thus resulting in the optimal solution = /5 = /5 =.0 5. a) minimize = (cost, $) + 0 (nitrogen, oz) + 3 (phosphate, oz) (potassium, oz), 0. a) maimize = (profit, $) + 0 (labor, hr) + 3 (wood, l (demand, chairs), 0 : = 0 = = : = = 5 = * : = = = Point is optimal 0 : = 0 = = 00 : = /7 = 3/7 =,7 * : = = 3. =,70 : = = 0 =,00 Point is optimal 0 7. In order to solve this problem, you must substitute the optimal solution into the resource constraint for wood and the resource constraint for labor and determine how much of each resource is left over. 7

3 TYLM0_0393.QX //09 :3 M Page Labor + 0 hr () + (3.) There is no labor left unused. Wood + 3 lb () + (3.) =. There is. lb of wood left unused.. The new objective function, = , is parallel to the constraint for labor, which results in multiple optimal solutions. Points ( = /7, = 3/7) and ( =, = 3.) are the alternate optimal solutions, each with a profit of $,000.. Sugar + (0) + () There is no sugar left unused. * : = 0 = 9 = 5 : = = 3 = : = = = Point is optimal 0 9. a) maimize = + 5 (profit, $) (flour, l + (sugar, l 5 (demand for cakes), 0 * : = 0 = = 0 : = = 3 = 7 : = 5 = 0 = 5 Point is optimal 0. In order to solve this problem, you must substitute the optimal solution into the resource constraints for flour and sugar and determine how much of each resource is left over. Flour lb 5(0) + 5() = 5 There are 5 lb of flour left unused.. a) minimize = 0 + (cost, $) 3 + (antibiotic, units) + (antibiotic, units) + (antibiotic 3, units), 0 : = 0 = = 0 * : = = = 3 : = 3 = = 90 : = = = 0 0 Point is optimal 0 3. a) maimize = (profit, $) 3 + (gold, oz) + 0 (platinum, oz) (demand, bracelets), 0

4 TYLM0_0393.QX //09 :3 M Page 9 : = 0 = =,00 : = = =,00 *: = = 3 =,00 : = = 0 =,00 Point is optimal 0 However, this creates a situation where both points and are optimal, ie., multiple optimal solutions, as are all points on the line segment between and.. a) maimize = + 0 (profit, $) (wool, yd ) + 00 (labor, hr), 0. The new objective function, = , is parallel to the constraint line for platinum, which results in multiple optimal solutions. Points ( =, = ) and ( =, = 3) are the alternate optimal solutions, each with a profit of $3,000. The feasible solution space will change. The new constraint line, 3 + = 0, is parallel to the eisting objective function. Thus, multiple optimal solutions will also be present in this scenario. The alternate optimal solutions are at =.33, = and =., = 3., each with a profit of $, a) Optimal solution: = necklaces, = 3 bracelets. The maimum demand is not achieved by the amount of one bracelet : = 0 = =,00 * : =.5 = 3.7 =,73 : = 0 = 0 =,000 Point is optimal The feasible solution space changes from the area 0 to 0'', as shown on the following graph. The solution point on the graph which corresponds to no bracelets being produced must be on the ais where = 0. This is point on the graph. In order for point to be optimal, the objective function slope must change such that it is equal to or greater than the slope of the constraint line, 3 + =. Transforming this constraint into the form y = a + b enables us to compute the slope: = 3 = 9 3/ From this equation the slope is 3/. Thus, the slope of the objective function must be at least 3/. Presently, the slope of the objective function is 3/: 00 = 0 = /00 3/ The profit for a necklace would have to increase to $00 to result in a slope of 3/: 00 = 00 = /00 3/ The etreme points to evaluate are now, ', and '. : = 0 = =,00 *': = 5. = 0.5 =, 9

5 TYLM0_0393.QX //09 :3 M Page. ': = = 0 =,00 Point ' is optimal 9. maimize = s + 0s + 0s 3 : 0. : = 0 = 5 = 5 * : = = = 7 : = = 0 = Point is optimal 0 + s = + s = + + s 3 = 5, 0 : s =, s =, s 3 = 0 : s = 0, s = 5, s 3 = 0 : s = 0, s =, s 3 =. maimize = s + 0s 3 + 0s : s = + + s = 0 + s 3 = + s =, 0 Point is optimal 0 0 : = 0 = = 0 * : = = 5. =. : = = 0 = : s = 0, s = 0, s 3 =, s = 0 : s = 0, s = 3., s 3 = 0, s =. : s =, s =, s 3 = 0, s = : = = = * : = = 5 = 5 : = 5 = 0 = 97.5 Point is optimal It changes the optimal solution to point ( =, =, = ), and the constraint, + 5, is no longer part of the solution space boundary.. a) Minimize = + (labor cost, $) + (claims) + 0 (workstations) (defective claims), 0 5 : =.5 = 0 =,00 * : = 0. : = 0 =.70 = 0 =, =, Point is optimal : = 5.55 = 3.5 =,37.9

6 TYLM0_0393.QX //09 :3 M Page 5. hanging the pay for a full-time claims processor from $ to $5 will change the solution to point in the graphical solution where =.5 and = 0, i.e., there will be no part-time operators. hanging the pay for a part-time operator from $ to $3 has no effect on the number of full-time and part-time operators hired, although the total cost will be reduced to $,7.95. The problem becomes infeasible. 3. * : =. =. =. : = =.5 = 3.5. Eliminating the constraint for defective claims would result in a new solution, = 0 and = 37.5, where only part-time operators would be hired. Feasible space Point is optimal 0 7. The solution becomes infeasible; there are not enough workstations to handle the increase in the volume of claims.. Point is optimal 0 : = = = 5 * : = = = : = = 0 = 3. 0 Point is optimal : = = 3.5 = 9 * : = 5 = 3 = : = = = : =.7 =.33 Point is optimal = : = () = 3 () = (3) * : = = () = : = 3.9 (5) = * : = 3. = = 37. : = 5.33 = 3.33 = 9.3 : = 9. =. = 79. Point is optimal

7 TYLM0_0393.QX //09 :3 M Page 3. a) Maimize = $ (profit, $) + 5 ( freezer space, gals.) ( budget,$) or 0 ( demand), No additional profit, freezer space is not a binding constraint. 3. a) minimize = (cost, $) + (high-grade ore, tons) + (medium-grade ore, tons) + (low-grade ore, tons), 0 * : =.9 : = 0 = = 90 * : = = 3. =.35 = 3 = 0 : = 3 = = 70 : = : = = 0 =,00 Point is optimal 0 = 0 Point is optimal = a) maimize = (profit, $) + (stamping, days) + (coating, days) + 9 (lots), 0 3. a) maimize = + 70 (profit, $) + 0 (assembly, hr) + (finishing, hr) + (inventory, units), 0 : = 3 = = 7,00 * : = 5 = 5 =,0 : = = 3 = 7,0 Point is optimal 0 : = 0 = = * : = 3.3 =.7 = 5 : = 5.3 =.7 = : = = 0 = 0 Point is optimal The slope of the original objective function is computed as follows: = = = /70 3/7 slope = 3/7

8 TYLM0_0393.QX //09 :3 M Page 3 The slope of the new objective function is computed as follows: = = 90 = /70 9/7 slope = 9/7 The change in the objective function not only changes the values but also results in a new solution point,. The slope of the new objective function is steeper and thus changes the solution point. : = 0 : = 5.3 = =.7 = = 0 : = 3.3 : = =.7 = 0 = 7 = a) Maimize = 9 + (profit, $,000s) + (grapes, tons) (storage space, yd 3 ) (processing time, hr) 7 (demand, Nectar) 7 (demand, Red), 0 Points and would be eliminated and a new optimal solution point at = 5.09, = 5.5, and =.7 would result.. a) maimize =. +.9 X cans, 0 : X=0 : X=9 : =$. *: X=3 *: X= *: =$. Point is optimal X : = 0 = 7 = : = = 7 = * : = = = : = 5.7 =. =.79 E : = 7 =.75 = 5.5 F : = 7 = 0 = 3 3. The model formulation would become, maimize = $ , 0 Optimal point The solution is = 3., = 57., and = $9.7 The discount would reduce profit. E F 0. a) 5() + () 0 hr = hr left unused. a) minimize = $ = 3,0,000, , 0 3

9 TYLM0_0393.QX //09 :3 M Page 00 X : X =,5.5 : X =,35. : =,3.7 Point is optimal : X =,95.05 *X =,000 * = $,70.7 +,90 3 +, , X 5. a) minimize = ,000,000, = 3,0, 0 X : X=0 : X=0 : = *: X=9 *: X=9 *: =590. Point is optional : X=0 *: X=0 *: = X X 0 7. new constraint is added to the model in 5. The solution is = 0, =.7, = $ : X=,000 : X=3,5.57 : =5.5 Point is optimal *: X=,95.05 *: X=,000 *: =5.05 X *: X=0.07 : X=.7 : =5 : X=0 *: X=0 *: = Point is optimal = 3 fewer defective items. a) maimize = $ X X

10 TYLM0_0393.QX //09 :3 M Page 5. a) maimize = (profit, $) + (available land, acres) (labor, hr) (fertilizer, tons) (shipping space, acres) 37 (shipping space, acres), ( , 0 : = 0 * : =. 0 = 37 =,0 =. = 7,3 0 : = 7.5 E : = 0 = 37 = 3.3 =,0 =,390 0 : =.7 F : = 0 = 33.3 = 0 =,0 =,00 0 E Point is optimal F The feasible solution space changes if the fertilizer constraint changes to tons. The new solution space is ''''. Two of the constraints now have no effect Optimal solution - = 0 = z = 7,0, a) = $ invested in stocks = $ invested in bonds maimize = $ (average annual return) + $70,000 (available funds) /( + ).5 (% of stocks) ,000 (total possible loss), The new optimal solution is point ': ': = 0 *': = 5.7 = 37 =.9 =,0 =,57 ': = 3 ': = = 37 = 0 =,0 =,00. a) Maimize = $7,00 +,0 + 3,0 /( + ).0 (00S) , optimal: = 37,70.59 = 33,5. z =, (00S) 5

11 TYLM0_0393.QX //09 :3 M Page X 5. = eams assigned to rad = eams assigned to Sarah minimize = = 0 (70/7.) or 0 (00/), 0 + 3,0 oz or ( gal. oz) (.0)(.05) + (.0)(.05) lbs. olombian (.35)(.05) + (.)(.05) lbs. Kenyan (.5)(.05) + (.)(.05) lbs. Indonesian / = 3/, 0 Solution: = 7.3 cups =.9 cups = $ X * : = 70 = = optimal : = 0 = 0 = : = 7.3 =.9 = X If the constraint for Sarah s time became 55 with an additional hour then the solution point at would move to = 5, = 55 and = 9.. If the constraint for rad s time became.33 with an additional hour then the solution point () would not change. ll of rad s time is not being used anyway so assigning him more time would not have an effect. One more hour of Sarah s time would reduce the number of regraded eams from to 9., whereas increasing rad by one hour would have no effect on the solution. This is actually the marginal (or dual) value of one additional hour of labor, for Sarah, which is 0.0 fewer regraded eams, whereas the marginal value of rad s is zero. 5. a) = # cups of Pomona = # cups of oastal maimize = $ X 55. a) The only binding constraint is for olombian; the constraints for Kenyan and Indonesian are nonbinding and there is already etra, or slack, pounds of these coffees available. Thus, only getting more olombian would affect the solution. One more pound of olombian would increase sales from $.09 to $3.0. Increasing the brewing capacity to 0 gallons would have no effect since there is already unused brewing capacity with the optimal solution. If the shop increased the demand ratio of Pomona to oastal from.5 to to to it would increase daily sales to $0.00, so the shop should spend etra on advertising to achieve this result.

12 TYLM0_0393.QX //09 :3 M Page Multiple optimal solutions; and alternate optimal * : = 0 = 0 = 0,000 * : = = = 0,000 : = =.7 =,9 : = 0 = 0 = 0,000 0 Multiple optimal solutions; and alternate optimal SE SOLUTION: METROPOLITN POLIE PTROL The linear programming model for this case problem is minimize = /0 + y/5 + y 5 + y y.5,y 0 The objective function coefficients are determined by dividing the distance traveled, i.e., /3, by the travel speed, i.e., 0 mph. Thus, the coefficient is /3 0, or /0. In the first two constraints, + y represents the formula for the perimeter of a rectangle. The graphical solution is displayed as follows. 5 y 70 Infeasible Problem Optimal point The optimal solution is =, y =.5, and = This means that a patrol sector is.5 miles by mile and the response time is 0.05 hr, or 3 min Unbounded Problem SE SOLUTION: THE POSSIILITY RESTURNT The linear programming model formulation is Maimize = = $ / 3/ or 3 0 /( + ). or The graphical solution is shown as follows. 7

13 TYLM0_0393.QX //09 :3 M Page Optimal point : = 3.3 =. = $77.3 * : = 0 = 0 = $00 optimal : = = 5 = $ hanging the objective function to = $ + would result in multiple optimal solutions, the end points being and. The profit in each case would be $90. hanging the constraint from to has no effect on the solution. SE SOLUTION: NNELLE INVESTS IN THE MRKET = no. of shares of inde fund = no. of shares of internet stock fund Maimize = (.7)(75) + (.)(0) = = $ 0, , > 0 = 03 = 0 = $9,9.37 Eliminating the constraint.33 will have no effect on the solution. Eliminating the constraint will change the solution to = 9, = 5.55, = $,73.5. Increasing the amount available to invest (i.e., $0,000 to $0,00) will increase profit from = $9,9.37 to = $9,9. or approimately $0.5. Increasing by another dollar will increase profit by another $0.5, and increasing the amount available by one more dollar will again increase profit by $0.5. This indicates that for each etra dollar invested a return of $0.5 might be epected with this investment strategy. Thus, the marginal value of an etra dollar to invest is $0.5, which is also referred to as the shadow or dual price as described in hapter 3.