Chapter Two: Linear Programming: Model Formulation and Graphical Solution
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1 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. Sensitivity analysis ( ) 0. Maimization, graphical solution. Standard form ( 0). Maimization, graphical solution 3. Standard form ( ). Maimization, graphical solution 5. onstraint analysis ( ). Minimization, graphical solution 7. Sensitivity analysis ( ). Sensitivity analysis ( ) 9. Sensitivity analysis ( ). Minimization, graphical solution 3. Minimization, graphical solution 3. Sensitivity analysis ( 3) 33. Minimization, graphical solution 3. Maimization, graphical solution 35. Minimization, graphical solution 3. Maimization, graphical solution 37. Sensitivity analysis ( 3) 3. Minimization, graphical solution 39. Maimization, graphical solution 0. Maimization, graphical solution. Sensitivity analysis ( 3). Maimization, graphical solution 3. Sensitivity analysis ( 0). Maimization, graphical solution 5. Sensitivity analysis ( ). Minimization, graphical solution 7. Sensitivity analysis ( ). Maimization, graphical solution 9. Sensitivity analysis ( ). Maimization, graphical solution 5. Sensitivity analysis ( ) 5. Maimization, graphical solution 53. Minimization, graphical solution 5. Sensitivity analysis ( 53) 55. Minimization, graphical solution 5. Sensitivity analysis ( 55) 57. Maimization, graphical solution 5. Minimization, graphical solution 59. Sensitivity analysis ( 5) 0. Maimization, graphical solution. Sensitivity analysis ( 5). Multiple optimal solutions 3. Infeasible problem. Unbounded problem PROLEM SOLUTIONS. a) # cakes # loaves of bread maimize $ cups of flour minutes, - opyright 03 Pearson Education, Inc. publishing as Prentice Hall
2 :.5 5 : *: optimal 0. a) Maimize (line, hr) (line, hr), : * : 3 7 : 3 Point is optimal 0 3. a) Minimize (cost, $) + (vitamin, mg) + (vitamin, mg), * :. : /5 /5. :.0 Point is optimal 0. The optimal solution point would change from point to point, thus resulting in the optimal solution /5 / a) Minimize (cost, $) + 0 (nitrogen, oz) + 3 (phosphate, oz) (potassium, oz), : : 5 * : Point is optimal 0. a) Maimize (labor, hr) + 3 (wood) (demand, chairs), - opyright 03 Pearson Education, Inc. publishing as Prentice Hall
3 : 0 : /7 3/7,7 * : 3.,70 :,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. Labor + 0 hr () + (3.) There is no labor left unused. Wood + 3 () + (3.) There is. lb of wood left unused.. The new objective function, 0 + 0, 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 $, a) Maimize (flour, l + (sugar, l 5 (demand for cakes), * : : 3 7 : 5 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() There are 5 lb of flour left unused. Sugar + (0) + () There is no sugar left unused.. * : 9 5 : 3 : Point is optimal 0. a) Minimize + (cost, $) 3 + (antibiotic, units) + (antibiotic, units) -3 opyright 03 Pearson Education, Inc. publishing as Prentice Hall
4 + (antibiotic 3, units), : * : 0 0 : : 3 90 Point is optimal a) Maimize (gold, oz) + 0 (platinum, oz) (demand, bracelets), :,00 :,00 Point is optimal 0 *: 3,00 :,00. 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 +, 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. 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 3/ 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 (wool, yd ) + 00 (labor, hr), :,00 * :.5 3.7,73 :,000 Point is optimal opyright 03 Pearson Education, Inc. publishing as Prentice Hall
5 7. The feasible solution space changes from the area 0 to 0'', as shown on the following graph The etreme points to evaluate are now, ', and '. :,00 *': 5..5, ':,00 Point ' is optimal. a) Maimize , optimal,, a) No, not this winter, but they might after they recover equipment costs, which should be after the nd winter. 55.5,5 No, profit will go down c) 5,35 Profit will increase slightly d) $,073 Profit will go down from (c) 0. : 5 5 * : 7 : Point is optimal 0. Maimize s + 0s + 0s 3 + s + s + + s 3 5, : s, s, s 3 : s, s 5, s 3 : s, s, s 3. : * : 5. Point is optimal. : opyright 03 Pearson Education, Inc. publishing as Prentice Hall
6 3. Maimize s + 0s 3 + 0s s + + s + s 3 + s, : s, s, s 3, s : s, s 3., s 3, s. : s, s, s 3, s. : * : 5 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), :.5 : ,00,37.9 * :. :.70,735.97, Point is optimal 7. 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, 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 $, Eliminating the constraint for defective claims would result in a new solution, and 37.5, where only part-time operators would be hired. 9. The solution becomes infeasible; there are not enough workstations to handle the increase in the volume of claims.. 3. : Point is optimal 5 * : : 0 :.7.33 Point is optimal : () 3 () (3) * : () : (5) The problem becomes infeasible opyright 03 Pearson Education, Inc. publishing as Prentice Hall
7 Feasible space * :... : Point is optimal 0 0 Point is optimal : * : 5 3 : Point is optimal * : : : a) Maimize $ (freezer space, gals.) (budget, $) or (demand), * : Point is optimal : 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), : 90 * : 3 : 3 70 :,00 Point is optimal 0-7 opyright 03 Pearson Education, Inc. publishing as Prentice Hall
8 39. a) Maimize (stamping, days) + (coating, days) + 9 (lots), : 3 7,00 * : 5 5,0 : 3 7,0 Point is optimal 0 0. a) Maimize (assembly, hr) + (finishing, hr) + (inventory, units), : * : : : Point is optimal 0 0. The slope of the original objective function is computed as follows: /70 3/7 slope 3/7 The slope of the new objective function is computed as follows: /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. : : : 3.3 : a) Maimize 9 + (profit, $,000s) + (grapes, tons) (storage space, yd 3 ) (processing time, hr) 7 (demand, Nectar) 7 (demand, Red), E : 7 : 7 * : Optimal point : E : F : 7 3 F 0 3. a) 5() + () 0 hr hr left unused - opyright 03 Pearson Education, Inc. publishing as Prentice Hall
9 Points and would be eliminated and a new optimal solution point at 5.09, 5.5, and.7 would result. 00 X 0. a) Maimize cans, : X,5.5 : X,35. :,3.7 Point is optimal : X,95.05 *X,000 * $,70.7 X X : X0 : X9 : $. *: X3 *: X *: $. Point is optimal 7. a) Minimize ,000,000, ,0, X The model formulation would become, maimize $ , The solution is 3., 57., and $9.7 The discount would reduce profit. 00 X : X,000 : X3,5.57 : 5.5 Point is optimal *: X,95.05 *: X,000 *: a) Minimize $ ,0,000, , fewer defective items. a) Maimize $ ,90 3 +, , X -9 opyright 03 Pearson Education, Inc. publishing as Prentice Hall
10 X : X0 : X0 : *: X9 *: X9 *: 590. Point is optional : X0 *: X0 *: new constraint is added to the model in X The solution is 0,.7, $5 *: X0.07 : X.7 : 5 : X0 *: X0 *: Point is optimal a) Maimize (available land, acres) (labor, hr) (fertilizer, tons) (shipping space, acres) 37 (shipping space, acres), X X : * :. 37.,0 7,3 : 7.5 E : ,0,390 :.7 F : ,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 The new optimal solution is point ': ': *': ,0,57 ': 3 ': 37,0,00 5. a) Maimize $7,00 +,0 + 3,0 /( + ) , - opyright 03 Pearson Education, Inc. publishing as Prentice Hall
11 Optimal solution - z 7,0, a) Minimize $(.05)() + (.)(.75) ,00, 9 $ The new solution is.7.7 $.7 If twice as many guests prefer wine to beer, then the Robinsons would be approimately bottles of wine short and they would have approimately 53 more bottles of beer than they need. The waste is more difficult to compute. The model in problem 53 assumes that the Robinsons are ordering more wine and beer than they need, i.e., a buffer, and thus there logically would be some waste, i.e., 5% of the wine and % of the beer. However, if twice as many guests prefer wine, then there would logically be no waste for wine but only for beer. This amount logically would be the waste from.7 bottles, or $0, and the amount from the additional 53 bottles, $3.9, for a total of $ a) Minimize (3 + ) / ( + ) ( + ).5 ( + ), optimal, $05, opyright 03 Pearson Education, Inc. publishing as Prentice Hall
12 5. a) No, the solution would not change X No, the solution would not change c) Yes, the solution would change to hina ( ).5, razil ( ).5, and $9, a) $ invested in stocks $ invested in bonds maimize $ (average annual return) + $70,000 (available funds) /( + ).5 (% of stocks) ,000 (total possible loss), 0 * : 70 optimal : X (00S) , optimal: 37, ,5. z, (00S) 5. eams assigned to rad eams assigned to Sarah minimize (70/7.) or 0 (00/), 59. 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. 0. a) # cups of Pomona # cups of oastal Maimize $ ,0 oz or ( gal. oz) (.0)(.05) + (.0)(.05) lbs. olombian (.35)(.05) + (.)(.05) lbs. Kenyan (.5)(.05) + (.)(.05) lbs. Indonesian / 3/, - opyright 03 Pearson Education, Inc. publishing as Prentice Hall
13 X Solution: 7.3 cups.9 cups $ * :,000 * :,000 : ,9 : 0, : X. a) The only binding constraint is for olombian; the constraints for Kenyan and Indonesian are nonbinding and there are 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. 0 Multiple optimal solutions; and alternate optimal Multiple optimal solutions; and alternate optimal Infeasible Problem Unbounded Problem opyright 03 Pearson Education, Inc. publishing as Prentice Hall
14 SE SOLUTION: METROPOLITN POLIE PTROL 5 3 y The linear programming model for this case problem is Minimize /0 + y/5 + y 5 + y y.5, y 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. Optimal point The optimal solution is, y.5, and.05. This means that a patrol sector is.5 miles by mile and the response time is 0.05 hr, or 3 min. SE SOLUTION: THE POSSIILITY RESTURNT The linear programming model formulation is Maimize $ / 3/ or 3 /( + ). or.90. The graphical solution is shown as follows Optimal point : 3.3. $77.3 * : $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.90. to.0.0 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 3 $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, $, opyright 03 Pearson Education, Inc. publishing as Prentice Hall
15 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 opyright 03 Pearson Education, Inc. publishing as Prentice Hall
Chapter Two: Linear Programming: Model Formulation and Graphical Solution
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.
More informationChapter Two: Linear Programming: Model Formulation and Graphical Solution
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