Problema 1. x 2. 4 Optimal Solution. x 1 = 3, x 2 = 1.5 (3,1.5) x 1 (0,0) (4,0) Optimal Solution x 1 = 0, x 2 = 3. Value of Objective Function = 18

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1 Problema a. 6 x Optimal Solution 3 2 x 1 = 3, x 2 = 1.5 Value of Objective Function = 13.5 (3,1.5) 1 (0,0) (4,0) x x 2 Optimal Solution x 1 = 0, x 2 = 3 Value of Objective Function = 18 1 (0,0) x 1 c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3). Problema a. Let F = number of tons of fuel additive S = number of tons of solvent base Max 40F + 30S 2/5F + 1 /2 S 200 Material 1 1 /5 S 5 Material 2 3 /5 F + 3 /10 S 21 Material 3 F, S 0 S

2 . Modelos de Decisiones Tarea 1 x Material 3 Tons of Solvent Base Material 2 20 Optimal Solution (25,20) 10 Feasible Region Material x 1 F Tons of Fuel Additive c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints. Problema a. Let E = number of shares of Eastern Cable C = number of shares of ComSwitch Max 15E + 18C 40E + 25C 50,000 Maximum Investment 40E 15,000 Eastern Cable Minimum 25C 10,000 ComSwitch Minimum 25C 25,000 ComSwitch Maximum E, C 0 ver gráfica en la sig. página c. There are four extreme points: (375,400); (1000,400);(625,1000); (375,1000) d. Optimal solution is E = 625, C = 1000 Total return = $27,375

3 C 2000 Minimum Eastern Cable Number of Shares of ComSwitch Maximum Comswitch Maximum Investment 500 Minimum Conswitch Number of Shares of Eastern Cable E Problema a. S x Tons of Solvent Base Feasible Region Tons of Fuel Additive x F 1

4 Yes. New optimal solution is F = 18.75, S = 25. Value of the new optimal solution is 40(18.75) + 60(25) = c. An optimal solution occurs at extreme point 3, extreme point 4, and any point on the line segment joining these two points. This is the special case of alternative optimal solutions. For the manager attempting to implement the solution this means that the manager can select the specific solution that is most appropriate. Problema a. S Tons of Solvent Base Points satisfying material requirements Minimum S Points satisfying minimum production requirements Minimum F F Tons of Fuel Additive There are no points satisfying both sets of constraints; thus there will be no feasible solution. Materials Minimum Tons Required for F = 30, S = 15 Tons Available Additional Tons Required Material 1 2/5(30) + 1/2(15) = Material 2 0(30) + 1/5(15) = Material 3 3/5(30) + 3/10(15) = Thus RMC will need 1.5 additional tons of material 3. Problema a. Let O1 = percentage of Oak cabinets assigned to cabinetmaker 1 O2 = percentage of Oak cabinets assigned to cabinetmaker 2 O3 = percentage of Oak cabinets assigned to cabinetmaker 3 C1 = percentage of Cherry cabinets assigned to cabinetmaker 1 C2 = percentage of Cherry cabinets assigned to cabinetmaker 2 C3 = percentage of Cherry cabinets assigned to cabinetmaker 3 Min 1800 O O O C C C3 50 O C1 40 Hours avail. 1 42O C2 30 Hours avail O C3 35 Hours avail. 3 O1 + O2 + O3 = 1 Oak C1 + C2 + C3 = 1 Cherry O1, O2, O3, C1, C2, C3 0

5 Note: objective function coefficients are obtained by multiplying the hours required to complete all the oak or cherry cabinets times the corresponding cost per hour. For example, 1800 for O1 is the product of 50 and 36, 1764 for O2 is the product of 42 and 42 and so on. Cabinetmaker 1 Cabinetmaker 2 Cabinetmaker 3 Oak O1 = O2 = O3 = Cherry C1 = C2 = C3 = Total Cost = $ c. No, since cabinetmaker 1 has a slack of hours. Alternatively, since the dual price for constraint 1 is 0, increasing the right hand side of constraint 1 will not change the value of the optimal solution. d. The dual price for constraint 2 is The upper limit on the range of feasibility is Therefore, each additional hour of time for cabinetmaker 2 will reduce total cost by $1.75 per hour, up to a maximum of hours. e. The new objective function coefficients for O2 and C2 are 42(38) = 1596 and 48(38) = 1824, respectively. The optimal solution does not change but the total cost decreases to $ Problema a. Let M 1 = units of component 1 manufactured M 2 = units of component 2 manufactured M 3 = units of component 3 manufactured P 1 = units of component 1 purchased P 2 = units of component 2 purchased P 3 = units of component 3 purchased Min 4.50 M M M P P P 3 2M 1 3M 2 4M 3 21,600 Production 1M 1 1.5M 2 3M 3 15,000 Assembly 1.5M 1 2M 2 5M 3 18,000 Testing/Packaging M 1 1P 1 = 6,000 Component 1 1M 2 1P 2 = 4,000 Component 2 1M 3 1P 3 = 3,500 Component 3 M 1, M 2, M 3, P 1, P 2, P 3 0 Source Component 1 Component 2 Component 3 Manufacture Purchase Total Cost: $73,550 c. Since the slack is 0 in the production and the testing & packaging departments, these department are limiting Benson's manufacturing quantities. Dual prices information: Production $0.906/minute x 60 minutes = $54.36 per hour Testing/Packaging $0.125/minute x 60 minutes = $ 7.50 per hour

6 d. The dual price is -$ this tells us that the value of the optimal solution will worsen (the cost will increase) by $7.969 for an additional unit of component 2. Note that although component 2 has a purchase cost per unit of $8.80, it would only cost Benson $7.969 to obtain an additional unit of component 2. Problema Let x m = number of units produced in month m I m = increase in the total production level in month m D m = decrease in the total production level in month m s m = inventory level at the end of month m where m = 1 refers to March m = 2 refers to April m = 3 refers to May Min 1.25 I I I D D D 3 Change in production level in March x 1-10,000 = I 1 - D 1 or x 1 - I 1 + D 1 = 10,000 Change in production level in April x 2 - x 1 = I 2 - D 2 or x 2 - x 1 - I 2 + D 2 = 0 Change in production level in May x 3 - x 2 = I 3 - D 3 or x 3 - x 2 - I 3 + D 3 = 0 Demand in March x 1 - s 1 = 12,000 or x 1 - s 1 = 9,500 Demand in April s 1 + x 2 - s 2 = 8,000 Demand in May s 2 + x 3 = 15,000 Inventory capacity in March s 1 3,000 Inventory capacity in April s 2 3,000 Optimal Solution: Total cost of monthly production increases and decreases = $2,500 x 1 = 10,250 I 1 = 250 D 1 = 0 x 2 = 10,250 I 2 = 0 D 2 = 0 x 3 = 12,000 I 3 = 1750 D 3 = 0 s 1 = 750 s 2 = 3000 Problema 9 EZ Windows se resuelve de forma similar al problema anterior y a la aplicación sobre planeación de la producción vista en clase.