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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Problema 1 12. a. 6 2 2 4 x₂ x₁ 1 3 5 (0,0) 1 3 5 6 (3,1.5) (4,0) 4 Optimal Solution x_{1 = 3,}x_{2 = 1.5}

Value of Objective Function = 13.5

b. 2 2 4 x₂ x₁ 1 3 (0,0) 1 3 5 6 7 8 9 10 Optimal Solution x_{1 = 0,}x_{2 = 3}

Value of Objective Function = 18

c. There are four extreme points: (0,0), (4,0), (3,1,5), and (0,3). Problema 2

21. a. Let F = number of tons of fuel additive S = number of tons of solvent base

Max 40F + 30S s.t. 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

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b. . x₂ x₁ 0 10 20 30 40 50 10 20 30 40 50 60 70 Feasible Region

Tons of Fuel Additive Material 2 Optimal Solution (25,20) M ate ria l 3 Mat_eria l 1

Tons of Solvent Base

c. Material 2: 4 tons are used, 1 ton is unused. d. No redundant constraints.

Problema 3

28. a. Let E = number of shares of Eastern Cable

C = number of shares of ComSwitch

Max 15E + 18C

s.t.

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 b. 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

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Problema 4 47. a. 0 10 20 30 40 50 10 20 30 40 50 60 70 Feasible Region

Tons of Fuel Additive

Tons of Solvent Base

1 5 2 3 4 x₂ x₁ 0 500 1000 500 1000 C E 1500 1500

2000 Minimum Eastern Cable

Maximum Comswitch

Minimum Conswitch Maximum Investment

Number of Shares of Eastern Cable

N u m b er o f S h ar es o f C o m S w it ch S F

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b. Yes. New optimal solution is F = 18.75, S = 25. Value of the new optimal solution is 40(18.75) + 60(25) = 2250.

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 5 48. a.

There are no points satisfying both sets of constraints; thus there will be no feasible solution.

b. Materials

Minimum Tons Required for F = 30, S = 15

Tons Available Additional Tons Required

Material 1 2/5(30) + 1/2(15) = 19.5 20

-Material 2 0(30) + 1/5(15) = 3 5

-Material 3 3/5(30) + 3/10(15) = 22.5 21 1.5

Thus RMC will need 1.5 additional tons of material 3. Problema 6

29. 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

Min 1800 O1 + 1764 O2 + 1650 O3 + 2160 C1 + 2016 C2 + 1925 C3

s.t.

50 O1 + 60 C1 ≤ 40 Hours avail. 1

42O2 + 48 C2 ≤ 30 Hours avail. 2

30 O3 + 35 C3 ≤ 35 Hours avail. 3

O1 + O2 + O3 = 1 Oak

C1 + C2 + C3 = 1 Cherry

O1, O2, O3, C1, C2, C3 ≥ 0 0

10 20 30 40

Tons of Fuel Additive

T o n s o f S o lv en t B as e 10 20 30 40 Points satisfying material requirements Points satisfying minimum production requirements Minimum F M in imu m S 50 S F

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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.

b.

Cabinetmaker 1 Cabinetmaker 2 Cabinetmaker 3

Oak O1 = 0.271 O2 = 0.000 O3 = 0.729

Cherry C1 = 0.000 C2 = 0.625 C3 = 0.375

Total Cost = $3672.50

c. No, since cabinetmaker 1 has a slack of 26.458 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 1.750. The upper limit on the range of feasibility is 41.143. Therefore, each additional hour of time for cabinetmaker 2 will reduce total cost by $1.75 per hour, up to a maximum of 41.143 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 $3552.50.

Problema 7

30. a. Let M1 = units of component 1 manufactured M2 = units of component 2 manufactured

M3 = units of component 3 manufactured

P₁ = units of component 1 purchased

P₂ = units of component 2 purchased

P₃ = units of component 3 purchased

Min 4.50 M₁  5.00M₂  2.75M₃  _6.50P₁ _{ 8.80P}₂ _{ 7.00P}₃ s.t. _2M₁  _3M₂  _4M₃ ≤ 21,600 Production 1M1  1.5M2  3M3 ≤ 15,000 Assembly 1.5M₁  2M₂  5M₃ ≤ 18,000 Testing/Packaging M₁  1P₁ = 6,000 Component 1 1M₂  1P₂ = 4,000 Component 2 1M3  1P3 = 3,500 Component 3 M1, M2, M3, P1, P2, P3 ≥ 0 b.

Source Component 1 Component 2 Component 3

Manufacture 2000 4000 1400

Purchase 4000 0 2100 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

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d. The dual price is -$7.969. 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 8

20. Let xm = number of units produced in month m

Im = increase in the total production level in month m Dm = decrease in the total production level in month m sm = 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 _{I1 + 1.25 I2 + 1.25 I3 + 1.00 D1 + 1.00 D2 + 1.00 D3} s.t.

Change in production level in March

x1 - 10,000 = I1 - D1 or x1 - I1 + D1 = 10,000 Change in production level in April

x2 - x1 = I2 - D2 or _{x2 - x1 - I2 + D2 = 0}

Change in production level in May

x3 - x2 = I3 - D3 or _{x3 - x2 - I3 + D3 = 0} Demand in March 2500 + _{x1 - s1 = 12,000} or _{x1 - s1 = 9,500} Demand in April s1 + x2 - s2 = 8,000 Demand in May s2 + x3 = 15,000

Inventory capacity in March

s1 ≤ 3,000

Inventory capacity in April

s2 ≤ 3,000

Optimal Solution:

Total cost of monthly production increases and decreases = $2,500

x1 = 10,250 I1 = 250 D1 = 0 x2 = 10,250 I2 = 0 D2 = 0 x3 = 12,000 I3 = 1750 D3 = 0 s1 = 750 s2 = 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.