Problem B.1, HR7E Solve the following LP graphically R. Saltzman

Maximize 4X + 6Y = Zsubject to: (1) X + 2Y <= 8

(2) 5X + 4Y <= 20(3) X >= 0(4) Y >= 0

Note: There is a typograhpical error in the book regarding the last 2 nonnegativity constraints.

Y

Corner Points

X Y Z

11 0 0 0

10 0 4 24

9 4 0 16

8 1.333 3.333 25.33 <-- optimal solution

7 and optimal Z

6

5

4 (2)

3

2

1 FR (1)

0

1 2 3 4 5 6 7 8 9 10 11 X

Feasible

Problem B.2, HR7E Solve the following LP graphically R. Saltzman

Maximize X + 10Y = Zsubject to: (1) 4X + 3Y <= 36

(2) 2X + 4Y <= 40(3) Y >= 3(4) X >= 0(5) Y >= 0

Y13 Corner Points12 X Y Z11 0 3 3010 0 10 100 <-- optimal solution

9 6.75 3 36.758 2.4 8.8 90.4765 FR43 (3)21 (1) (2)0

1 3 5 7 9 11 13 15 17 19 X

Feasible

Problem B.4, HR7E R. Saltzman

Maximize 30X1 + 10X2 = Zsubject to: (1) 3X1 + X2 <= 300

(2) X1 + X2 <= 200(3) X1 <= 100(4) X2 >= 50(5) X1 - X2 <= 0

a) Solve the problem graphically:

X2

300 (3) X1 X2 Z275 0 50 500250 0 200 2000225 (1) 50 50 2000200 50 150 3000 <-- optimal solution175 75 75 3000 <-- optimal solution150125 (5)100 FR

75 (2)50 (4)25

05 0 10 0 15 0 20 0 X1

b) Is there more than 1 optimal solution? Yes. Also, all the points between (50, 150) and (75, 75) have the same optimal value of 3000.

FeasibleCorner Points

Problem B.6, HR7E Ed Solver Dog Food Co. R. Saltzman

a) Formulation: Let C = # of chicken-flavored biscuits per packageLet L = # of liver-flavored biscuits per package

Mininize .02*C + .01L = Zsubject to: (1) C + L >= 40 (Nutrient A requirement)

(2) 4C + 2L >= 60 (Nutrient B requirement)(3) L <= 15(4) C >= 0(5) L >= 0

b) Solve the problem graphically:L

40C L Z

35 40 0 0.8025 15 0.65 <-- optimal solution & cost

30

25

20

15 (3)

10(1) FR

5 (2)

05 10 15 20 25 30 35 40 C

Corner PointsFeasible

Problem B.8, HR7E Optimal Mix of Bathtubs R. Saltzman

a) Formulation: Let A = # of model A bathtubsLet B = # of model B bathtubs

Maxinize 90A + 70B = Z Total Profitsubject to: (1) 125A + 100B <= 25000 Steel Availability

(2) 20A + 30B <= 6000 Zinc Availability(3) A >= 0(4) B >= 0

b) Solve the problem graphically:Corner Points

B A B Z0 0 0

250 200 0 18000 <-- optimal solution(1) 0 200 14000

200 85.7 142.9 17714

150

100

50 FR (2)

050 100 150 200 250 300 A

Feasible

Problem B.9, HR7E Mattresses & Box Springs R. Saltzman

a) Formulation: Let M = # of mattresses to produceLet B = # of box springs to produce

Mininize 20M + 24B = Z Total Costsubject to: (1) M + B >= 30 Minimum production requirement

(2) 1M + 2B >= 40 Stitching machine requirement(3) M >= 0(4) B >= 0

b) Solve the problem graphically:Corner Points

B M B Z40 0 800

50 0 30 72020 10 640 <-- optimal solution

40

30(1) FR

20

10(2)

010 20 30 40 50 M

Feasible

Problem B.10, HR7E Making Computers R. Saltzman

a) Formulation: Let A = # of Alpha 4 minicomputers to produceLet B = # of Beta 5 minicomputers to produce

Maxinize 1200A + 1800B = Z Total Profitsubject to: (1) 20A + 25B = 800 Full employment

(2) A >= 10 Minimum Alpha 4 production(3) B >= 15 Minimum Beta 5 production

b) Solve the problem graphically:Corner Points

B A B Z10 24 55200 <-- optimal solution

50 21.25 15 52500

40 (2) (FR is the line segment between these 2 corner points)

30

20 (1)

10 (3)

010 20 30 40 A

Feasible

Problem B.12, HR7E Krista's LP R. Saltzman

Mininize X1 + 2X2 = Z Total Costs.t. (1) X1 + 3X2 >= 90

(2) 8X1 + 2X2 >= 160(3) 3X1 + 2X2 >= 120(4) X2 <= 70

Solve graphically: CostX1 X2 Z

X2 A 90 0 90.00B 25.71 21.43 68.57 <-- optimal

80 C 8 48 104.00D 2.5 70 142.50

70 D

60

50C FR

40

30

20 B

10

0 A10 20 30 40 50 60 70 80 90

X1minimum iso-cost line = 68.57

FeasibleCorner Points

Problem B.16, HR7E Busing Students R. Saltzman

Superintendent must assign students living in 5 geographic sectors to 3 schools.

1. Different numbers of students live in each sector2. Each high school has a capacity of 900 students3. Some students must be bused - distances are shown in the table4. Students living in a sector where there is a school walk (0 bus miles)

Goal: Find assignment that minimizes the total # of student miles traveling by bus to school.

Let Xij = Number of students from sector I bused to school in sector j

Data Mileage SupplyFrom \ To School-in-Sector B School-in-Sector C School-in-Sector E (Students)Sector A 5 8 6 700Sector B 0 4 12 500Sector C 4 0 7 100Sector D 7 2 5 800Sector E 12 7 0 400(Fake) F 0 0 0 200Demand 900 900 900 2700 \ 2700

Allocations Optimal Solution (found using Solver)From \ To School-in-Sector B School-in-Sector C School-in-Sector E Row TotalSector A 400 0 300 700Sector B 500 0 0 500Sector C 0 100 0 100Sector D 0 800 0 800Sector E 0 0 400 400(Fake) F 0 0 200 200

Col. Total 900 900 900 2700 \ 2700

Total Cost 5400

Problem B.18, HR7E Restaurant Scheduling R. Saltzman

* Open 24 hours a day* Servers work 8 hour shifts, reporting for duty at beginning of one of 6 time periods:

Period Time # of Servers Requiredi = 1 3 am - 7 am 3i = 2 7 am - 11 am 12i = 3 11 am - 3 pm 16i = 4 3 pm - 7 pm 9i = 5 7 pm - 11 pm 11i = 6 11 pm - 3 am 4

Goal: Find minimum # of servers required to cover the schedule.

Let Xi = # of servers who begin work at start of period i, i = 1, 2, 3, 4, 5, 6.

X1 X2 X3 X4 X5 X6 ΣNo. of Servers 3 14 2 7 4 0 30 <-- optimal solutionCost of Server 1 1 1 1 1 1 (via Solver)

Period Time X1 X2 X3 X4 X5 X6 LHS RHS1 3 am - 7 am 1 1 3 >= 32 7 am - 11 am 1 1 17 >= 123 11am - 3 pm 1 1 16 >= 164 3 pm - 7 pm 1 1 9 >= 95 7 pm - 11 pm 1 1 11 >= 116 11 pm - 3 am 1 1 4 >= 4

That is: Minimize X1 + X2 + X3 + X4 + X5 + X6 = Z subject to: X1 + X6 >= 3

X1 + X2 >= 12 X2 + X3 >= 16 X3 + X4 >= 9 X4 + X5 >= 11 X5 + X6 >= 4All Xj >= 0, for j = 1, 2, 3, 4, 5, 6

Problem B.19, HR7E Birdhouse Builder R. Saltzman

a) Formulation: Let W = # of Wren Birdhouses to buildLet B = # of Bluebird Birdhouses to build

Maxinize 6W + 15B = Z Total Profitsubject to: (1) 4W + 2B <= 60 Labor availability

(2) 4W + 12B <= 120 Lumber availability(3) W >= 0(4) B >= 0

b) Solve the problem graphically:Corner Points

B W B Z15 0 90

50 0 10 15012 6 162 <-- optimal solution

40

30

20(1)

10 (2)

0 FR5 10 15 20 25 30 W

Feasible

Problem B.25, HR7E Advertising Agency R. Saltzman

a) Formulation: Let T = # of TV spots to runLet S = # of Sunday newspaper ads to run

Maxinize 35T + 20S = Z Total Exposure (in 1000's)subject to: (1) 3000T + 1250S <= 100000 Advertising Budget

(2) T >= 5 Minimum # of TV spots(3) T <= 25 Maximum # of TV spots(4) S >= 10 Minimum # of Sunday ads

b) Solve the problem graphically:Corner Points

S T S Z5 10 375

80 5 68 1535 <-- optimal solution25 10 1075

70 25 20 1275

60

50

40 (1)(2) (3)

30FR

20

10(4)

05 10 15 20 25 30 T

Feasible

Problem B.26, HR7E Factories & Warehouses R. Saltzman

Unit Shipping Costs & Capacities

To Warehouse ProductionFrom A B C Capability

Factory 1 6$ 5$ 3$ 6Factory 2 8$ 10$ 8$ 8Factory 3 11$ 14$ 18$ 10Capacity 7 12 5

a) Write the objective function and constraints:

Objective: Minimize 6X1A + 5X1B + 3X1C + 8X2A + 10X2B + 8X2C + 11X3A + 14X3B + 18X3C

Constraints: X1A + X1B + X1C = 6X2A + X2B + X2C = 8X3A + X3B + X3C = 10

X1A + X2A + X3A = 7X1B + X2B + X3B = 12X1C + X2C + X3C = 5

Plus 9 nonnegativity constraints: all variables (cells) must be at least 0.

Problem C.1, HR7E Transportation Problem R. Saltzman

From Los Angeles Calgary Panama City SupplyMexico City 6$ 18$ 8$ 100

Detroit 17$ 13$ 19$ 60Ottawa 20$ 10$ 24$ 40

Demand 50 80 70

a) Find an initial solution using the northwest-corner method:

From Los Angeles Calgary Panama City SupplyMexico City 50 50 100

Detroit 30 30 60Ottawa 40 40

Demand 50 80 70

b) Find an initial solution using the lowest-cost method:

From Los Angeles Calgary Panama City SupplyMexico City 50 50 100

Detroit 40 20 60Ottawa 40 40

Demand 50 80 70

c) The total cost of the northwest-corner solution = 3,120$ The total cost of the lowest-cost solution = 2,000$

To

To

To