Click here to load reader
Upload
raquel-simoes
View
214
Download
0
Embed Size (px)
Citation preview
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