answers to set 3
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Answers to set 3
• Set 4 due Thurs March 14
• Memo C-2 due Tues March 19
• Next exam Thurs April 4
• Tutor Mon 4-6PM, Thurs 2-4PM, BB 4120
Problem 5
• X1=ingred 1
• X2=ingred 2
• Objective: MIN cost = 3x1+ 5x2
• Constraints
• 1)nitro 10x1+2x2 > 20
• 2)phos 6x1 + 6x2 > 36
• 3)potas x2 > 2
5(cont’d)
Nitr
X1 X2
0 10
2 0
5(cont’d)
Phos
X1 X2
0 6
6 0
5(cont’d)
Potas
X1 X2
0 2
Infinity 0
x2
0,10
2,0
0,6
6,0
0,2
A
B
C
infeasibleinfeasible
infeas
(1) (2)
(3)
feasible
Corner pt B: nitro, phos
• 10x1+2x2=20
• 6x1+6x2=36
• X1=1
• X2=5
Corner pt C: Phos,potas
• 6x1+6x2=36
• X2=2
• X1=4
Feasible corner points
Corner pt x1 X2
A 0 10
B 1 5
C 4 2
MIN 3x1+5x2
X1 x2 3x1+5X2
0 10 3(0)+5(10)=50
1 5 3(1)+5(5)=28
4 2 3(4)+5(2)=22=min cost
Exam format
• Buy 4 pounds of ingredient #1 and 2 pounds of ingred #2 for $22 cost
Problem 6
• X1=chairs
• X2=tables
• Obj max profit = 400x1+100x2
• Constr
• (1)labor 8x1+10x2 < 80
• (2) wood 2x1 + 6x2 < 36
• (3) demand x1 < 6
x2
0,8
10,0
0,6
18,06,0
feasible
A
B
C
D
infeasible
infeasible
infeasible
Corner points
X1 X2
0 6 A
4.3 4.6 B
6 3.2 C
6 0 D
MAX 400x1+100x2
X1 X2 400x1+100x2
0 6 600
4.3 4.6 2171
6 3.2 2720=max
6 0 2400
Exam Format
• Make 6 chairs and 3.2 tables per day for $2720 profit
Problem 7
• Plug optimal solution to 6 into labor constraint: 8*6+10*3.2=80=avail, so no labor slack
• Wood 2*6+6*3.2=31.2<36 avail, so wood slack = 36-31.2=4.8
Problem 8
• Re-do 6 with new table profit = $500
• NEW objective function = 400x1+500x2
MAX 400x1+500x2
X1 X2 400x1+500x2
0 6 3000
4.3 4.6 4000=max
6 3.2 4000=max
6 0 2400
Exam Format
• Sensitive to change since two corner points are now equally optimal
Problem 24
• X1=permanent
• X2=temp
• Obj min cost =64x1+42x2
• Constr
• 1) claims 16x1+12x2 > 450
• 2) computer station x1 + x2 < 40
• 3) defective .5x1 +1.4x2 < 25
x2
0,40
0,37
0,18
28,0 40,0 50,0
(1)(2)
(3)
Infeasible
infeasible
A
B
C
D
Corner points
x1 x2
28.1 0 A
20.1 10.7 B
34.4 5.6 C
40 0 D
Min cost
x1 x2 64x1+42x2
28.1 0 1800
20.1 10.7 1736=min
34.4 5.6 2438
40 0 2560
Exam format
• Hire 20.12 permanent operators and 10.67 temporary operators for cost of $1736
• Real world: probably round up to 21 and 11 in anticipation of sick leave, attrition, etc
Problem 25a
• Sensitivity on previous problem
• change objective function coefficient
• a: change x1 coefficient
New cost = 54x1+42x2
X1 X2 Cost
28.1 0 1519
20.1 10.7 1535
34.4 5.6 2094
40 0 2160
New min =1519
• X1>0 only
• previous problem: x1>0 and x2>0
• different corner pt
• no more temps
• solution sensitive to change
25b: change x2 coefficient
• Min cost = 64x1 + 36x2
X1 X2 64x1+36x2
28.1 0 1800
20.1 10.7 1672=min
34.4 5.6 2404
40 0 2560
25b: insensitive
• Problem 24: corner pt B optimal
• Problem 25b: corner pt B optimal
• same mix as 24
26: Remove constraint from 24
• Remove constraint 3 (defective)
x2
0,40
0,37
0,18
28,0 40,0 50,0
(1)(2)
Infeasible
infeasible
A
B
C
D
Infeasible
Problem 26
x1 x2 Cost= 64x1+42x2
28 0 1792
40 0 2560
0 37 1554=min
0 40 1680
Problem 26: sensitive
• Problem 24: permanent and temp
• Problem 26: temp only
• New corner point
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