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