Apr 4 Exam: Two hours

Apr 4 Exam: Two hours

• Everyone takes exam 4:00 to 6:00PM Thur

• Memo C-2 due Tues

• Set 5 due Thur March 21

Answers to set 4

• P 91, 5a

• X1=basketballs

• X2=footballs

• MAX 12x1 + 16x2

• Consr (1)(rubber) 3x1+2x2 < 500

• (2) (leather)4x1+ 5x2 < 800

P 91, Problem 6

• Continuation of problem 5

• Intercepts: (1) 0,250 and 167,0

(2) 0,160 and 200,0

x2

x1

0,250

167,0

0,160

200,0(1)

(2)128,57

A

B

C

infeasible

infeasible

Corner pts

pt x1 x2 12x1+16x2

A 0 160 2560=max

B 128.5 57.2 2457

C 167 0 2004

Exam format

• Make 160 footballs for $2560 profit

Part (a): slack

• CORNER PT A

(1)(rubber)3*0+2*160=320<500, so slack=500-320=180 (2)(leather)4*0+5*160=800,no slack

CORNER PT B: zero slack

CORNER PT C(1)no slack (2)4*167=668<800, slack =800-668=132

Part(b): Sensitivity

• Change in objective function

• Section 1: new basketball profit=13

Corner pts

pt x1 x2 13x1+16x2

A 0 160 2560

B 128.5 57.2 2590=max

C 167 0 2171

Section 1: sensitive

• Old optimum at pt A

• New optimum at pt B

• Start making basketballs

6b:Section 2

• New football profit = 15

Corner pts

pt x1 x2 12x1+15x2

A 0 160 2400=max

B 128.5 57.2 2400=max

C 167 0 2004

6b: Section 2: tie

• Sensitive since optimum at A or B

6c: constraint change

• Right-hand side of rubber constraint

• Section 1:

• Old : 500

• New: 1000

• New (1) 3x1+2x2< 1000

• New intercepts: 0,500 and 333,0

x2

0,160

200,0

(2)A

infeasible0,500

333,0

New(1)

infeasible

infeasible

Max profit

x1 x2 12x1+16x2

0 160 2560=max

200 0 2400

6c: Section 1:insensitive

• Original solution: 160 footballs

• New: 160 footballs

• Same optimum

6c:Section 2

• Right hand side of leather constraint

• Old: 800

• New: 1300

• New (2) 4x1+5x2 < 1300

• New intercepts: 0,260 and 325,0

x1

0,250

167,0 325,0(1)

C

infeasible

0,260

New(2)

infeasible

6c: Section 2

x1 x2 12x1+16x2

0 250 4000=max

167 0 2004

6c:Section 2:sensitive

• While output mix is same as original (footballs only), we must check slack to see if input sensitive

• Original problem: pt A was optimum

Rubber slack>0, no leather slack

New problem: optimum on (1), so no rubber slack, but leather slack>0, so input sensitive

Set 4(3)Sensitivity range(Algebra)

• C1=basketball profit

• objective function

• Z= c1x1+16x2

• X2= (1/16)Z –(C1/16)x1

Constraint (1)

• (1) 3x1+2x2 = 500

2x2 = 500-3x1

x2 = 250 –1.5x1

• Set objective function coef of x1 = constr (1) coef of x1

C1/16 = 1.5, so C1 =24

• Old C1 = 12, so C1 < 24

Constraint (2)

• (2) 4x1+5x2 = 800

X2 = 800/5 –(4/5)x1

• Set objective function coef of x1 =

constr (2) coef of x1

• C1/16 = 4/5 =.8

• C1=.8*16= 12.8

• Old C1=12, so C1 < 12.8

Answer: C1 < 12.8

• Answer must be in intersection of both ranges: C1 < 24 and C1 < 12.8

• ------------------------------------24

• ------------------12.8

• -------------12=old C1

interpretation

• Make footballs only if unit basketball profit is less than $12.80

• But if unit basketball profit exceeds $12.80, start making basketballs

• Since current profit is $12.00, big change if unit profits increases by 80 cents