apr 4 exam: two hours
DESCRIPTION
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. - PowerPoint PPT PresentationTRANSCRIPT
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