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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.

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Apr 4 Exam: Two hours

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  1. 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

  2. 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

  3. P 91, Problem 6 • Continuation of problem 5 • Intercepts: (1) 0,250 and 167,0 (2) 0,160 and 200,0

  4. x2 infeasible 0,250 0,160 A (2) 128,57 B infeasible (1) 200,0 167,0 x1 C (1

  5. Corner pts

  6. Exam format • Make 160 footballs for $2560 profit

  7. 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

  8. Part(b): Sensitivity • Change in objective function • Section 1: new basketball profit=13

  9. Corner pts

  10. Section 1: sensitive • Old optimum at pt A • New optimum at pt B • Start making basketballs

  11. 6b:Section 2 • New football profit = 15

  12. Corner pts

  13. 6b: Section 2: tie • Sensitive since optimum at A or B

  14. 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

  15. x2 infeasible 0,500 0,160 A (2) New(1) infeasible infeasible 200,0 333,0 (1

  16. Max profit

  17. 6c: Section 1:insensitive • Original solution: 160 footballs • New: 160 footballs • Same optimum

  18. 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

  19. 0,260 infeasible 0,250 New(2) infeasible (1) 325,0 167,0 x1 C (1

  20. 6c: Section 2

  21. 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

  22. Set 4(3)Sensitivity range(Algebra) • C1=basketball profit • objective function • Z= c1x1+16x2 • X2= (1/16)Z –(C1/16)x1

  23. 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

  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

  25. Answer: C1 < 12.8 • Answer must be in intersection of both ranges: C1 < 24 and C1 < 12.8 • ------------------------------------24 • ------------------12.8 • -------------12=old C1

  26. 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

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