# Answers to exam #2 Ver III 90-97n=16 80-89n=42 70-79n=45 60-69n=20 <60n=4 Total = 127.

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Answers to exam #2 Ver III 90-97n=16 80-89n=42 70-79n=45 60-69n=20 <60n=4 Total = 127

1)obj fn: c1+28x2 = z 28x2 = z –c1x1 x2 = z/28 –(c1/28)x1 Constr 1) 2x1 + x2 = 200 x2 = 200 –2x1 c1/28 = 2, so c1 < 56 since old c1=5

Constr 2) 4x1+7x2=600 7x2=600-4x1 x2 = 600/7 –(4/7)x1 obj fn: c1/28 = 4/7 c1 < 16 since old c1=5

Short answer: 3 pt each 2) data envelopment analysis (p 135) 3) parameters (p 73) 4) Minnesota (p 42)

5) formulation(31 pt) B)obj fn MIN 10000x1 + 4000x2 Constr#1) x2 2, or x1> 2x2, or x2/x1 <.5 #2) 140x1 + 70x2 > 100 #3) 90x1+40x2 > 2 50x1 + 30x2

#4) 40x1+35x2 <.20 (at most) 140x1+70x2 This problem was inspired by homework problem 31, Taylor p 155

(6)orig problem intercepts 40 pt x1X2 0-4.5 30 01.33333 20

(0,1.333) (2,0) (0,-4.5) (3,0) infeasible (0,0)

6a) feasible pts x1x23x1+7x2 01.33333339.3 206=min 309

Exam format 6b: make 2 brooms 6c: cost = 6

6d) new intercepts x1X2 03.333333 50

(0,3.333) (0,-4.5) (3,0) infeasible (0,0) (5,0(5,0 (5,0) infea infeasible (3.6,.9)

6d) feasible pts x1x23x1+7x2 3.6.917=min 03.33323

6) Exam format E) make 3.6 brooms,.9 rakes F) min cost =17 G) orig: no rakes, new: makes rakes, so output sensitive H) orig: slack >0 (optimum not on both constr), new: slack = 0, so input sensitive

IV(1) formulation(31 pt) B)obj fn MIN 12000x1 + 5000x2 Constr#1) x2 2, or x1> 2x2, or x2/x1 <.5 #2) 180x1 + 95x2 > 100 #3) 110x1+50x2 > 2 70x2 + 45x2

#4) 60x1+45x2 <.20 (at most) 180x1+95x2

IV(2)orig problem intercepts 40 pt x1X2 0-4.5 30 01 1.50

(0,1) (1.5,0) (0,-4.5) (3,0) infeasible (0,0)

IV(2)(a) feasible pts x1x23x1+7x2 017 1.504.5=min 309

IV(2)(b)make 1.5 brooms IV(2)© cost = \$ 4.50

IV(2)(d): new intercepts x1X2 03 4.50

(0,3) (0,-4.5) (3,0) infeasible (0,0) (4.5,0) (3.5,.7) infeasible

IV(2)(d):feasible pts x1x23x1+7x2 0321 3.5.715=min

IV(2)(e) Make 3.5 brooms,.7 rakes (f) 15 (g) output sensitive (new: rakes>0) (h) input sensitive (new slack =0)

IV(3)obj fn: c1+28x2 = z 28x2 = z –c1x1 x2 = z/28 –(c1/28)x1 Constr 1) 2x1 + x2 = 200 x2 = 200 –2x1 c1/28 = 2, so c1 < 56 since old c1=5

Constr 2) 4x1+6x2=600 7x2=600-6x1 x2 = 600/7 –(6/7)x1 obj fn: c1/28 = 6/7 c1 < 24 since old c1=5

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