Department

ManagementCourse Code

MGOC10H3Professor

Vinh QuanStudy Guide

MidtermThis

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MGOC10 Analysis for Decision Making – L01/L02 – Midterm Solution

Q1.

(a) a = 0, b = 10000, c = 6000, d = 0

(b)

6– 6 ≤ range coeff x5 ≤ 6 + ∞ so 0 ≤ range coeff x5 ≤ ∞

New Z = old Z – reduction due to coefficient of x5 dropping

= 150000 - (6 - 4)*(x5) = 150000 - (6 - 4)*(5000) =140000

(c)

10000 -5000 ≤ RHS R5 ≤ 10000 +333 get 5000 ≤ RHS R1 ≤ 10333

Dual price R5 = -12 means for each unit increase in RHS, Z worst off by 12.

Worst off means Z goes up. So new Z=150000 + 12(10200 - 10000) = 152400

Q2.

P1, P2 = the number of model 1,2 motors to produce

AA = the number of hours allocated to area A

AB = the number of hours allocated to area B

AC = the number of hours allocated to area C

TBA = the number of hours transferred from B to A

TCA = the number of hours transferred from C to A

TCB = the number of hours transferred from C to B

OTA = OT hours used in area A OTB = OT hours used in area B OTC = OT hours used in area C

Min 80P1 + 65P2 + 55OTA + 55OTB + 55OTC

s.t. 15P1 + 3P2 AA

4P1 + 10P2 AB

4P1 + 8P2 AC

AA = 10000 + OTA + TBA + TCA

AB = 10000 + OTB + TCB - TBA

AC = 18000 + OTC - TCA - TCB

TBA 12000

TCA + TCB 20000

P1 + P2 ≥ 50,000

4P1 ≤ P2

OTA ≤ 5000, OTB ≤ 6000, OTC ≤ 7000

all variables 0

Q3.

From graph, X1 = 100, X2 = 100 is optimal.

Z = 30(100) + 10(100) = 4000.

From graph, if if X2 ≥ 50 became X2 = 50, then the

problem is infeasible.

1

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