Department

ManagementCourse Code

MGOC10H3Professor

Vinh QuanStudy Guide

QuizThis

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Department of Management, UTSC

MGOC10 Analysis for Decision Making – L01/L02 - Assignment #2 – Solution

Problem 1

(a)

B = thousands of $ invest in Bond fund, S = thousands of $ invest in Stock fund

Max Z = 0.18S + 0.06B

S + B ≤ 720 (money available)

S ≤ 0.65(S + B) (max stock))

0.22S + 0.05B ≤ 100 (max loss)

S, B ≥ 0

(b)

S + B ≤ 720 → (0, 720) (720, 0)

S ≤ 0.65(S + B) → 0.35S ≤ 0.65B → (0,0) (325, 175)

0.22S + 0.05B ≤ 100 → (300, 680) (400, 240)

Z1 = 0.18S + 0.06B = 18 → (0, 300) (100, 0)

Z2 = 0.18S + 0.06B = 36 → (0, 600) (200, 0)

To obtain optimal solution solve: 0.22S + 0.05B = 100 and S + B = 720 get

S = 376.47058 or $376,470.58 , B = 343.52941 or 343,529.41, Z = 88.37647 or $88,376.47

Max amount can lose is 0.22S + 0.05B = 0.22(376.47058) + 0.05(34352941) = 100 or $100,000.

(c)

From graph, current solution is optimal if slope OBJ steeper than CON1 and flatter than CON3

Objective: Z = Cs S + .06B get B = Z/.06 – Cs/.06 S, slope = – Cs/.06

CON1: S + B ≤ 720 get B =720 – S, slope = -1

CON3: 0.22S + 0.05B = 100 → B = 2000 – 4.4S, slope = -4.4

Want slope OBJ steeper than CON1 : – Cs/0.06 ≤ -1 → 1 ≤ Cs/.06 get 0.06 ≤ Cs

Want slope OBJ flatter than CON3 : – Cs/0.06 ≥ -4.4 → 4.4 ≥ Cs/.06 get 0.264 ≥ Cs

Therefore range is 0.06 ≤ Cs ≤ .264

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