Principles Of Operations Management
Principles Of Operations Management
11th Edition
ISBN: 9780135173930
Author: RENDER, Barry, HEIZER, Jay, Munson, Chuck
Publisher: Pearson,
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Chapter 5, Problem 24P
Summary Introduction

To identify: The design that will yield the highest profit.

Introduction:

Expected monetary value (EMV):

Expected monetary value (EMV) is a calculation system for expected returns for the certain decision made by a company.

Expert Solution & Answer
Check Mark

Answer to Problem 24P

The design K1 can be used because of the highest profit.

Explanation of Solution

Given information:

Design Cost Probability Number of non-defective (out of 100)
K1 $100,000.00 0.8 90
0.2 70
K2 $130,000.00 0.85 90
0.15 75
K3 $180,000.00 0.9 95
0.1 80

Numberofsandwichesproduceddaily=500Costofeachsandwich=$1.30Sellingpriceofeachsandwich=$2.50Numberofdaysof operation=300

Calculation of EMV:

Principles Of Operations Management, Chapter 5, Problem 24P

Calculation of expected outcome and EMV:

Design K1:

Probability 0.80

=$100,000(10100×500×300×$1.30)+(90100×500×300×$1.20)=$100,000$19,500+$162,000=$42,500

The expected outcome is calculated by adding the cost for the design K1, $100,000 and the cost of 10 unsold sandwiches ($1.20) and subtracting the resultant with the profit of 90 sandwiches sold ($2.50-$1.20=$1.30). The expected outcome is $42,500.

Probability 0.20

=$100,000(30100×500×300×$1.30)+(70100×500×300×$1.20)=$100,000$58,500+$126,000=$32,500

The expected outcome is calculated by adding the cost for the design K1, $100,000 and cost of 30 unsold sandwiches ($1.20) and subtracting the resultant with the profit of 70 sandwiches sold ($2.50-$1.20=$1.30). The expected outcome is -$32,500.

Calculation of EMV:

EMV is calculated by multiplying the payoff with the respective probabilities.

=(42,500×0.80)+(32,500×0.20)=$27,500 (1)

The expected outcome for probability 0.80, $42,500 is multiplied and the expected outcome for probability 0.20, -$32,500 is multiplied and the resultant of both is added to give the EMV for design K1 is $27,500.

Design K2:

Probability 0.85

=$130,000(10100×500×300×$1.30)+(90100×500×300×$1.20)=$130,000$19,500+$162,000=$12,500

The expected outcome is calculated by adding the cost for the design K2, $130,000 and the cost of 10 unsold sandwiches ($1.20) and subtracting the resultant with the profit of 90 sandwiches sold ($2.50-$1.20=$1.30). The expected outcome is $12,500.

Probability 0.15

=$130,000(25100×500×300×$1.30)+(75100×500×300×$1.20)=$130,000$48,500+$135,000=$43,750

The expected outcome is calculated by adding the cost for the design K2, $130,000  and the cost of 25 unsold  sandwiches ($1.20) and subtracting the resultant with the profit of 75 sandwiches sold ($2.50-$1.20=$1.30). The expected outcome is -$43,750.

Calculation of EMV:

EMV is calculated by multiplying the payoff with the respective probabilities.

=(12,500×0.85)+(43,750×0.15)=$4,062.50 (2)

The expected outcome for probability 0.85, $12,500 is multiplied and the expected outcome for probability 0.15, -$43,750 is multiplied and the resultant of both is added to give the EMV for design K2 is $4,062.50.

Design K3:

Probability 0.90:

=$180,000(5100×500×300×$1.30)+(95100×500×300×$1.20)=$180,000$9,750+$171,000=$18,750

The expected outcome is calculated by adding the cost for the design K3, $180,000 and the cost of 5 unsold sandwiches ($1.20) and subtracting the resultant with the profit of 95 sandwiches sold ($2.50 - $1.20=$1.30). The expected outcome is - $18,750.

Probability 0.10:

=$180,000(20100×500×300×$1.30)+(80100×500×300×$1.20)=$180,000$39,000+$144,000=$75,000

The expected outcome is calculated by adding the cost for the design K3, $180,000  and cost of 20 unsold  sandwiches ($1.20) and subtracting the resultant with the profit of 80 sandwiches sold ($2.50-$1.20=$1.30). The expected outcome is - $75,000.

Calculation of EMV:

EMV is calculated by multiplying the payoff with the respective probabilities.

=(18,750×0.90)+(75,000×0.1)=$24.375 (3)

The probability 0.90 and its expected outcome, -$18,750 is multiplied and the expected outcome for probability 0.10, -$75,000 is multiplied and the resultant of both is added to give the EMV for design K3 is - $24.375.

From the calculated EMV of three designs K1, K2 and K3, it can be found that EMV for design K1 is higher than other two (refer equation (1),(2) and (3)). Therefore, it is recommended to use design K1.

Hence, the design K1 can be used because of the highest profit.

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Chapter 5 Solutions

Principles Of Operations Management

Ch. 5.S - Prob. 4PCh. 5.S - Prob. 5PCh. 5.S - Prob. 6PCh. 5.S - Prob. 7PCh. 5.S - Prob. 8PCh. 5.S - Prob. 9PCh. 5.S - Prob. 10PCh. 5.S - A Southern Georgia school district is considering...Ch. 5.S - Prob. 12PCh. 5.S - Prob. 13PCh. 5.S - Prob. 14PCh. 5.S - Prob. 15PCh. 5.S - Prob. 16PCh. 5.S - Prob. 17PCh. 5.S - Prob. 18PCh. 5.S - Prob. 19PCh. 5.S - Prob. 1.1VCCh. 5.S - Prob. 1.2VCCh. 5.S - Prob. 1.3VCCh. 5.S - Prob. 2.1VCCh. 5.S - Prob. 2.2VCCh. 5.S - Prob. 2.3VCCh. 5.S - Prob. 3.1VCCh. 5.S - Prob. 3.2VCCh. 5.S - Prob. 3.3VCCh. 5.S - Prob. 3.4VCCh. 5 - Prob. 1EDCh. 5 - Prob. 1DQCh. 5 - What techniques do we use to define a product? ...Ch. 5 - Prob. 3DQCh. 5 - Prob. 4DQCh. 5 - Prob. 5DQCh. 5 - Prob. 6DQCh. 5 - Describe four organizational approaches to product...Ch. 5 - Explain what is meant by robust design.Ch. 5 - Prob. 9DQCh. 5 - What information is contained in a bill of...Ch. 5 - What information is contained in an engineering...Ch. 5 - What information is contained in an assembly...Ch. 5 - Prob. 13DQCh. 5 - Explain how the house of quality translates...Ch. 5 - Prob. 15DQCh. 5 - Prob. 16DQCh. 5 - Why are the direct interaction and surrogate...Ch. 5 - Prob. 18DQCh. 5 - Prob. 1PCh. 5 - Prob. 2PCh. 5 - Prob. 3PCh. 5 - Construct a house of quality matrix for a...Ch. 5 - Prob. 5PCh. 5 - Prob. 6PCh. 5 - Prob. 7PCh. 5 - Prob. 8PCh. 5 - Prepare a bill of material for (a) a pair of...Ch. 5 - Prob. 10PCh. 5 - Prepare a script for telephone callers at the...Ch. 5 - Prob. 12PCh. 5 - Prob. 13PCh. 5 - Prob. 14PCh. 5 - Prob. 15PCh. 5 - Prob. 16PCh. 5 - Prob. 17PCh. 5 - Prob. 18PCh. 5 - Prob. 19PCh. 5 - Residents of Mill River have fond memories of ice...Ch. 5 - Prob. 21PCh. 5 - Prob. 22PCh. 5 - Prob. 23PCh. 5 - Prob. 24PCh. 5 - Prob. 25PCh. 5 - Prob. 1CSCh. 5 - Prob. 2CSCh. 5 - Prob. 3CSCh. 5 - Prob. 1VCCh. 5 - Prob. 2VCCh. 5 - Prob. 3VCCh. 5 - Prob. 4VC
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