Operations Management
Operations Management
13th Edition
ISBN: 9780136860419
Author: Lee Krajewski
Publisher: Pearson Education
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Chapter A, Problem 2P

a)

Summary Introduction

To determine: The decision table for the decision.

Introduction: Decision table is formats or visual representations were data is expressed arranged, determined and calculated to make a effective decision making. A decision table is a tabular representation that is used to analyze decision alternatives and states of nature.

b.

Summary Introduction

To determine: Maximax decision

Introduction:

Maximax is the decision making method which come decision making under uncertainty. This method finds an alternative that maximizes the maximum outcome of each alternative or we can say that calculating the maximum outcome within every alternatives.

c.

Summary Introduction

To determine: The minimax decision.

Introduction

Maximin is the decision making method which makes decision making under uncertainty. This method will find an alternative that maximizes the minimum outcome of every alternative or we can say that calculating the minimum outcome within the each alternative.

d.

Summary Introduction

To determine: Equally likely decision

Introduction

Equally likely is the decision method which come decision making under uncertainty. Under this condition, equal probability is assigned under each uncertainty state of nature.

e.

Summary Introduction

To draw: Decision tree. Assume each outcome is equally likely, and find the highest EMV.

Introduction:

Decision tree is graphical representation of decision making process which has state of nature, alternative, payoffs and their probabilities of outcomes.

Operations Management, Chapter A, Problem 2P

EMV: It is expected value or payout that has different possible state of nature, each with their associated possibilities.

Formula:

EMV=((Payoff of 1st state of nature)×(Probability of 1st state of nature)+(Payoff of 2nd state of nature)×(Probability of 2ndstate of nature)+...+(Payoff of last state of nature)×(Probability of laststate of nature))

Here probabilities are equal likely in each case. So probabilities of the be 1/3= 0.3333

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