Question

Fuzzy logic in supply chain management. A branch of mathematics known as fuzzy logic was used to improve customer service in supply chain management. (Decision Analytics, February 2014.) Customers rate the importance of one service factor relative to another using the following numerical scale: 1 = service factors are equally important, 3 = one factor is moderately more important, 5 = one factor is strongly more important, 7 = one factor is very strongly more important and 9 = one factor is extremely more important. Fuzzy numbers were developed to allow for variation in customer responses. For example, the fuzzy number 1∼represents an actual response of either 1 or 3; the fuzzy number 7∼represents a response of 5, 7, or 9. Consider the probabilities of the actual responses for each fuzzy number shown in the table.

Fuzzy Response	Probabilities of Actual Responses
$\stackrel{\mathbf{~}}{\mathbf{1}}$	$\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}$
$\stackrel{\mathbf{~}}{\mathbf{3}}$	$\mathbf{P}\mathbf{\left(}\mathbf{1}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}$
$\stackrel{\mathbf{~}}{\mathbf{5}}$	$\mathbf{P}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}$
$\stackrel{\mathbf{~}}{\mathbf{7}}$	$\mathbf{P}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{9}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}$
$\stackrel{\mathbf{~}}{\mathbf{9}}$	$\mathbf{P}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{=}\mathbf{1}\mathbf{/}\mathbf{3}\mathbf{,}\mathbf{P}\mathbf{\left(}\mathbf{9}\mathbf{\right)}\mathbf{=}\mathbf{2}\mathbf{/}\mathbf{3}$

a. If a customer gives a fuzzy response$\stackrel{\mathbf{~}}{\mathbf{7}}$, what is the probability that the actual response is not a 7?

b. If both$\stackrel{\mathbf{~}}{\mathbf{5}}$$\stackrel{\mathbf{~}}{\mathbf{9}}$ represent a possible fuzzy response of a customer, what are the possible actual responses for this customer?

Short answer

1. 2/3
2. 1 & 1

Step-by-step solution

Introduction

The possibility that an event will occur is referred to as its probability. The probability of an event occurring, or P (E), is defined as the ratio of the number of favourable outcomes to the total number of outcomes.

The formula represents as:

$\mathbf{P}\mathbf{\left(}\mathbf{E}\mathbf{\right)}\mathbf{\text{}}\mathbf{=}\frac{\mathbf{Favourable}\mathbf{}\mathbf{out}\mathbf{}\mathbf{come}}{\mathbf{Total}\mathbf{}\mathbf{out}\mathbf{}\mathbf{come}}$

Find the actual probability response is not 7

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\mathbf{not}\mathbf{7}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{9}\mathbf{\right)}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{1}}{\mathbf{3}}\\ \mathbf{=}\frac{\mathbf{2}}{\mathbf{3}}\end{array}$

Hence, the probability of actual response is not 7 is 2/3.

Find the probability of both 5~ and 9~

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\stackrel{\mathbf{~}}{\mathbf{5}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{3}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{5}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\\ \mathbf{}\mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{1}}{\mathbf{3}}\\ \mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{1}\mathbf{+}\mathbf{1}}{\mathbf{3}}\\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{3}}\\ \mathbf{=}\mathbf{1}\end{array}$

Hence, the probability $\stackrel{\mathbf{~}}{\mathbf{5}}$ is 1.

$\begin{array}{c}\mathbf{P}\mathbf{\left(}\stackrel{\mathbf{~}}{\mathbf{9}}\mathbf{\right)}\mathbf{=}\mathbf{P}\mathbf{\left(}\mathbf{7}\mathbf{\right)}\mathbf{+}\mathbf{P}\mathbf{\left(}\mathbf{9}\mathbf{\right)}\\ \mathbf{=}\frac{\mathbf{1}}{\mathbf{3}}\mathbf{+}\frac{\mathbf{2}}{\mathbf{3}}\\ \mathbf{=}\frac{\mathbf{1}\mathbf{+}\mathbf{2}}{\mathbf{3}}\\ \mathbf{=}\frac{\mathbf{3}}{\mathbf{3}}\\ \mathbf{=}\mathbf{1}\end{array}$

Hence, the probability$\stackrel{\mathbf{~}}{\mathbf{9}}$is1.