EBK PRODUCTION AND OPERATIONS ANALYSIS
7th Edition
ISBN: 8220102480681
Author: Olsen
Publisher: WAVELAND
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Chapter 13.7, Problem 28P
Summary Introduction
Interpretation:
The optimal time for replacement:
Concept Introduction:
The objective of optimal policy is to determine the value of 't’ that minimizes the total cost of maintenance and replacement over an infinite horizon.
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A product engineer has developed the following equation for the cost of a system component: C = (10P) 2, where C is the cost in dollars and P is the probability that the component will operate as expected. The system is composed of two identical components, both of which must operate forthe system to operate. The engineer can spend $173 for the two components. To the nearest two decimal places, what is the largest component probability that can be achieved?
a factory that produces a small number of items per day. In about 30% of working days the factory produces 7 units of the product, in 45% of working days it produces 8 units, and the rest of working days it produces 9 units. After production is completed, each unit is thoroughly inspected. Each unit fails inspection with probability ??. If two or more units fail inspection on the same day, the factory closes for a week to re-calibrate equipment. Say the factory opens today after being closed for a week, write a mathematical expression to calculate the probability that the factory will remain open at least 30 days before closing again.
A product engineer has developed the following equation for the cost of a system component: C = (10P)2, where is the cost in dollars and Pis the probability that the component will operate as expected. The system is composed of two identical components, both of which must operate for the system to operate. The engineer can spend $173 for the two components. To the nearest two decimal places, what is the largest component probability that can be achieved?
Chapter 13 Solutions
EBK PRODUCTION AND OPERATIONS ANALYSIS
Ch. 13.1 - Prob. 3PCh. 13.1 - Prob. 4PCh. 13.1 - Prob. 5PCh. 13.1 - Prob. 6PCh. 13.2 - Prob. 7PCh. 13.2 - Prob. 9PCh. 13.3 - Prob. 13PCh. 13.3 - Prob. 14PCh. 13.4 - Prob. 15PCh. 13.4 - Prob. 16P
Ch. 13.4 - Prob. 17PCh. 13.4 - Prob. 18PCh. 13.4 - Prob. 19PCh. 13.4 - Prob. 20PCh. 13.6 - Prob. 21PCh. 13.6 - Prob. 22PCh. 13.6 - Prob. 23PCh. 13.6 - Prob. 24PCh. 13.6 - Prob. 25PCh. 13.7 - Prob. 26PCh. 13.7 - Prob. 27PCh. 13.7 - Prob. 28PCh. 13.7 - Prob. 30PCh. 13.7 - Prob. 31PCh. 13.7 - Prob. 32PCh. 13.7 - Prob. 33PCh. 13.7 - Prob. 34PCh. 13.8 - Prob. 35PCh. 13.8 - Prob. 36PCh. 13.8 - Prob. 37PCh. 13.8 - Prob. 38PCh. 13.8 - Prob. 39PCh. 13.8 - Prob. 40PCh. 13.8 - Prob. 41PCh. 13 - Prob. 42APCh. 13 - Prob. 43APCh. 13 - Prob. 44APCh. 13 - Prob. 45APCh. 13 - Prob. 46APCh. 13 - Prob. 48APCh. 13 - Prob. 49APCh. 13 - Prob. 51APCh. 13 - Prob. 52APCh. 13 - Prob. 53AP
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