Practical Management Science
6th Edition
ISBN: 9781337406659
Author: WINSTON, Wayne L.
Publisher: Cengage,
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Chapter 6.5, Problem 30P
Summary Introduction
To determine: The optimal solution using solver.
Introduction: The variation between the present value of the
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99. A plastics manufacturer has 1,200 boxes of transparent wrap in stock at one factory
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Retailer 1
Retailer 3
Factory 1
Factory 2
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13
13
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A retailer has 7 stores in different cities in northern Ontario. The weekly demand for each store follows the normal distribution with a mean value of 60 and a standard deviation of 10. The retailer is planning to use a single centralized store in northern Ontario and they think the demand in the single store will be equal to the sum of the demand in all 7 cities. It takes 3 weeks for the supplier to send the products to the retailer. Calculate the safety stock for the centralized and decentralized cases if the correlation of demands between different cities is 0.3. Assume the planned customer service level is 98%.
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Dr. Thompson makes three bean mixes for coffee shops located in the Austin. The three mixes, referred to as the Tameka Mix, the Christian Mix, and the Tamia Mix, are made by mixing different percentages of five types of beans. In preparation for the fall season, Dr. Thompson just purchased the following shipments of beans at the prices shown:
Type of Beans
Shipment Amount in pounds
Cost Per Shipment in dollars
Arabica
6000
7500
Robusta
7500
7125
Liberica
7500
6750
Excelsa
6000
7200
Juno
7500
7875
The Tameka Mix consists of 15% Arabica, 25% Robusta beans, 25% Liberica, 10% Excelsa, and 25% Juno. The Christian Mix consists of 20% of each type of beans, and the Tamia Mix consists of 25% Arabica, 15% Robusta beans, 15% Liberica, 25% Excelsa, and 20% Juno.
Dr. Thompson analyzed the cost to get the beans ready and determined that the profit contribution per pound is $1.65 for the Tameka Mix, $2.00 for the Christian Mix, and $2.25 for the…
Chapter 6 Solutions
Practical Management Science
Ch. 6.3 - Prob. 1PCh. 6.3 - Prob. 2PCh. 6.3 - Solve Problem 1 with the extra assumption that the...Ch. 6.3 - Prob. 4PCh. 6.3 - Prob. 5PCh. 6.3 - Prob. 6PCh. 6.3 - Prob. 7PCh. 6.3 - Prob. 8PCh. 6.3 - Prob. 9PCh. 6.3 - Prob. 10P
Ch. 6.4 - Prob. 11PCh. 6.4 - Prob. 12PCh. 6.4 - Prob. 13PCh. 6.4 - Prob. 14PCh. 6.4 - Prob. 15PCh. 6.4 - Prob. 16PCh. 6.4 - Prob. 17PCh. 6.4 - Prob. 18PCh. 6.4 - Prob. 19PCh. 6.4 - Prob. 20PCh. 6.4 - Prob. 21PCh. 6.4 - Prob. 22PCh. 6.4 - Prob. 23PCh. 6.5 - Prob. 24PCh. 6.5 - Prob. 25PCh. 6.5 - Prob. 26PCh. 6.5 - Prob. 28PCh. 6.5 - Prob. 29PCh. 6.5 - Prob. 30PCh. 6.5 - In the optimal solution to the Green Grass...Ch. 6.5 - Prob. 32PCh. 6.5 - Prob. 33PCh. 6.5 - Prob. 34PCh. 6.5 - Prob. 35PCh. 6.6 - Prob. 36PCh. 6.6 - Prob. 37PCh. 6.6 - Prob. 38PCh. 6 - Prob. 39PCh. 6 - Prob. 40PCh. 6 - Prob. 41PCh. 6 - Prob. 42PCh. 6 - Prob. 43PCh. 6 - Prob. 44PCh. 6 - Prob. 45PCh. 6 - Prob. 46PCh. 6 - Prob. 47PCh. 6 - Prob. 48PCh. 6 - Prob. 49PCh. 6 - Prob. 50PCh. 6 - Prob. 51PCh. 6 - Prob. 52PCh. 6 - Prob. 53PCh. 6 - Prob. 54PCh. 6 - Prob. 55PCh. 6 - Prob. 56PCh. 6 - Prob. 57PCh. 6 - Prob. 58PCh. 6 - Prob. 59PCh. 6 - Prob. 60PCh. 6 - Prob. 61PCh. 6 - Prob. 62PCh. 6 - Prob. 63PCh. 6 - Prob. 64PCh. 6 - Prob. 65PCh. 6 - Prob. 66PCh. 6 - Prob. 67PCh. 6 - Prob. 68PCh. 6 - Prob. 69PCh. 6 - Prob. 70PCh. 6 - Prob. 71PCh. 6 - Prob. 72PCh. 6 - Prob. 73PCh. 6 - Prob. 74PCh. 6 - Prob. 75PCh. 6 - Prob. 76PCh. 6 - Prob. 77PCh. 6 - Prob. 78PCh. 6 - Prob. 79PCh. 6 - Prob. 80PCh. 6 - Prob. 81PCh. 6 - Prob. 82PCh. 6 - Prob. 83PCh. 6 - Prob. 84PCh. 6 - Prob. 85PCh. 6 - Prob. 86PCh. 6 - Prob. 87PCh. 6 - Prob. 88PCh. 6 - Prob. 89PCh. 6 - Prob. 90PCh. 6 - Prob. 91PCh. 6 - Prob. 92PCh. 6 - This problem is based on Motorolas online method...Ch. 6 - Prob. 94PCh. 6 - Prob. 95PCh. 6 - Prob. 96PCh. 6 - Prob. 97PCh. 6 - Prob. 98PCh. 6 - Prob. 99PCh. 6 - Prob. 100PCh. 6 - Prob. 1CCh. 6 - Prob. 2CCh. 6 - Prob. 3.1CCh. 6 - Prob. 3.2CCh. 6 - Prob. 3.3CCh. 6 - Prob. 3.4CCh. 6 - Prob. 3.5CCh. 6 - Prob. 3.6C
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