Answered: a. The data from 10 samples with five…

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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11. Dave would prefer to order every week but needs to justify any change in order size. How much would ordering cost be reduced to justify on
every week?
111. What would be the weekly holding cist if Dave orders every two weeks?
iv. What would be the maximum inventory level if Dave orders every two weeks?

Transcribed Image Text:a. The data from 10 samples with five pieces in each sample in a rolling process showed that the parts had an average thickness of .009 inch, and an average range of .0035 inch. Determine the upper and lower control limits that should be set for the mean and range charts. b. Dave's Auto Shop services and repairs a particular brand of foreign automobile. Dave uses oil filters throughout the year. The shop operates 52 weeks per year, and weekly demand is 200 filters. Dave estimates that it costs $50 to place an order and his annual holding cost rate is $5 per oil filter. Currently, Dave orders in quantities of 800 filters. What penalty cost is Dave's Auto Shop incurring by its current replenishment policy? 11. Dave would prefer to order every week but needs to justify any change in order sıze. How much would ordering cost be reduced to justify ordering

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Hello! As you have posted 2 different questions, we are answering the first question. In case you require the unanswered question also, kindly re-post them as separate question.

From the given information, there are 10 samples with five pieces.

The average thickness of 0.009 inch and an average range of 0.0035 inch.

From the X-bar and R chart table, for sample size, n=5,

A₂=0.577, D₃=0 and D₄=2.114.

The upper and lower control limits for the mean chart are calculated below:

$U C L = x - b a r + A_{2} \bar{R} = 0.009 + (0.577 * 0.0035) = 0.011$

$L C L = x - b a r - A_{2} \bar{R} = 0.009 - (0.577 * 0.0035) = 0.0070$

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