A cost of quality model proposed by Juran & Gryna defined the expected costs of

Question

A cost of quality model proposed by Juran & Gryna defined the expected costs of three different product evaluation options:

Expected cost of no Inspection: NpA

Expected cost of sampling: nI + (N - n)pAPa + (N - n)(1- Pa)I

(N - n)(1- Pa)I ---> Cost of examining every item if the lot is rejected (100% sorting, no rework).

Expected cost of 100% Inspection: NI

Where:

N =         number of items in lot

n =          number of items in sample

p =          proportion defective in lot

A =         damage cost incurred if defective slips through inspection

I =           inspection cost per unit

Pa =         probability the lot will be accepted (use binomial distribution to estimate)

Use the above model to estimate the costs associated with assessing a production lot containing 150 golf balls. If acceptance sampling is to be used, the engineer wants to employ a (20, 1) sampling plan. Marketing estimates if a bad golf ball slips through to the customer it will cost the company $0.25. Inspectors can check 500 golf balls per hour, and their average hourly wage (including benefits) is $18.00. The proportion defective has historically been around 5%.

A) What are the expected costs of the no inspection, sampling, and 100% inspection alternatives?

B) If a bad golf ball slipping through to the customer actually costs the company $0.50, and the acceptance number is tightened to 0 (keeping all other parameters the same), what are the expected costs of the three alternatives?

C) Answer the same question assuming the proportion defective jumps to 10%, but the acceptance number is loosened to 2 (but still assume a $0.50 damage cost).

Explanation / Answer

N =         number of items in lot =150

n =          number of items in sample = 20

p =          proportion defective in lot = .05

A =         damage cost incurred if defective slips through inspection =$ 0.25

I =           inspection cost per unit =$0.036

Pa =         probability the lot will be accepted (use binomial distribution to estimate) = 0.3774

A. Expected cost of no Inspection: NpA = 150*.05*.25 = 1.875

Expected cost of sampling: nI + (N - n)pAPa + (N - n)(1- Pa)I = 0.036*20 + (150-20)*.05*.25*0.3774 +(150-20)(1-0.3774)*.036 =.72+ 130*.05*.25*.3774+130*0.6226*0.036 =0.72+0.61+2.91 = 4.247

Expected cost of 100% Inspection: NI =150*0.036 = 5.4

B. Changing A=$ 0.50 and Pa = .5199

Expected cost of no Inspection: NpA = 150*.05*.5 = 3.75

Expected cost of sampling: nI + (N - n)pAPa + (N - n)(1- Pa)I = 0.036*20 + (150-20)*.05*.5*0.5199 +(150-20)(1-0.5199)*.036 =.72+ 130*.05*.5*.5199+130*0.4880*0.036 =0.72+1.69+2.283 = 4.693

Expected cost of 100% Inspection: NI =150*0.036 = 5.4

C. Changing p=0.1   A=$ 0.50 and Pa = 0.0300     

Expected cost of no Inspection: NpA = 150*.1*.5 = 7.5

Expected cost of sampling: nI + (N - n)pAPa + (N - n)(1- Pa)I = 0.036*20 + (150-20)*.1*.5*0.0300 +(150-20)(1-0.0300)*.036 =.72+ 130*.1*.5*.0300+130*0.9699*0.036 =0.72+0.195+4.53 = 5.454

Expected cost of 100% Inspection: NI =150*0.036 = 5.4

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