bits 1stsem08 09 comprehensive exam

8/9/2019 BITS 1stSem08 09 Comprehensive Exam

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Question 1 (a)(i) Pareto Analysis reveals that the following are the important issues:

Pizza is late

Topping stuck to box lid

Wrong topping or combination


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Question 1 (a)

(ii) C&E Diagram should be used in the analysis stage after data analysis. Here, Pareto

Analysis has been done to identify the top 3 problems which contribute to nearly

80% of the problems. The C&E diagram is drawn thus.

Topping stuck to box lid Pizza is late

Wrong topping or combination

Pizza late delivery

OR

Wrong pizza delivery


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Question 1 (b)

Lot size = 1000; n=100; c=2

AQL = 0.01 ; Producers risk==0.05

LTPD=0.05; Consumers risk==0.10

We accept batches that are 2% nonconforming. The Probability of accepting a batch which is 2% (p=0.02) nonconforming is

Pa=0.6766. Producers risk here stands at 0.323324. At p=0.03, lot is rejected. Consumers risk (of accepting) is 0.42319.

AQL = 0.01 implies the lot is accepted/considered satisfactory upto p=0.01. Pa=.919699 which means producers risk is 1-

.919699= 0.080301 (=0.05 which is lesser than obtained value this implies producers risk is higher than specified)

LTPD=0.05 implies for p=0.05, the lot is rejected and Pa=.124652. The consumers risk (of accepting this bad lot) is .124652

(=0.10 which is lesser than obtained value which implies consumers risk is also higher)


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Question 2(b)

(i) Main problems are (accounting for 80% of issues)

Missing items

Furniture rubbed together while on truck

Late arrival for pickup

Broken Glass

Nicks and scratches from rough handling


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Question 2(b)

From the bar chart also, Missing Items, Furniture rubbed together, Late arrival and

broken glass account for majority of complaints


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Question 2(b)

(ii) The C&E Diagram can be drawn similar to the

solution in Q1(a) (ii).

Only the main causes (identified after paretoanalysis) must be included in the C&E Diagram

Missing Items Furniture rubbed together

Late arrival

Complaints

Broken Glass


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Question 3(a)

March Data: 8 out of 100 = 0.08 which is within control limits


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Question 3(b) (i)

A2, D3 and D4 can be obtained from the table given below


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Question 3 (b) (ii)

Long term process capability is

determined by Pp and Ppk

Process capability is lesser than the

targeted Ppk of 1.33


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Question 4(b) (i)


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Question 4 (b) (ii)

Cp=(USL-LSL)/6sigma

Here Cp=(.650-.550)/6*.012 = 1.388889

This implies the sigma level is 4.16 (Cp*3) Hence the process is capable of producing the

bottle.

bits 1stsem08 09 comprehensive exam

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