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3. The operation manager of a body and paint shop has five cars to schedule for repair. He would like to minimize throughput
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Answer #1

(a) Job Sequence using Johnson Rule.

CAR

BODY WORK(HOURS)

PAINT(HOURS)

A

12

6

B

7

4

C

9

10

D

3

4

E

12

5

1. The smallest time is located in Car D i.e. 3 Hours. Since the work is in Body Work, schedule this as first job. Now eliminate Job D from further consideration.

D

2. Now the smallest time is located in Car B i.e. 4 Hours. Since it is in the Paint, schedule this as last job and now eliminate Job B from consideration.

D

B

3. Now the smallest time is located in Car E i.e. 5 Hours. Since it is in the Paint, schedule this as second last job and now eliminate Job E from consideration.

D

E

B

4. Now the smallest time is located in Car A i.e. 6 Hours. Since it is in the Paint, schedule this as third car and now eliminate Job A from consideration.

D

A

E

B

5. Now the smallest time is located in Car C i.e. 9 Hours. This is the last car.

D

C

A

E

B

Thus sequence is D-C-A-E-B

(b) Idle time related to painting

Total hours in Painting = D(4)+C(10)+A(6)+E(5)+B(4)=29 Hours

Total hours Body work = A(12)+B(7)+C(9)+D(3)+E(12)=43 Hours

Idle time = 14 Hours

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