A manager wants to assign tasks to workstations as efficiently as possible, and

Question

A manager wants to assign tasks to workstations as efficiently as possible, and achieve an hourly output of 4 units. The department uses a working time of 56 minutes per hour. Assign the tasks shown in the accompanying precedence diagram (times are in minutes) to workstations using the following rules: a. In order of most following tasks. Tiebreaker: greatest positional weight. Work Station D,B,C B,E b. In order of greatest positional weight. Work Station Tasks F, G E, H D, C, F c. What is the efficiency? (Round your answer to 2 decimal places. Omit the "No" sign in your response.) Efficiency 80.36 %

Explanation / Answer

Given,

Demand = 4 units / 1 hr day

Average throughput rate = 1/4 hrs = 15 mins

So, cycle rate = 15 mins

Minimum number of workstations= sum of total task times / cycle time

                        = (3+2+4+7+4+5+6+9+5)/15 = 45/15 = 3 workstations

Task Table

Task

Task time

Followers

Followers #

Precedence

A

3

B,C,H,I

4

-

B

2

C,H,I

3

A

C

4

H,I

2

B

D

7

E,H,I

3

-

E

4

H,I

2

D,G

F

5

G,E,H,I

4

-

G

6

E,H,I

3

F

H

9

I

1

E,C

i

5

-

0

H

The total station time is nothing but the cycle time. So, total station time of each station is 15 mins.

Workstation 1:

                        First task =F

                        Time left= 15-5 = 10 mins

                        Second task = A

                        Time left= 10-3 = 7 mins

                        Third task = D

                        Time left= 7-7 = 0 mins

                       

So, workstation 1: F->A->D

Workstation 2:

                        First task =G

                        Time left= 15-6 = 9 mins

                        Second task = B

                        Time left= 9-2 = 7 mins

                        Third task = C

                        Time left= 7-4 = 3 mins

                       

So, workstation 2: G->B->C

Workstation 3:

                        First task=E

                        Time left= 15-4= 11 mins

                        Second task=H

                        Time left=11-9 = 2 mins

So, workstation 3: E->H

No workstation is left for assignment of task I, if following the theoretical minimum rule.

Workstation 4 (going against the minimum workstation rule):

                        First task= I

                        Time left=15 – 5= 10 mins

                       

So, workstation 4: I

Work Station

Tasks

Workstation Time

Idle Time

1

F->A->D

15 mins

0 mins

2

G->B->C

12 mins

3 mins

3

E->H

13 mins

2 mins

4

I

10 mins

5 mins

B) In order of Greatest positional weight : H->D->G->F->I->E->C->A->B

Workstation 1:

                        First task = D

                        Time left= 15-7 = 8 mins

                        Second task = F

                        Time left= 8-5 = 3 mins

                       

                       

So, workstation 1: D->F

Workstation 2:

                        First task =G

                        Time left= 15-6 = 9 mins

                        Second task = E

                        Time left= 9-4= 5 mins

                        Third task = A

                        Time left = 5-3 =2 mins

                       

So, workstation 2: G->E-> A

Workstation 3:

                        First task=B

                        Time left= 15-2= 13 mins

                        Second task=C

                        Time left=13-4 = 9 mins

                        Third task = H

                        Time left = 9-9 =0 mins

So, workstation 3: B->C->H

No workstation is left for assignment of task I, if following the theoretical minimum rule.

Workstation 4 (going against the minimum workstation rule):

                        First task= I

                        Time left=15 – 5= 10 mins

                       

So, workstation 4: I

Work Station

Tasks

Workstation Time

Idle Time

1

D->F

12 mins

3 mins

2

G->E-> A

13 mins

2 mins

3

B->C->H

15 mins

0 mins

4

I

10 mins

5 mins

C) Efficiency with 4 workstations = (sum of all tasks) / (no of workstations *   Cycle time)

        = (45)/ (4*14) = 45/60 = 0.75 = 75%

Task

Task time

Followers

Followers #

Precedence

A

3

B,C,H,I

4

-

B

2

C,H,I

3

A

C

4

H,I

2

B

D

7

E,H,I

3

-

E

4

H,I

2

D,G

F

5

G,E,H,I

4

-

G

6

E,H,I

3

F

H

9

I

1

E,C

i

5

-

0

H

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