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1. Using the precedence diagram shown below and the task time described in table

ID: 367842 • Letter: 1

Question

1. Using the precedence diagram shown below and the task time described in table below for three products A, B and C respectively, answer the questions j through vi below Assume daily demands of 4515 and 30respectively. Workday is 480-minutes long Product Task 2.00 2.25 4.00 3.50 4.85 5.10 1.75 2.05 3.75 1.95 4.15 2.20 4.45 3.05 2.15 2.85 2.70 2.20 3.90 4.50 4.05 2.25 3.95 2.70 i What is the average CT for the three products. ii. Compute the weighted average task times for each task. iii. Using the longest operation time rule, assign the tasks and determine the minimum number of workstations required iv. Compute the efficiency for this line v.What is the potential maximum CT for this line? What daily output can be expected for each of the three products at this CT? To increase the output further, what efforts are needed? Discuss. vi. Precedence Diagram forproblem 1

Explanation / Answer

1.

(i)

Average Cycle Time for product A, CTA = Production Time per day / Required Output per day = 480 / 45 = 10.67 minutes.

Average Cycle Time for product B, CTB = Production Time per day / Required Output per day = 480 / 15 = 32 minutes.

Average Cycle Time for product C, CTC = Production Time per day / Required Output per day = 480 / 30 = 16 minutes.

(ii)

Proportion of product A, qA = Demand of product A / Total Demand = 45 / 90 = 0.5

Proportion of product B, qB = Demand of product B / Total Demand = 15 / 90 = 0.17

Proportion of product C, qC = Demand of product C / Total Demand = 30 / 90 = 0.33

Task

Product

Weighted Average

A

B

C

1

2.00

3.75

2.70

2.53

2

2.25

1.95

2.20

2.18

3

4.00

4.15

3.90

3.99

4

3.50

2.20

4.50

3.61

5

4.85

4.45

4.05

4.52

6

5.10

3.05

2.25

3.81

7

1.75

2.15

3.95

2.54

8

2.05

2.85

2.70

2.40

Total

25.59

(iii)

Average Cycle Time of all the products taken together = 480 / 90 = 5.33

Theoretical minimum number of workstations, Nt = 25.59 / 5.33 = 4.80 = 5 (approx.).

Table showing Number of Following Tasks:

Task

Number of following tasks

1, 3

4

2,4

3

5,6

2

7

1

8

0

Table showing assignment of tasks to work stations:

Station

Task

Time (in minutes)

Unassigned time (in minutes)

Feasible remaining work task(s) (for same station)

Task (s) with most followers

Task (s) with Longest Operation Time

Station 1

---

3

0

3.99

5.33

1.34

3

---

3

---

3

---

Station 2

---

1

2

0

2.53

2.18

5.33

2.80

0.62

1, 2

2

---

1

2

---

1

2

---

Station 3

---

5

0

4.52

5.33

0.81

5

---

5

---

5

---

Station 4

---

4

0

3.61

5.33

1.72

4

---

4

---

4

---

Station 5

---

6

0

3.81

5.33

1.52

6

---

6

---

6

---

Station 6

---

7

8

0

2.54

2.40

5.33

2.79

0.39

7, 8

8

---

7

8

---

7

8

---

(iv)

Actual number of workstations, Na = 6 (from preceding table).

Efficiency of the assembly line using 6 workstations = T / (Na x C) = 25.59 / (6 x 5.33) = 0.8002 = 80.02%.

(v)

Potential maximum CT = Sum of Task Times = 25.59 minutes.

Daily Output expected for each of the three products = 480 / 25.59 = 18.76 19 units.

(vi)

The output can be increased further by using task for one or two products at a time and not all the three products together.

Task

Product

Weighted Average

A

B

C

1

2.00

3.75

2.70

2.53

2

2.25

1.95

2.20

2.18

3

4.00

4.15

3.90

3.99

4

3.50

2.20

4.50

3.61

5

4.85

4.45

4.05

4.52

6

5.10

3.05

2.25

3.81

7

1.75

2.15

3.95

2.54

8

2.05

2.85

2.70

2.40

Total

25.59