Question1 Consider the following serial process that produces shirts. All steps

Question

Question1 Consider the following serial process that produces shirts. All steps (A, B, C, D, and E) are necessary to create each finished shirt. Each step employs a single worker. Task times are shown on each step 4 min Stations A: Prepare Front B: Prepare Back C: Stitch Front & Back D: Iron E: Pack 8 min 6 min 5 min 3 min Note: A and B are done simultaneously. The front and back portion of the shirt arrive at A and B respectively. After preparation they are assembled in C before going on to D and E. Times shown at each station are per unit. a. What is the throughput time of the very first unit? b. What is the cycle time achieved by this process? What is the bottleneck operation? Working eight hours a day what is the daily capacity of the process? d. c. What is the capacity utilization of each operation in the above process? e. What is the average capacity utilization of the above process? Question 2 For this question consider hiring another worker and duplicating Operation C as shown in the diagranm below. The two workers work on successive products; they don't work on the same product at the same time. The product still requires the five Steps (A, B, C, D, E) 4 min 8 min 6 min 5 min 3 min 8 min is the throughput t very first unit? b. Working 8 hours a day, what is the daily capacity of the process? What is the cycle time? Where is the bottleneck now?

Explanation / Answer

1.(a) A and B occur together, hence their times will overlap. The throughput time of the first unit is sum of A, C,D and E, which is 23.

(b) Cycle time is the time between two successive output units.which is 8 minutes. The bottleneck operation is C with lowest output of 8 minutes taken per unit.

(c) Capacity of the process is equal to the slowest process which is C.

Output of C = 60/8 units /hour = 7.5

Capacity per 8 hours = 60

(d) Capacity of B = 60/3 = 20 units /hour. Utilisation =7.5/20 = 37.5%

Capacity of A = 60/4 =15 units /hour. Utilisation = 7.5/15 = 50%

Capacity of C = 60/8 =7.5 units / hour Utilisation = 7.5/7.5 = 100%

Capacity of D = 60/6 =10 units hour Utilisation = 7.5 /10 = 75%

Capacity of E = 60/5 =12 units /hour. Utilisation = 7.5/12 = 62.5%

Avg capacity utilisation = 4+8+6+5 /8+8+8+8 = 23/32 = 71.87

Q2 -

(a)Throughput time of the first unit is 4+8+6+5 =23

(b) Daily capacity of the process is equal to the bottleneck capacity which is now D with 6 min per unit or 10 units per hour. Cycle time is 6 min.

Capacity per 8 hours = 8x10 =80

(c) If one person does all operations, the time needed to produce one unit is

Sum of time taken in A, B,C,D and E = 4+3+8+6+5 =26 min.

(d) Capacity of A = 60/4 =15 units Utilisation = 10/15 = 66.66%

Capacity of B = 60/3 =20 units Utilisation = 10/20 =50%

Capacity of C = 60/4 = 15 units Utilisation = 10/15 =66.66%

Capacity of D = 60/6 =10 units Utilisation = 10/10 =100%

Capacity of E = 60/5 =12 units Utilisation = 10/12 = 83.3%

(e) Overall capacity utilisation

= 4+4+6+5 /6+6+6+6 = 19/24 = 79.16%

3. Combining A and B as one unit will make the process longer as combined operation of A and B will take 8 min.

In this, new process (A+B) will become a bottleneck now, with output of 60/8 = 7.5 units /hour

Cycle time = time between successive units will now be 8 min.

Capacity in 8 hours day will be 8x7.5 = 60 units

Capacity utilisation = 8+4+6+5 /8+8+8+8 = 23/32 = 71.87

So, the new change is not advisable.

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