QUESTION 5 Suppose there are four steps in a process with steps A and B are done
ID: 385290 • Letter: Q
Question
QUESTION 5
Suppose there are four steps in a process with steps A and B are done in parallel. The output of A and B is assembled in step C followed by a quality check in step D. The capacities of steps A, B, C, and D are 20, 25, 14, and 45 units/day. What is the capacity of the process in units/hour?
QUESTION 6
Suppose there are four steps in a process with steps A and B are done in parallel. The output of A and B is assembled in step C followed by a quality check in step D. The activity times of steps A, B, C, and D are 9, 15, 18, and 10 minutes. What is the minimum flow time of the process?
QUESTION 7
Suppose there are four steps in a process with steps A and B are done in parallel. The output of A and B is assembled in step C followed by a quality check in step D. The activity times of steps A, B, C, and D are 10, 12, 14, and 15 minutes. Suppose Step E is added following step D with activity time of 12 minutes and it has one worker. Which of the following is true?
The capacity of the process will reduce since 12 additional minutes will be needed.
The minimum flow time will increase by 12 minutes.
The capacity of the process will increase since now there is an additional worker.
Both b. and c. are correct.
A.The capacity of the process will reduce since 12 additional minutes will be needed.
B.The minimum flow time will increase by 12 minutes.
C.The capacity of the process will increase since now there is an additional worker.
D.Both b. and c. are correct.
Explanation / Answer
(5)
A and B are parallel process, so, the capacity of A and B combined in 20+25 = 45 per day
D has the same capacity i.e. 45 per day
C is having the minimum capacity of 14 per day.
So, the capacity of the entire line = capacity of C = 14 per day.
(6)
Minimum flow time is the minimum time spent on a single unit from start to end.
There are two possibilities - Following path A-C-D or following B-C-D
Note that Time required in A < Time required in B
So, minimum flow time will be obtained in A-C-D = 9+18+10 = 37 min
(7)
Minimum flow time earlier was through A-C-D = 10+14+15 = 39 min
With the addition of E, this will definitely increase by 12 min.
In 60 min time span, A will produce 60/10 = 6 parts, B will produce 60/12 = 5 parts. So, combined capacity of A+B is 6+5=11 units per hour
Similarly, capacity of C = 60/14 = 4.3 per hour and capacity of D = 60/15 = 4 per hour
When E is added, its capacity is 60/12 = 5 per hour. So, still, the minimum capacity station is D with a capacity of 4 per hour. So, nothing will change the line as the capacity is concerned. The capacity of the line will still be 4 per hour with D being the bottleneck.
So, only (b) is correct.