Exam one for SCM

SCM 424 Answers to Practice Problems for Exam 1

1. Process Flow Diagram for Making a Shirt

2. 5 minutes/unit: If task A operated without interruption, its cycle time—the average time between successive units—would be 5 minutes. Therefore, the cycle time is 5 minutes/unit.
3. 12 minutes/unit: The entire process can run only as fast as its bottleneck. The bottleneck of the process is task D because it has the longest task time. Therefore, the cycle time for the entire process is 12 minutes/unit.
4. 31 minutes: We assume for a one-unit rush order that it will never be “blocked” as it moves from process to process. Because the unit can move through step A faster than through the parallel step B, the front will have to wait for its companion back coming out of step B, so the fastest time for a rush order is: 10 + 4 + 12 + 5 = 31 minutes
5. 40 units/day: The longest cycle time and bottleneck of this process is 12 minutes/unit at process D. Therefore, the daily capacity of this process is: (8 hr/day)(60 min/hr)/(12 min/unit) = 40 units/day.
6. 42%: Labor utilization is the time during which the employee is actively working on the task divided by the time that the employee is available. Because the process is running at bottleneck pacing, the cycle time for the entire process is 12 minutes. The employee at task A is available for 12 minutes but is actually working on the task for 5 minutes. Thus, labor utilization at task A is 5 minutes/12 minutes = approximately 42%.
7. Process Flow Diagram for Making a Shirt with additional worker

MTT = 31 minutes

A one-unit rush order still requires each of the four process steps. Although tasks D is duplicated, one unit still requires 12 minutes to be processed. We assume for a one-unit rush order that it will never be “blocked” as it moves from task to task. Because the unit can move through task A faster than through the parallel task B, the front will have to wait for its companion back coming out of task B, so the fastest time for a rush order is: 10 + 4 + 12 + 5 = 31 minutes.

8. 60%: The cycle time Task D is the average of two tasks, each running at 12 minutes/unit, or 6 minutes per unit. At full capacity, the entire system is running at bottleneck pace, 10 minutes/unit. Therefore, the capacity utilization for ironing is 6/10 = 60%. The ironing workers are busy 60% of the time and idle 40% of the time.
9. 48 units/day: At task D, two units can be processed in 12 minutes, so the average cycle time is 6 minutes. The bottleneck is task B at 10 minutes/unit. The daily capacity is: (8 hours/day × 60 minutes/hour)/(10 minutes/unit) = 48 units/day
10. 4 minutes/unit: The cycle time for each individual ironing worker is 12 minutes/unit. For three workers in parallel, thus for the entire ironing task, the cycle time is 12/3 = 4 minutes/unit
11. 10 minutes/unit: The system’s cycle time is 10 minutes/unit determined by task B which is the bottleneck.
12. 1 unit/minute or 6 units/hour: Capacity is the inverse of cycle time. Therefore, system’s capacity is 1/10 unit per minute, or 0.1 units per minute. It can also be expressed in units per hour: 0.1 units/minute = 6 units/hour.
13. 100%: Task B is the bottleneck process. If the process is operating at full capacity, capacity utilization at the bottleneck process is always 100%.
14. 33 batches/day: Batch process time for Step 1 is 30+2(10)=50minutes, for Step 2 it is 50+3(10)=80 minutes and for Step 3 it is 40+5(10)=90 minutes. Bottleneck is Step 3 for a batch size of 10 units. So cycle time of the entire process is 90 minutes. There are 480 minutes is an 8 hour work day. One batch takes 90 minutes so 480/90 = 5.33 batches can be processed in one day.
15. 220 mins: Throughput time is the total time a batch spends in the process. Adding up the batch process times for all 3 steps, it is 50+80+90=220 minutes
16. 6 minutes: 5 batches are made in a day. Cycle time is 90 minutes. The total time for 5 batches is 5*90=450 minutes. That means 480-450=30 minutes in a day is not utilized. Since there are 5 batches (cycles) 30/5 =6 minutes is not used in each cycle.
17. 3 applications/hour: Because federal filing has two employees, their equivalent time per applicant is 15 minutes, so their capacity is 4 applicants per hour. The state filing process is the bottleneck because it has a capacity of only three applicants per hour. Thus, the entire process has a capacity of 3 applicants/hour
18. 1 application/hour: Adding a second person to the state filing department reduces the equivalent time per applicant to 10 minutes, resulting in a capacity for that department of 6 applications per hour. Doing so, however, shifts the bottleneck department to a tie between the federal filing and the billing and delivery departments, which have a capacity of 4 applicants each. Thus, adding the additional person to state filing increases the capacity of the process by 1 applicant per hour.
19. 75 minutes: Throughput time is the total time a application spends in the process. Adding up the process times for all 4 steps, it is 10+20+30+15=75 minutes
20. 25 sandwiches/hour: If we consider the “load ingredients” (Step 1) and “slice and box” (Step 3) tasks are carried out by Zeynep and Bobby respectively, then Zeynep’s cycle time = 2 minutes/ sandwich, so she has a capacity of 30 sandwiches /hour. Bobby can slice and box 60 sandwiches /hour. The maximum output of the oven is 5 sandwiches in 12 minutes; per hour, it is (60/12 minutes) × 5 sandwiches = 25 sandwiches/hour. The process is thus constrained by the oven so the process capacity = 25 sandwiches/hour.
21. 83%: Time for 5 sandwiches in Step 1 is 5*2min = 10 minutes. Cycle time due to bottleneck step (oven) is 12 minutes. Therefore only 10 minutes out of 12 is utilized at Step 1 giving a utilization of  (10/12)*100 =83%
22. 15 mins: 2+12+1 =15 minutes
23. All measures will be affected by the change in the cycle time of the bottleneck. If cycle time of the bottleneck decreases (increases), process capacity will increase (decrease), process cycle time will decrease (increase), and process throughput time will decrease (increase)
24. 4:59 p.m.: The 60 people in line will take 30 minutes to process (60 people × ½ minute/person), and then you’ll need another 30 seconds to buy your ticket, so from the time you arrive until the time you enter, it will take 30.5 minutes. Thus, you should arrive by 4:59 p.m.
25. Yes it is possible. There is a tradeoff between the setup time and the unit process time*batch size. Remember the Stonehaven Case.