Quiz 2 Sample problems

Answers: 1. R2 is the slowest process at 50 units/minute. R2 limits the output. 2. Decoupling inventory builds between R1 and R2 at the rate of 10 units/minute. 3. TT = 1 minute /60 units + 1min /50 units + 1 minute/70 units = 0.051 minutes/unit Notice that TT is the sum of the cycle time. Output rate = 1/cycle time. If we had wait time, we would add it here. Technically, we have wait time at the decoupling inventory, but I asked you to ignore it for this question. 4. Based on the bottleneck speed, the output rate (processing rate) is 50 units/minute 5. The bottleneck, R2. 6. WIP = TT * PR = 0.051 m/u * 50 units/min = 2.55 units (always make sure the units are the same e.g., minutes and units for this example problem). 7. The setup time is part of the throughput time because we must wait for the machine to be set up prior to processing any units. a. TT = 1m/60 units + 1 min/50 units + 10 minutes + 1/70 units = 10.051 minutes/unit b. Output rate is not affected. The bottleneck remains unchanged. c. WIP is impacted since TT increased by 10 minutes. WIP = TT * PR = 10.051 min/unit x 50 units/min = 502.55 units d. Setup time is used to convert the resource from one product to another product. To reduce setup, we have the following choices: i. Eliminate unneeded steps. ii. Move some steps to prior to the process to setup (called external setup). iii. Reduce the time to complete one or more steps (e.g., become more efficient, called internal setup). iv. Improve maintenance and housekeeping (to enable quickly finding required tools). e. Q=1,000 units will minimize setup time to once = 1 setup Some resources (e.g., a machine) require time to convert the resource for the next type of product. For example, you have a setup for each case project. (So, three setups for the semester). If instead, you combined the EV and BD projects on the same case, you’d have one setup instead of two setups. f. The number of setups is 1000 units/100 batch size= 10 setups Setup time = 10 setups * 10 minutes = 100 minutes 8. With a batch size of 100, the throughput time will increase because we need to process all 100 units before moving the batch to the next resource. We use 1/output rate = cycle time for each resource. a. TT = 100 * (1/60) + 100 * (1/50) + 100 * (1/70) = 100*0.051 = 5.1 minutes/batch (or minutes/unit). In this case, it takes one unit the same amount of time as one batch. b. Cycle stock = Q/2 or 100/2 = 50 units on average. If you’d like to reflect on the process, consider this revised flow diagram, where C = cycle stock; D = decoupling inventory. C  R1  C  D  C  R2  C  C  R3  C  output. For 8a, I asked you to assume no wait time (e.g., decoupling inventory) 9. Just add the 10 minutes to the answer in 8a = 5.1 minutes + 10 minutes = 15.1 min/batch