A process has two steps A and B. Two identical machines are operating in Step A
ID: 326962 • Letter: A
Question
A process has two steps A and B. Two identical machines are operating in Step A (A1 and A2), and one machine in Step B. Cycle times for A1, A2, and B are 17 hours. If the average time that an item spends in system is 6 hours, how many work-in-process inventories are in the process on average?
Answer is 0.35 +/- 0.02 but I need help getting there. I posted this the other day but it wasn't solved correctly to give me the right answer. I'll give the full amount of points to whoever can answer this correctly for me.
Explanation / Answer
Little’s law
WIP = Throughput (flow rate)* Lead Time (flow time/average time spent in the system)
STEP 1
Calculate throughput
station1(units per hour)
station 2(units per hour)
capacity of each station
= number of resources/processing time
0.118
0.059
capacity of the system
= capacity of bottle neck
0.059
flow rate
=Min(Demand, Capacity)
0.059
Station 1 has 2 resources A1 and A2
Station 1, Capacity= 2/17 =0.118 units per hour
Station 2, Capacity= 1/17 =0.059 units per hour
Station 2 has the lowest capacity and is the bottleneck
The capacity of the system is determined by the capacity of the bottleneck
The capacity of system =0.059 units per hour
Throughput rate = Min(Demand, Capacity)
Demand is not given. so we take the capacity of the system as the throughput
Throughput / flow rate = 0.059 units per hour
STEP 2
Average time spent in the system(flow time or lead time) = 6 hrs
Throughput flow rate = 0.059 units per hour
Work in process = Throughput * Lead Time =0.059* 6 = 0.354 units
station1(units per hour)
station 2(units per hour)
capacity of each station
= number of resources/processing time
0.118
0.059
capacity of the system
= capacity of bottle neck
0.059
flow rate
=Min(Demand, Capacity)
0.059
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.