Consider a process with three activities in sequence, each with one resource, each with a processing time of one hour per unit, and each with a coefficient of variation equal to one (i.e., moderate variability in processing times). The coefficient of variation of inter-arrival times to each step of the process is also one (i.e., moderate variability in inter-arrival times). If we operate the process as a push system, we can use the waiting time equation to compute the average WIP level that would result from any given release rate of jobs (we'll assume the release rate is strictly less than process capacity and therefore the release rate is the flow rate (R)). To do this, first note that the utilization of all stations is R R UTIL = = = R. capacity 1 We can compute the flow time (Tstation) at any of the three activities as the sum of waiting time (using the waiting time equation) and process time: Tstation = 1x UTIL 1-UTIL X 12 + 12 +1= 2 1- UTIL Using Little's Law, we can convert this into the WIP at each station: WIP station =R x Tstation = UTIL x T station = UTIL 1 - UTIL Finally, because the three stations are identical, the total WIP in the process is given by 3UTIL 3R 3UTIL WIP = 1 - UTIL 3R 1-R If, instead, we were to operate this same process as a CONWIP line, we can compute the flow rate that would result from any choice of WIP by using the practical worst-case formula WIP R= WIP + CWIP - 1 WIP WIP + 2 because we know that CWIP is three for a balanced three-station process. Question 3. Suppose in the push system we set the release rate of jobs to R = 0.9 jobs per hour. What is the resulting average WIP level? Question 4. Suppose in the CONWIP system we set WIP equal to the answer from Question 3. What is the resulting flow rate? What does this show us about the relative efficiency of the push and CONWIP systems