Application of Little’s Law:
The data for the first 50 days indicated that: No. of Jobs arrived till date = 153. No. of Completed Jobs = 152 Since there were no jobs in the order queue, the inventory of the system, taken as a whole, was 1 i.e. 60 kits in total. This number included the kits waiting in the machine queues and jobs that are currently being processed on the machines. The average arrival rate of the jobs per day was 3.06. This was taken as the initial throughput for the system. Applying Little’s law,Flow time=InventoryThroughput ∴Flow time=0.3267 hrs.
This value matched with the average lead time of 0.39 hrs.
Initially we had one machine for every station and the throughput rate was assumed to be equal to the average job arrival rate per day i.e. 3.06 jobs/day. The average utilization for each machine was calculated from the given data. After knowing the average utilization, service rate (capacity) for each machine was calculated using the following formula: Service Rate capacity=Average ThroughputNo.of Machines*Average Utilization Average Utilization| 0.43188| 0.31582| 0.2037|
No. of Machines| 1| 1| 1|
Service Rate (Jobs/Day)| 8.335648791| 11.3989| 17.67305|
Machine 2 is getting the inputs from machines 1 & 3 and hence, assuming that the throughput rate of machine 2 is 3.06 jobs/day is not right. From the above table it can be seen that, the machine 1 is the bottleneck. Waiting Time in the Queues:
We assumed that the job arrival rate and the job processing rate followed a Memory less pattern and we used the PK formula to calculate the waiting times for the different queues. We made the following assumptions: * The average job arrival rate at station 1 is equal to the average job arrival rate in the system and the same holds for the standard deviation. This can serve as a good first approximation as station1 was the first activity to be done on a job. * Assuming steady state operations, the job arrival rates at the stations 2 & 3 were approximated as the average of the kits present in the respective station queues.
The waiting times were then deduced and the same are presented below. | Station 1 = Average Job arrival Rate/day| Station 2 =Avg Queue Length| Station 3 =Avg Queue Length| Average Arrival Rate| 3.060| 5.717| 0.673|
Standard Deviation| 2.045| 13.605| 1.296|
C.V.| 0.668| 2.380| 1.925|
No. Of Machines| 1.000| 1.000| 1.000|
Average Utilization (ρ)| 0.432| 0.316| 0.204|
Waiting Time (days)| 0.048| 0.078| 0.161|
From the demand data of the first 50 days we calculated the demand forecast by running a regression of jobs arrival rate v/s the no. of days. We kept improving the regression by including the actual demand as the situation evolved and we got more actual points. The demand forecast was really important as the machine purchase decision is based on that.
Break- Even Analysis:
The break even analysis for the different machines is shown below. The table represents the no. of units that have to be sold in the best case to just break even the cost of the machines. | Machine 1| Machine 2| Machine 3|
Contract Type| C1| C2| C1| C2| C1| C2|
Purchase Cost| 90000| 80000| 100000|
Max. Revenue/Job| 1000| 1500| 1000| 1500| 1000| 1500| Cost/Job| 600| 600| 600| 600| 600| 600|
Max Profit/Job| 400| 900| 400| 900| 400| 900|
Breakeven Units| 225| 100| 200| 89| 250| 112|
The above analysis was very important with respect to two aspects: * It gave a rough idea which machine will take longer to breakeven. * It also revealed that the profit on contract 2 was more than twice to that of on contract 1. So to maximize the total cash we should switch to contract 2. Strategy
Since our strategy was to switch to contract 2 as soon as possible, an average lead time of 0.37 days was not acceptable....