Little Field
Initial Analysis
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.
Capacity Determination:
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
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.
Capacity Determination:
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