Littlefield Case Essay

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Littlefield Case Capacity Assessment: how many machines to buy at each stage Based on the data for 50 days of simulated operations following observations were made: Average Demand: 12.22 orders per day, standard deviation: 3.55 orders per day Average Jobs completed: 11.875 orders per day Average Utilization of machine 1: 0.896 Average Utilization of machine 2: 0.93 Average Utilization of machine 3: 0.898 Average Capacity = Average Jobs completed / (Average Utilization * # of machines) Therefore, average Capacity of machine 1: 4.42 orders per day Average Capacity of machine 2: 12.77 orders per day Average Capacity of machine 3: 13.22 orders per day CVa = Std. Dev of Demand / Mean Demand = 0.29 CVp is assumed to be zero Therefore, average waiting time in the queue turned out to be 0.6 hours, 0.68 hours, and 0.66 hours respectively at stage 1, stage 2 and stage 3. Considering these waiting times, the new throughput rate at each stage came to be 11.93, 9.37 and 9.71 orders per day respectively at stage 1, stage 2 and stage 3. If one more machine is bought at each stage the capacity will increase and the average waiting time in the queue would get reduced. The new throughput rate (including the average queue waiting time) at each stage came to be 16.37, 21.78 and 22.55 orders per day respectively at stage 1, stage 2 and stage 3. Since now we could effectively meet the demand of 12.22 orders per day, we bought one machine at each stage. Since the average utilization of machine 2 was highest, the attempt was made to buy the machine 2 first. Lot size at machine 1: We know that @ Batch size = 60, Capacity of M/C 1 = 4.42 orders per day 4.42 orders per day per machine means total processing time of 326 minutes per order Setup time = 84 minutes, Actual process time = 242 minutes for 60 units Capacity = Batch Size / (Setup time +

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