Littlefield Lab We began our analysis by searching for bottlenecks that existed in the current system. It was easily identified that major issues existed in the ordering process. Without calculations, you could tell the reorder point was too low since the historical plots showed inventory levels at zero for two or more days at a time. The number of jobs in customer orders showed correlating spikes at the same time of the inventory outages. We reviewed the utilization and queues of the other stations in the system but were hesitant to make in immediate changes since we were not entirely certain the effects of correcting the inventory policy. To correct the inventory policy, we want to find the optimal ordering quantity based on the calculation EOQ=
√2 DS H
. Demand was calculated by taking the average
number of customer orders per day over the first 50 days. We came up with a figure near 12.24 orders per day which we multiplied by 268 days for the entire simulation. Setup cost was provided for us at $1,000 per order and holding cost was generated by multiplying the cost per unit by the interest rate which gave us a yield of 60. Based on this information, EOQ was 331 units after rounding. From the history, this was the second change we made to the inventory policy, up from 299. This was a result of a discussion in demand, where we had calculated demand based on 218 days for the simulation instead of the 268 days of the actual simulation. Next we wanted to fix the reorder point so we could eliminate inventory shortages and determined we would like to do so while maintaining a 95% service level. Since R=mean DDLT+z*std. dev. of DDLT, we needed to calculate Mean DDLT, z & std. deviation of demand. For mean DDLT, we used the 12.24 * 4 day lead time which gave us a rounded value of 49. Excel was utilized to figure z=1.64 for the service level of 95%. We found the std. dev. of DDLT by multiply the sq. rt. of the lead time by the std. deviation of demand which gave us a rounded figure of 61. After addressing the issues with the inventory policy, we decided to wait to see how the effects of removing that particular bottleneck would affect the system. The data on utilization and time for a sample to make it through a station was difficult as a result of the frequent large number of customer orders being released after inventory came in. FRANK’S PARAGRAPH
At station two, we decided not to change the ordering policy from FIFO. Writing a case analysis on the Manzana Insurance reading, we recalled the significant struggles and bottleneck caused by prioritizing one type of policy over the other. Our primary concern was to reduce the overall lead time of the system and
prioritizing station two or four over the other was going to increase the time at that station causing lower efficiencies. While the utilization of the machines would be high, the time to process units would ultimately be higher and the affects would be seen at the overall lead time. With regard to choosing the contracts, we initially remained at contract one. The lead times at that point were still far too high to switch to the other contracts and we would the revenue would likely be zero since the current system was not producing a lead time less than four days and the maximum allowed on contract two was three days. As improvements were made to the system, the lead time dropped drastically in a very short period of time. As we approached to having a lead time of close to one day we switched to contract two. The change was not made as we approached a lead time of three days since the penalties were so extreme over the span from one to two days, our revenues would have been far less than our performance under contract three. This same logic was applied on the transition from contract two to contract three which only happened seven days apart due to the extreme improvements that our change in machinery made to the system. Since the lab shuts down on day 268, we wanted to modify our last order to leave as little inventory as possible yet still meet demand. We recalculated the average arrival rate over the entire simulation to be approximately 11.8 orders per day, which we rounded to 12 for a small buffer. Based on this daily consumption, we found what day the next reorder point would be which was day 222. We wanted to find the demand for the final 46 days with an average of 12 orders per day came up with a figure of 552 units needed to meet demand over the last 46 days. Since we already had 61 units at the time of ordering, we subtracted that figure from the 552 and came up with a final order value of 491. This value should allow us to have a low inventory level at the end of the simulation. To achieve this, we had to change the reorder point to zero. This strategies hinges heavily on the lab not being in existence past day 268, if that were not the case, we would not recommend this plan.
How to reach our optimal decisions and what is the operations statistics after optimization? 1 )
Rough approximation of the machine service rate μ based on the statistics from first 5 days: Station 1 2 3 12.2 12.2 12.2 Arrival Rate = Th = λ (jobs per day) 4 4 4 Machines 3 1 1 Utilization of station 90% 94% 70%
Machine service rate μ = λ/ (m*Utilization) (jobs per day)
4.53 3
13.0 21
17.4 86
2 )
Approximate lead time improvements in optimized plan (via lowest M increase for max Lq improvement) Original Plan Optimized Plan Station 1 2 3 1 2 3 12.2 12.2 12.2 12.2 Arrival Rate = Th = λ (jobs per day) 4 4 4 4 12.24 12.24 Machine service rate μ (jobs per day), 4.53 13.0 17.4 4.53 13.02 17.48 from step 1 3 21 86 3 1 6 Machines 3 1 1 5 2 2 7.3 18.0 1.63 0.1 Lq (as per Cheat-cheat table) 54 50 0 98 0.277 0.098 2.70 0.94 0.70 2.70 λ/μ 0 0 0 0 0.940 0.700 0.60 1.47 0.13 0.01 Wait time = Wq = Lq/λ (days) 1 5 3 6 0.023 0.008 0.22 0.07 0.05 0.22 Service time = 1/μ (days per job) 1 7 7 1 0.077 0.057 0.82 1.55 0.19 0.23 Ws = Wq + 1/μ (days) 1 1 0 7 0.099 0.065 Lead Time = Total Ws = Ws1+2*Ws2+Ws3 (days) 4.115 0.501
3 )
Post change verification of the Lead Time for our contract 3 Station 1 12.2 Arrival Rate = Th = λ (jobs per day) 4 Machines 5 Utilization of station 50% Machine service rate μ = λ/ (m*Utilization) 4.89 (jobs per day) 6 2.50 λ/μ 0 0.13 Lq (as per Cheat-cheat table) 0 0.01 Wait time = Wq = Lq/λ (days) 1 0.20 Service time = 1/μ (days per job) 4 0.21 Ws = Wq + 1/μ (days) 5 Lead Time = Total Ws = Ws1+2*Ws2+Ws3 (days)
optimal configurations for fulfilling 2 12.2 4 2 45%
3 12.2 4 2 45%
13.6 00 0.90 0 0.22 9 0.01 9 0.07 4 0.09 2
13.6 00 0.90 0 0.22 9 0.01 9 0.07 4 0.09 2
0.492
