Question 1
What would be the bottleneck resource?
The bottleneck resource is identified below:
BOTTLENECK IDENTIFICATION | |||
Average Truck Delivery | 75 | bbls | |
Barrels of berries at peak harvest | 19000 | ||
Number of trucks | 253.3 | ||
Total Hours | 12 | Hours | |
Trucks per hour | 21.1 | ||
Minutes per truck arrival | 2.8 | Minutes | |
Barrels per minute | 26.4 | ||
26.4 barrels per minutes x 70% wet berry ratio | 18.5 | ||
26.4 barrels per minutes x 30% dry berry ratio | 7.9 | ||
Plant Capacity (wet berries ) | 600BBS/hour= | 10 | bbls/min |
Plant Capacity (Dry berries) | 1500bbs/hour= | 25 | |
Dry berries arrival | 7.92 | < 25 | No Bottleneck |
Wet berries arrival | 18.47 | > 10 | Bottleneck |
Total wet berries | 13300 | ||
Wet berries plant capacity | 600 | bbls/hour | |
Total time required in 1 day to complete 19000 barrels | 22.17 | hours |
Question 2
What would be the plant’s process capacity?
The process capacity for the shown below:
PLANT'S PROCESS CAPACITY | ||
RESOURCE POOL | EFFECTIVE CAPACITY,BBLS/HR | |
Dumping Capacity | ||
Kiwanee dumpers | 5 | |
Avg time to dump | 7.5 | minutes |
Avg weight in trucks | 75 | bbls |
Total dumping capacity | 3000 | |
Holding Capacity | ||
Total holding (1-16) | 4000 | |
Total Holding (17-24) | 2000 | |
Total Holding (25-27) | 1200 | |
DE stoning Capacity | ||
Total de-stoning capacity | 4500 | |
DE chaffing capacity | ||
Total de-chaffing capacity | 4500 | |
Separator Capacity | ||
No. of Separator units | 3 | |
Total separator capacity | 1200 | |
Bagging station capacity | ||
Maximum output each day | 8000 | bbls |
Total capacity | 667 | bbls/day |
Bulk bin loading capacity | ||
No. of loaders for bulk bin | 4 | |
Total capacity | 800 | bbls/day |
Bulk truck loading capacity | ||
No. of loaders of bulk trucks | 2 | |
Total capacity | 2000 | bbls/day |
TOTAL PLANT PROCESS CAPACITY | 23867 | bbls/day |
Question 3
Assuming that 12,600 wet berries arrive per day (over a 12 hour day from 7am to 7pm) – does the time at which the plant finishes processing the last wet berry change? Why?
If we assume that the total wet berries arriving per day is reduced to 12600 from 13300 as previously calculated, then based upon the plant capacity of 600 bbls per hour, the time when the last wet berry is processed is changed and it is decreased as seen below.
Total wet berries arriving per day (assumed) | 12600 | ||
Wet berries plant capacity | 600 | ||
Total time required in 1 day to complete 19000 barrels | 21.00 | hours |
The reason for this is that the bottleneck has been reduced from the process to a certain extent. It is the bottleneck of the process, which is causing delay in the processing of the wet berries at peak time period. The management needs to take steps to remove this bottleneck from the process for instance, the holding bin at level 1 (17-24) should be converted into the wet berries and should not include the dry berries. Another recommendation for the management is to invest over the temporary storage and in this manner to reduce the truck wait time by 0-20 minutes. The mangement needs to improve the scheduling of the process so that the bottleneck is reduced and utilization is increased for the complete process.
Escort Service
Question 1
On average, what fraction of the time is the escort busy? That is, if a student calls, what is the likelihood that the escort is currently occupied with another student?
Standard Deviation for Call Receiving | 15 | Minutes |
Standard Deviation for picking & dropping a student | 15 | Minutes |
Total Available Time | 1440 | Minutes |
Escort Busy Time | 15 | Minutes |
Maximum fraction of escort busy time of total time | 2.08% | |
Minimum fraction of escort busy time of total time | 0.00% | |
Average fraction of escort busy time of total time | 1.04% |
Question 2
How much time does it take, on average, from the moment a student calls for an escort to the moment the student arrives at her/his destination?
Escort Busy Time | 15 | Minutes |
Average time to pick & drop a student | 15 | Minutes |
Standard deviation for picking & dropping a student | 15 | Minutes |
Maximum average time from student calling & arriving at his destination | 45 | Minutes |
Minimum average time from student calling & arriving at his destination | 15 | Minutes |
Average time from student calling & arriving at his destination | 30 | Minutes |
Question 3
Suppose that the escort service purchases a scooter that the escort can use to pick-up and transport students. This reduces the average amount of time it takes for him to pick up and transport a student to 5 minutes, although the coefficient of variation of this time remains 1. Under this new system, how much time does it take, on average, from the moment a student calls for an escort to the moment the student arrives at her/his destination?
Operations Management Home Work Case Solution
Escort Busy Time | 5 | Minutes |
Average time to pick & drop a student | 5 | Minutes |
Standard deviation for picking & dropping a student | 5 | Minutes |
Maximum average time from student calling & arriving at his destination | 15 | Minutes |
Minimum average time from student calling & arriving at his destination | 0 | Minutes |
Average time from student calling & arriving at his destination | 15 | Minutes |
Operations in Workplace
Question 1
Briefly describe your current job or most recent job/position – what are your responsibilities and how would you describe your contribution in the ‘value added transformations’ within your organization..............
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