Rubric for Online Discussion and discussion 6 answers
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Purdue University *
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Course
57000
Subject
Industrial Engineering
Date
Feb 20, 2024
Type
docx
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5
Uploaded by AdmiralWolverine1375
Question 1
(10 points)
No initial posting exists to evaluate. 0 Points
Your final answer is incorrect but the work shows one or more steps that are correct. 1-6 Points
The final answer is incorrect but most of the work is correct. 7 Points
The final answer is incorrect with only a minor calculation error, or the final answer is correct but the work needs more detail. 8 Points
The final answer is correct and the work is complete. 10 points
Question 2 (10 points)
No initial posting exists to evaluate. 0 Points
Your final answer is incorrect but the work shows one or more steps that are correct. 1-6 Points
The final answer is incorrect but most of the work is correct. 7 Points
The final answer is incorrect with only a minor calculation error, or the final answer is correct but the work needs more detail. 8 Points
The final answer is correct and the work is complete. 10 points
Participation in Discussion (10 points)
No responses to other classmates were posted in this discussion forum.
0 Points
May include one or more of the following: 1-5 Points
Comments to only one other student's post.
Comments are not substantive, such as just one line or saying, “Good job” or “I agree.
Comments are off topic.
Comments to two or more classmates’ initial posts but only on one day of the week. Comments are substantive, meaning they reflect and expand on what the other student wrote.
6-7 Points
Comments to two or more classmates’ initial posts on more than one day. Comments are substantive, meaning they reflect and expand on what the other student wrote.
8-9 Points.
Comments to two or more classmates’ initial posts and to the instructor's comment (if applicable) on two or more days. Responses demonstrate an analysis of peers’ comments,
building on previous posts. Comments extend and deepen meaningful conversation and may include a follow-up question.
10 Points
Example 14
Management has decided to consider a number of different scenarios in looking for a solution to their inventory problem. These scenarios are summarized by the rows in the following table.
Order Size Reorder Point
35 25
45 25
55 25
65 25
75 25
For each scenario in the previous table, run 100 simulations and report the overall mean service level and the estimated total cost. Does this new reorder point achieve the desired service level of .9? If so, which scenario would you recommend? Yes, the new reorder point achieves the desired service level of .9. I would recommend the second scenario with order size 45 and reorder of 25, service level is 0.96 with the lowest estimated total cost of
$1,399. As the order size increases, What happens to the ordering and holding costs? As the order size increases, the ordering cost decreases and the holding cost increases. Order Size
(35)
Reorder Point (20)
Mean Service
Level
Estimated Total cost
Ordering Cost
Holding Cost
35
25
0.95
$859 + $569.10 = $1,428.10
($100 per order)*(8.59 orders) = $859
(60 days)*($.50 per day per
computer) *(2.21 + 16.76 computers) = $569.10
45
25
0.96
$694 + $705 = $1,399
($100 per order)*(6.94 orders) = $694
(60 days)*($.50 per day per
computer) *(5.19 + 18.31 computers) = $705
55
25
0.96
$589 + $849 = $1,438
($100 per order)*(5.89 orders) = $589
(60 days)*($.50 per day per
computer) *(8.93 + 19.37 computers) = $849
65
25
0.97
$498 + $995.40 = $1,493.40
($100 per order)*(4.98 orders) = $498
(60 days)*($.50 per day per
computer) *(13.14 + 20.04 computers) = $995.40
75
25
0.98
$430 + $1,179 = $1,609
($100 per order)*(4.30 orders) = $430
(60 days)*($.50 per day per
computer) *(18.33 + 20.97 computers) = $1,179
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Example 11
Consider the bank process described in Example 11. Run 100 simulations and report the overall mean service levels and the cycle times in Line when there are one, two, and three tellers.
Interval (minutes) Frequency Relative frequency
1-2 15 19%
2-3 5 6%
3-4 25 31%
4-5 30 38%
5-6 5 6%
Step 1: I used the Bank 1_Variation 1 excel sheet. I added teller 3 to the work station and changed the working time to Dis(1). Step 2: Under buffers I added teller 3 under output. Added “interval” and relative frequency data under discrete distribution. Then clicked “Run Simulation”.
The overall mean cycle time of the Line plus the Tellers is .59 + 2.4 = 2.99. Observe that the overall mean service level is 1.00. Also notice that the overall mean inventory of Line is .30 objects. We see that the overall mean fraction time working of each teller (.72, .60, and .46 time
units). Finally, the overall mean final inventory of Served Customers, or the throughput, is 239.18.