ooo TFW 9:31 PM 74% K Back Homework 2.docx 3. (28 points) A factory consists of
ID: 3367332 • Letter: O
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ooo TFW 9:31 PM 74% K Back Homework 2.docx 3. (28 points) A factory consists of three workstations where each customer order must proceed through the workstations in sequence. Suppose that the workloads are balanced, which means that, on average, each order requires equal amounts of work at each of the three stations. Assume that the average operation time is 60 minutes. Now suppose that six orders will be scheduled tomorrow. Using this average time for each order at each station, we would expect that completion times would look like the following table, which is an ideal scenario with an expected schedule length of 480 minutes Order 1Order 2Order 3 Order 4Order 5 Order 6 Station 1 60 120 180 240 300 360 Station 2 120 180 240 300 360 420 Station 3 180 240 300 360 420 480 However, the actual operation times are random and follow a triangular distribution with a minimum of 30 minutes, most likely of 60 minutes, and maximum of 90 minutes. Build a simulation model with the forecast set as the time at which all work is completed (schedule length). To make this work efficiently, build two tables. One table will contain the minutes of work for each Order at each Station, with the triangular probability distributions set for each cell in the table. For example the corresponding table for the example provided above is that each job takes exactly 60 minutes, therefore the first table would contain values of 60 in each cell. The second table will utilize the values from the first table to track the time at which work is completed for each order at each station (this is the table shown above). Note that for Order 1, the work proceeds directly from Station 1 to Station 2 to Station 3 with no potential for delays. For Station 1, the next order can start immediately after the work on the current order is completed, again no potential for delays. However, starting with Order 2 at Station 2, work can only beginExplanation / Answer
we generate 18 random values form a triangle distribution using
rtriangle(18, a=30, b=90, c=60)
command in R
table 1
for 2nd row we take the max of time taken from s2 o1 to d2 o2 and s1 o2 to s2 o2
similar comparisons are made for other cells as well
so, the wait process time is easily greater than 480 mins.
O1 O2 O3 O4 O5 O6 S1 69.19273 72.8228 56.08978 58.40369 73.30195 73.51006 S2 53.22114 86.0842 48.85015 70.43972 57.52364 48.20782 S3 57.54573 72.96642 61.7409 76.45202 43.78542 71.01743Related Questions
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