SMA 6304 / MIT 2.853 / 2.854
Fall 2004
Solutions to Homework #4
(Totally 100 points)
[A - A Project with Random Task Duration] (50 points)
Q1. (15%)
a. Model description
Create work flow according to the given conditions. Here we make several assumptions
- Exponential time distribution for all tasks
- When we process multiple tasks simultaneously, the working force is unlimited
which means the average durations of tasks are independent to each other.
- All tasks will be processed immediately if all preceding tasks are completed, or no
preceding task is required. In this sense, task A, D and E will start simultaneously.
b. Forecast expression
In this model, we need to forecast the total time to complete the final report. Based on
above assumptions, it can be expressed in the formula
Dtotal = max( DA + DB + DC , max( DD , DE ) + DF ) + DG
Q2. (15%)
a. Experiment
We build the Crystal Ball model and set up the time distribution for each task cell. Notice
that the rate is the inverse of average duration. Define forecast cell use formula in
question 1.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
Run the simulation with 10,000 trials and 98% precision. Set the filter from 20 days to
+infinity. The result shows that the probability of taking more than 20 days is 21.14%.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
b. Confidence interval
In order to get a 98% confidence interval of the probability (total duration > 20 days), we
first extract the forecast cell data and setup indicator cells to justify whether the event
occurs or not.
The logic of the indicator cell is
- If forecast value > 20, set indicator to 1
- Otherwise, set indicator to 0
Afterwards, we use formula to calculate the confidence interval of indicators.
Yn  z 1−/2
S 2 n
n
Finally, we get the 98% confidence interval of the probability
- Mean: 21.15%
- Lower bound: 20.2020%
- Upper bound: 22.1022%
Q3. (20%)
We are allowed to assign the intern to two tasks. In other words, we need to pick up two
tasks whose durations will be reduced by 1 day.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
a. Intuition
We first analyze the flow model intuitively. Before the merge point task G in the flow
model, there are two branches, A-B-C and (D,E)-F. Apparently, the upper branch’s total
average duration is 3 days longer than that of the lower branch. Therefore, it will better to
improve any machine in the upper branch than any one in the lower branch.
Among the upper branch, we pick up the task with largest average duration because since
all task durations are exponentially distributed, larger average duration reflects larger
average variations. In order to gain the most benefit, we assign the intern to task B.
When we consider the second task to assign, we still focus on the upper branch, including
the last task G. We found that the second largest average duration time is 3 days of task A
and G. Comparing these two options, the reduction on task G will make more impact on
the total duration. The reason is that, suppose at some probability, the total duration (not
total average duration) of lower branch is larger than that of the upper branch, shorter
duration of task G will reduce the total time while shorter duration of task A won’t make
any difference. Therefore, let’s put the second assignment at task G.
B. Experiment
By observing the sensitivity analysis, we can easily find out that the most two sensitive
tasks are B and G, which exactly meet our intuitive explanation.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
[B – BigBank ATM] (50 points)
Q1. (25%)
a. Model description
The BigBank case is presented in the figure. The routing out of the queue is a little bit
complicated since there is a jump rule for Cash only customers.
In Simul8 model, there are two options to represent the logic.
Option I: (BIGBANK_OPTION1.S8)
As the routing discipline is the key issue in the logic, we can logically separate customers
into two queues. Then, use the time stamp to identify its position in the physical queue, in
order to decide the routing-in sequence.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
Key points when building the model
Labels and Visual logic at entry point
1. Set Label ‘work type’ to differentiate ‘All operations’ or ‘Cash only’ customers
a. All operations customer: work type = 1
b. Cash only customer: work type = 2
2. After set the value of Label ‘work type’, use visual logic to set Label ‘work item
image’ according to the value of ‘work type’.
a. All operations customer: Green ball
b. Cash only customer: Red ball
3. Also use visual logic to set Label based distribution ‘process rate’ to identify the
right process distributions at ATMs
a. All operations customer: distribution AO, N(4,1)
b. Cash only customer: distribution CO, N(3, 0.75)
Arrival distribution
1. Create new time dependent distribution with 3 time slots
2. Label action to set ‘work type’ value according to probability profile
a. All operations: 30%, set value = 1
b. Cash only: 70%, set value = 2
All operations ATMs
1. Use label based distribution ‘process rate’ to set the distribution
2. Link routing-in to queues of all operations and cash only customers
3. Set routing-in principle as ‘oldest’, to sort the customer according to time stamp
Cash only ATMs
1. Use label based distribution ‘process rate’ to set the distribution
2. Link routing-in only to queue of cash only customers
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
Option II: (BIGBANK_OPTION2.S8)
If we choose to use one logic queue, then we need to use the ‘segregated results’ option
in the queue property and the ‘Batch by type’ option in the work centers’ routing-in
setup.
In addition to the key points in Option I, we need to setup the storage area and work
centers routing-in correctly.
Storage area property:
Select the segregate results option and choose the Label ‘work type’ to do
segregation.
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
Cash only ATMs：
In routing-in setup -> options, select ‘Batch by type’ and use Label ‘work type’ at
fixed value ‘2’. Then, the Cash Only ATMs will only route-in the Cash Only
customers.
Below is the simul8 model of Option II.
Q2, Q3. (10%)
Before we run the simulation, we should be aware of these points
Set clock
1. Unit: minute
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SMA 6304 / MIT 2.853 / 2.854
Fall 2004
2. Running time:
a. Start time: 11:30
b. Time in each day: 1:30
3. Warm up period: we don’t need to consider warm up so set it as 0
4. Data collection period: 90 min (from 11:30 – 1:00)
Set run and trial
1. Number of runs in a trial: 50
3. Confidence range: 95%
Results of trial run show below
Average waiting time of
Cash Only customer (min)
Average waiting time of
All Operations customer
(min)
Average customers
waiting in line
Option I
2.03 ± 0.43
Option II
2.01 ± 0.42
3.72 ± 0.61
3.63 ± 0.62
2.72 ± 0.6
2.68 ± 0.55
We can see that, since Option I and Option II are equivalent, the results of two models
are pretty close.
Q4. (15%)
By using duplication wizard, we add one ‘All Operations’ ATM. Keep other setup and
run the simulation. We get results in the table.
Average waiting time of
Cash Only customer (min)
Average waiting time of
All Operations customer
(min)
Average customers
waiting in line
Option I
0.54 ± 0.14
Option II
0.52 ± 0.13
1.09 ± 0.22
1.05 ± 0.23
0.76 ± 0.2
0.74 ± 0.18
The average waiting time for Cash Only customer is reduced by 73.4%, while the average
waiting time of All Operations customer is reduced by 70.7%.
We observe that the impact is not linear with the increase of capacity. As we only
increase 25% capacity when we add one ATM, we get more than 70% reduction on both
‘Cash only’ and ‘All operations’ average waiting time. The impact is significant.
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