Problem Set 3 Hints
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Rutgers University, Newark *
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Course
22:839:670
Subject
Industrial Engineering
Date
Dec 6, 2023
Type
docx
Pages
3
Uploaded by MajorTigerPerson969
Boston University
Metropolitan College
AD 690 Problem Set 3
Transportation Manager’s Dilemmas
Hints
Hints
for Question 1:
Set up a Problem Solution Table.
Below is a format that you can use.
Problem Solution Table
Box Trucks
Shipment size for each supplier to each store = X cartons
Number of shipments per year
= X cartons per year/X per truck = X per year
Annual shipping costs using box trucks
= X shipments/year x ? stores x ? suppliers x $? = $?
Average inventory at each store
= X/2 = X cartons
Annual inventory holding cost
= X cartons x 0.40 x 10 x 5 = $?
Total annual cost
= $? + $? = $?
Vans
Shipment size for each supplier to each store = X cartons
Number of shipments per year
= X cartons per year/X per van= X per year
Annual shipping costs using large trucks
= X shipments/year x ? stores x ?
suppliers x $? = $?
Average inventory at each store
=X/2 = X cartons
Annual inventory holding cost
= X cartons x 0.40 x 10 x 5 = $?
Total annual cost
= $? + $? = $?
The solution is to use ? for the deliveries. The difference is $? – $? = $? in favor of ?
.
Hint:
The SCM expects a written report outlining the logic of the recommendation. The
recommendation might suggest that potential savings is sufficient to make a change. There may
be other costs that have not been considered like training and operational aspects.
For
example, if you recommend vans, they are smaller and more compact than trucks, so it may be
safer for drivers during extreme weather conditions. However, remember that vans do not carry
as much load as box trucks, so the capacity of the vans is less.
Hints
for Question 2:
Note that the Poisson distribution is a model that describes the probability of obtaining X
successes of a Poisson process based on an expected value. The textbook suggests the following
model:
P
(
X
=
x
)=
λ
x
e
−
λ
x!
,x
=
0,1,2,...
The transportation manager’s situation per a Poisson Distribution (calculations) is as follows:
The expected value of X is: µ (mean) = λ = 2
The standard deviation of X is: σ = λ
½
= 2
½
=1.4
2σ = 2.8; µ ± 2σ=2 ± 2.8 or 0-4.8; This would include 95% of the possibilities. Therefore, you
should consider 0 through 5 in your analysis.
Boston University
Metropolitan College
Hints
for Question 2 (continued):
Set up a Problem Solution Table.
Below is a format that you can use. You will need to calculate
values for the empty cells (Columns 2, 4, 6, 8, 10).
ƛ = 2
Expected Fees based on Number of Trucks
No.
P(x≥)
P(x)
0
2
4
6
8
10
0
0.1353
0.1353
0
1
0.4060
0.2707
54.14
2
0.6767
0.2707
108.28
3
0.8571
0.1804
108.24
4
0.9473
0.0902
72.16
5
0.9834
0.0361
36.1
0.9834
$379.92
Storage Fee
$100.00
Truck Fee
$200.00
Hint:
Fees for storage are equal to P(x) times the number of trucks times $100.
Fees for trucks are equal to P(x) times the number of trucks times $200.
Hints
for Question 3:
The learning curve concept is interesting. It is important to note that the time to accomplish a
task or a series of tasks declines as more experience (learning) is gained. The concept suggests
that the required time decreases at a given rate as cumulative experience doubles. The
phenomenon is attributable to the fact that it often takes less time to accomplish tasks as the
workers gain more and more experience in performing their tasks. Less time means lower costs.
Hint:
You might consider using the following variable names and equations:
Y = ax
n
x = the unit number
Y = number of hours to achieve the x
th
unit
a = number of hours to achieve the first unit
n= log b/log 2, b = the learning rate (based on
doubling)
a = number of hours for the first day (trip) = 12 hours
n= log b/log 2, b = 0.85
Y(x) = (12 hours) * (x)
log 0.85/log2
; Log 0.85/log 2=
-?
Y(x) = 12 hours * x
-?
Hint:
Set up a Problem Solution Table.
Below is a format that you can use. You will need to
calculate values for the empty cells (Columns for Hours, Training Hours, Costs and Cost Sum).
.
Y(x)
x
x
n
Δ = Y(x) - 8
Hours
Δ * 20 Drivers =
Training Hours
Overtime rate = $40 x 1.5=$60
Costs = $60 * Training Hours
12.0
1
1
4.0
80.0
$4800.00
10.2
2
0.8
5
9.2
3
0.7
7
8.6
4
0.7
2
Boston University
Metropolitan College
8.2
5
0.6
8
7.9
6
0.6
6
Sum
Cost Sum = $?
Hints
for Question 4:
This seems to be an LP cost minimization problem that can e solved with Excel or Excel Solver.
The
Cost (min) objective function
is:
Z= 5x
1A
+ 5x
1B
+ 8x
1C
+ 9x
1D
+ 7x
2A
+ 11x
2B
+ 6x
2C
+ 11x
2D
+ 6x
3A
+ 6x
3B
+ 7x
3C
+ 8x
3D
+ 4x
4A
+ 5X
4B
+ 6x
4C
+ 12x
4D
,
Subject to the following
constraints
:
x
1A
+ x
1B
+ x
1C
+ x
1D
= 300
x
2A
+ x
2B
+ x
2C
+ x
2D
= 350
x
3A
+ x
3B
+ x
3C
+ x
3D
= 400
x
4A
+ x
4B
+ x
4C
+ x
4D
= 550
x
1A
+ x
2A
+ x
3A
+ x
4A
= 200
x
1B
+ x
2B
+ x
3B
+ x
4B
= 500
x
1C
+ x
2C
+ x
3C
+ x
4C
= 500
X
1D
+ x
2D
+ x
3D
+ x
4D
= 400
X
ij
=˃ 0 where i = 1, 2, 3 and 4 (suppliers) and j = A, B, C and D (customers)
All numbers are in tons.
There are 9 constraints and 16 variables. There may be more constraints that you may wish to add. From
a mathematical perspective, the author feels that there are 7 missing constriant equations. You may
decide not to add more constraints.The solution is tested per equation Z to find the minimum cost.
Determine the minimum cost and provide a matrix that displays your solution of how many units of
product should be transported from suppliers to customers.
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