Tarlan Saderi
IE 416
HW #7
Page 403, #5:
Cars are produced at plants, then shipped to warehouses, and finally shipped to
customers. Detroit can produce 150 cars per week, and Atlanta can produce 100 cars per
week. Los Angeles requires 80 cars per week; Chicago, 70; and Philadelphia, 60. It costs
$10,000 to produce a car at each plant. And the cost of shipping a car between 2 cities is
given in the table below. Determine how to meet General Ford’s weekly demands at
minimum cost.
Plants: Detroit and Atlanta
Warehouses: Denver and New York
Customers: Los Angeles, Chicago, and Philadelphia
Plant
Detroit
Atlanta
Production
Capacity per week
150
100
Distribution
Center
Los Angeles
Chicago
Cost of shipping a car between two cities:
From
Denver
Detroit
1253
Atlanta
1398
From
Denver
New York
Los Angeles
1059
2786
Demand Forecast
Per week
80
70
New York
637
841
Chicago
996
802
It costs $10,000 to produce a car at each plant.
Philadelphia
1691
100
Output Data (Range of Optimality)
Output Data (Range of Feasibility)
Sensitivity Analysis:
Basic Variable: I choose the shipment cost from Atlanta to Denver, $1,398 units. My
motivation to choose Atlanta to Denver is because it has the highest value for the unit
cost of shipment. So it will have a higher impact on optimal solution when there are any
changes. If the unit cost changes in the range of $1253 and $1457, the number of unit to
shipment will be the same. But OF will change correspondingly, and it won’t be $338610
any more.
Performing my parametric analysis, I will choose a “starting u” of –145 to an “ending u”
of +145, and a “step of u” of 20. Therefore, my analysis will range from $1253 to $1543
per unit cost. This table shows the constant value for OF from the unit cost of 1473.
When unit cost is less than 1373, the OF will also decrease.
Non Basic:
I choose the shipment cost from Atlanta to NY since it has the lowest unit cost of $841.
Currently there is not any shipment from Atlanta to NY, if we want to ship any units we
need to reduce cost by at least RC, which is $59. That is a cost of shipment for less than
$782.
Performing my parametric analysis, I will choose a “starting u” of –60 to an “ending u”
of +60, and a “step of u” of 10. Therefore, my analysis will range from $781 to $901 per
unit cost. This table shows the constant value of $338610 for OF when the unit cost is
more than 791. Once the unit cost is less than 791, the OF will also decrease.
The following sensitivity analysis is by Ms. Robles:
Sensitivity analysis on increasing the supply of the Detroit plant
Because of the cost shipping for the Detroit plant is less than shipping from the Atlanta
plant, it is desirable to determine how the total cost would change if the available supply
from the Detroit plant is increased.
As the graph and table indicates, if the amount of the available supply from the Detroit
plant is increased, there is a decrease in total cost if at least 110 vehicles are shipped.
Report to the manager:
From
Detroit
Atlanta
Denver
To
Amount
Unit Cost ($)
Total Cost ($)
Denver
NY
Denver
LA
Chicago
Philadelphia
20
130
60
80
70
60
1,253
637
1,398
1,059
802
100
25,060
82,810
83,880
84,720
56,140
6,000
Total
338,610
The minimal transportation cost for General Ford is $33,610, and the number of
shipments should be as following:
20 units with the cost of $1,253 from Detroit (plant) to Denver (warehouse)
130 units with the cost of $637 from Detroit (plant) to NY (warehouse)
60 units with the cost of $1,398 from Atlanta (plant) to Denver (warehouse)
80 units with the cost of $1,059 from Denver (warehouse) to LA (customer)
70 units with the cost of $802 from NY (warehouse) to Chicago (customer)
60 units with the cost of $100 from NY (warehouse) to Philadelphia (customer)
The total costs for each shipment will be:
$25,060 to ship from Detroit to Denver
$82,810 to ship from Detroit NY
$83,880 to ship from Atlanta to Denver
$84,720 to ship from Denver to LA
$56,140 to ship from NY to Chicago
$6,000 to ship from NY to Philadelphia
We will have 40 units of unused supply from Atlanta, the recommendation is that
either we produce 60 units instead of 100 units, or find a new customer for an unused
supply.
The required capacities for our warehouses are: Denver 80 cars, New York 130
cars.
Add about your conclusions from sensitivity analysis.
Pg. 403 #6
A company must meet the following demands for cash at the beginning of each of
the next 6 months:
Month
1
2
3
4
5
6
Demand ($)
200
100
50
80
160
140
A. Assuming that all bills must be paid on time, formulate a balanced
transportation problem that can be used to minimize the cost of meeting the
cash demands for the next 6 months.
B. Assume that payment of bills can be made after they are due, but a penalty of
$0.05 per month is assessed for each dollar of cash demands that is postponed
for 1 month. Assuming all bills must be paid by the end of month 6, develop a
transshipment model that can be used to minimize the cost of paying the next
6 months’ bills.
(Hint: Transshipment points are needed in the form Ct = cash available at
beginning of month t after bonds for month t have been sold, but before month
t demand is met. Shipments in Ct occur from bond sales and Ct-1. Shipments
out of Ct occur to Ct+1 and demands for months 1, 2, …t).
Month of Sale ($)
Bond
1
2
3
1
2
3
4
5
6
0.21
0.50
1.00
0.19
0.50
1.00
0.17
0.50
1.00
0.13
0.33
1.00
0.09
0
1.00
0.05
0
0
The problem can be input as below as well (style 2):
Output (style 2):
Input (Style 1):
Output (Style 1):
Output Data (Range of Optimality)
Output Data (Range of Feasibility)
Sensitivity Analysis:
Basic Variable:
I choose Bond 3 to Month 2 because it has the highest number of shipment of 90 and it
also has the highest number for the penalty cost of $1. Therefore, any changes can have a
biggest change to OF which is minimizing the cost. If the penalty cost changes
somewhere between the range of $.98 and $1.02, the number of shipment will stay the
same but the OF will change from 166.2 correspondingly.
Performing my parametric analysis, I will choose a “starting u” of –$.5 to an “ending u”
of $.5, and a “step of u” of .10. Therefore, my analysis will range from $.5 to $1.05 per
unit cost. This table shows the constant value of $169.6 for OF when the penalty cost is
more than and equal $1.1. Once the penalty cost is less than $1, the OF will also
decrease.
Non Basic Variable:
I choose Bond 2 to Month 1 because it has the highest penalty cost of $.5 and smaller RC
of $.4. Currently, Bond 2 is not being used. If we want to use Bond 2, we need to go
under the penalty cost of $.1.
Performing my parametric analysis, I will choose a “starting u” of –$.3 to an “ending u”
of $.53and a “step of u” of .10. Therefore, my analysis will range from $.2 to $.8 per unit
cost. This table shows the constant value of $166.2 for OF.
The following sensitivity analysis is by Ms. Robles:
Perform Sensitivity Analysis to the amount of cash available at the beginning of month 1
Because there is no penalty incurred in using cash to satisfy the monthly demand, it is
desirable to determine how an increase in the available amount of cash changes the total
cost at the end of month 6. At can be see in the information provided, the more available
cash, the less the total cost incurred in satisfying the demand for all 6 months.
Table Summery:
From
Bond 1
Bond 2
Bond 3
Cash
To Month
Amount
Penalty Cost ($)
Total Cost ($)
2
3
4
5
5
1
2
6
1
10
50
80
60
100
50
90
140
150
0.19
0.17
0.13
0.09
0
1
1
0
0
Total
1.90
8.50
10.40
5.40
0
50.00
90.00
0
150
316.20
Report to the manager:
The minimum cost of company will be $316.2 and the available units and penalty cost
will be as following:
10 units at $0.19 penalty cost for Bond 1 used in Month 2
50 units at $0.17 penalty cost for Bond 1 used in Month 3
80 units at $0.13 penalty cost for Bond 1 used in Month 4
60 units at $0.09 penalty cost for Bond 1 used in Month 5
100 units at no penalty cost per for Bond 2 used in Month 5
50 units at $1.00 penalty cost for Bond 3 used in Month 1
90 units at $1.00 penalty cost for Bond 3 used in Month 2
140 units at no penalty cost for Bond 3 used in Month 6
$150 cash for the first month
The total penalty costs will be as following:
$1.90 for Bond 1 used in Month 2
$8.50 for Bond 1 used in Month 3
$10.40 for Bond 1 used in Month 4
$5.40 for Bond 1 used in Month 5
$0.00 for Bond 2 used in Month 5
$50.00 for Bond 3 used in Month 1
$90.00 for Bond 3 used in Month 2
$1.40 for Bond 3 used in Month 6
We will use all $200 of available Bond 1 and all $100 of available Bond 2, but
only $280 of available Bond 3. Bond 3 will be the only Bond with an unused amount of
$120 and it is desirable, because it has the highest penalty cost at $1.
Add from sensitivity analysis
b) Solve part b as a transshipment problem using WinQSB (Network modeling >>>
Network Option).
Assume that payment of bills can be made after they are due, but a penalty of $0.05 per
month is assessed for each dollar of cash demands that is postponed for one month.
Assuming all bills must be paid by the end of month 6, develop a transshipment model
that can be used to minimize the cost of paying the next six months’ bills.
Input:
Output:
Report to Manager:
The company’s financial problem was evaluated in two different ways.
In the first evaluation, the company’s demands were met in the month that they
were due. In this manner, the company would have to pay $166.20 in penalties for
cashing in bonds before they were due. We will fulfill the demand for month 1 by cash
and bond 3, demand for month 2 by bond 1 and 3, demand for months 3 and 4 by
bond 1, demand for month 5 by bond 1 and 2, and demand for month 6 by bond 3.
At the end of the six months, the company still has $120 in bond 3 still available.
In the second evaluation, bill could be paid after they were due, but the company
would assess a late fee. In this manner, the company would have to pay only $48 in fees.
We will fulfill the demand for month 1 on time by cash and bond 1, demand for
month 2 on time by bond 1. Bills would be paid late in the month of three. They will
be paid in month 5 by bond 2. Bills would be paid late in the month four. They will
be paid in months 5 and 6 by bonds 2 and 3. Bills would be paid one month late in
the month five by bond 3. At the end of month 6, the company had $70 in Bond 3 (as
C6 from Bond 3) and $50 in Bond 1.