USC Viterbi School of Engineering
Daniel J. Epstein Department of Industrial and Systems
Engineering
ISE 530: Optimization Methods for Analytics
Title
SSH Company: Coal Production Strategy
Prepared by: Han Ding [email protected]
Xue Yang [email protected]
Siddhi Nimhan [email protected]
Presented to: Dr. S. Parisay
Teaching Assistant: Yifan Liu
Date: December 2, 2014
1
Table of Content
1.
2.
3.
4.
Final Problem Statement
Analysis of Problem as a System
Problem Summary as a Table
Linear Programming
4.1 Problem formulation
4.2 Solution Summary as a Table
4.3 Analysis of the Solution
4.4 Sensitivity Analysis
4.5 Report to the Manager
5. Goal Programming
5.1 Additional Information for GP
5.2 Problem Formulation for GP
5.3 Solution for GP and Analysis
5.4 Preemptive GP and Sensitivity Analysis
5.5 Report to the Manager
6. Integer Linear Programming Formulation
6.1 Motivation for ILP
6.2 ILP Problem Solving
6.3 Solution Summary as a Table
6.4 Analysis of the Solution
6.5 Report to the Manager
Appendix A: Solution to the LP Problem
Appendix B: Sensitivity Analysis of BVs for LP
Appendix C: Sensitivity Analysis of Binding Constraint for LP
Appendix D: Solution to the GP Problem
Appendix E: Preemptive Goal Programming
Appendix F: Solution to the ILP Problem
Appendix G: Flow of Information
Appendix H: Comparison of WinQSB, LINDO and Excel Solver
2
1. Final Problem Statement
SSH company is a world famous coal production industry of producing different types of coal.
However, the production process releases contaminants into the atmosphere as particulates,
aerosols, vapors, or gases, which are dangerous to human health and surrounding environment.
In order to meet current allowable emission limits, SSH decides to install several kinds of add-on
pollution control devices in the ductwork (or flues) leading to the smoke stack. Their goal is to
achieve the pollution reduction requirements in the most economical way.
The main five kinds of pollutant generated in the production process are carbon oxides (COx),
sulfur dioxide (SO2), nitrogen oxides (NOx), hydrogen sulfide (H2S) and particulate matter (PM).
The monthly emission requirements are listed in the table 1.1 below.
Table 1.1: Monthly emission requirement of pollutant
Pollutant
Monthly emission requirement (ton)
Cox
<=24000
SO2
<=25000
NOx
<=23000
H2 S
<=35000
PM
<=65000
SSH produces 4 types of coal. Least monthly demand and profit of each type of coal are listed in
table 2. Production of each type of coal will generate those five pollutants but with different
amounts. Detailed information is also presented below in table 1.2.
Table 1.2: Information of 4 types of coal
Pollutant generated when producing 1 ton of coal
Coal
type
Profit
($/ton)
Demand
(ton/month)
COx
SO2
NOx
H2 S
PM
(ton)
(ton)
(ton)
(ton)
(ton)
1
1600
1000
2
9
12
5
6
2
1200
1200
7
1
4
6
8
3
1400
1300
11
5
3
7
15
4
1500
1500
5
7
10
4
7
3
The environmental issues are creating problems for company’s growth. Therefore it is important
to control pollution. The industry uses 3 types of controlling devices to reduce emission of
poisonous pollutants in air. The devices can be used alone or together to reduce pollution and the
company is able to buy multiple of each type of device. The number of hour’s labor works on
individual devices is shown in table 3. The prices of the devices and their ability to reduce the
effect of pollutants are given below in table 1.3. The maintenance cost associated with each
device is $400/month. And considering useful life for these devices, it is better to replace old
devices with new ones every month.
Table 1.3: Information of 3 types of devices
Cost
($)
COx
(ton/month)
1
1000
1
6
5.3
2.4
5.2
2
1500
4
6
8
6.7
2.3
3
2000
4
7.6
5.1
7.8
4.2
Device
SO2
NOx
H2 S
PM
(ton/month) (ton/month) (ton/month) (ton/month)
The goal is to fulfill all the requirements while maximize the profit after subtracting costs related
to reduction of pollutants.
4
2. Analysis of the Problem as a System
a) Define the needs and objectives
 The need is to produce the demanded coal as efficient as possible.
 The objective will be to maximize the profit of coal production by reducing the
cost of pollution reduction process. This objective is measured as $ amount.
b) Define input and output of the system
Input
 Information on amount of coal required and profit of coal according to type of
coal.
 Information of amount of pollutants produced and their limitation.
 Information of sales prices of devices and amount it reduces pollutants.
 Information on maintenance cost of devices.
Output





Information on amount of each type of coal produced.
Information on number of devices used for reducing pollution.
Information on how constraints are met.
Information on total maximum profit.
Information that will assist in sensitivity analysis.
Please refer to Appendix G for more information
c) Define controllable and uncontrollable factors
Controllable factors:
 Profit of each type of coal
 Maximum of coal produced.
 Maintenance cost of each type of device.
Uncontrollable factors:



Amount of pollutants produced.
Amount of pollutants reduced by devices.
Purchasing price of each type of device
d) Define a suitable model
Linear Programming (LP) model is suitable by creating linear constraints and linear
objective function. Goal Programming (GP) model can be used for fulfilling the
requirements of profit and cost of reducing pollutants. Integer Linear Programming (ILP)
can be used to make the float values into integer.
5
3. Problem Summary as a Table
Table 3.1: Problem Summary (1)
Coal
type
Weight
of coal
(ton)
Profit
($/ton)
Demand
(ton/month)
Pollution generated when producing 1 ton of coal
COx
(ton)
2
7
11
5
<=2400
0
1
X1
1600
>=1000
2
X2
1200
>=1200
3
X3
1400
>=1300
4
X4
1500
>=1500
Requirement for each kinds of pollution
(ton)
Sox
(ton)
9
1
5
7
<=2500
0
NOx
(ton)
12
4
3
10
<=2300
0
H2S
(ton)
5
6
7
4
<=3500
0
PM
(ton)
6
8
15
7
<=6500
0
Table 3.2: Problem Summary (2)
Device
type
Number Labor
of
Hours
device
Cost for Pollution decreased by using one device
each
COx(ton) SOx(ton) NOx(ton) H2S(ton) PM(ton)
one
1
2
3
Y1
Y2
Y3
1000
1500
2000
5
3
2
1
4
4
6
6
7.6
5.3
8
5.1
2.4
6.7
7.8
5.2
2.3
4.2
*For each devise, there is a monthly $400 maintenance fee.
6
4. Linear Programming
4.1 Problem Formulation for LP
Decision variables:
Xi = tons of production of coal i per month (i=1, 2, 3, 4)
Yi = number of devices i installed per month (i=1, 2, 3)
O.F.: maximize profit of producing coal after subtracting cost spend on devices
O.F.: Z = profit of coal – cost of buying devices – maintenance fee
= 1600X1 + 1200X2 + 1400X3 + 1500X4 – 1000Y1 – 1500Y2 – 2000Y3 – 400(Y1 + Y2 +
Y3)
Maximize Z = 1600X1 + 1200X2 + 1400X3 + 1500X4 – 1400Y1 – 1900Y2 – 2400Y3
Constraints for monthly demand of different types of coal:
C1 X1 >= 1000
C2 X2 >= 1200
C3 X3 >= 1300
C4 X4 >= 1500
Constraints for monthly emission requirements on different types of pollutant:
For example: for pollutant COx, amount of COx generated – amount of COx reduced by devices
should be less or equal than the emission requirement
C5 2X1 + 7X2 + 11X3 + 5X4 – Y1 – 4Y2 – 4Y3 <= 24000
C6 9X1 + X2 + 5X3 + 7X4 – 6Y1 – 6Y2 – 7.6Y3 <= 25000
C7 12X1 + 4X2 + 3X3 + 10X4 – 5.3Y1 – 8Y2 – 5.1Y3 <= 23000
C8 3X1 + X2 + 5X3 + 2X4 – 2.4Y1 – 6.7Y2 – 7.8Y3 <=35000
C9 6X1 + 8X2 + 15X3 + 7X4 – 5.2Y1 – 2.3Y2 – 4.2Y3 <= 65000
Sign restrictions:
Xi and Yi must be non-negative. Yi must be integers.
7
4.2 Solution Summary as a Table
The solution obtained by WinQSB is in Appendix A.
Below is a summary of this solution.
Table 4.1: Summary of Optimal Solution
Variables
Solution
Description
Variables
Weight of coal 1 produced
X1
1,462.5
Weight of coal 2 produced
X2
1,200
Weight of coal 3 produced
X3
1,300
Weight of coal 4 produced
X4
1,500
Number of device 1 used
Y1
0
Number of device 2 used
Y2
2,282*
Number of device 3 used
Y3
0
*Since Yi need to be integers, values of Yi are rounded. This may cause failure to satisfy some
of the constraints, but since the deviations would be very small, they can be ignored given certain
flexibility in the real situation.
Max profit: Z=3,514,200. (Difference with the solution is caused by the rounding of variable
Y2)
A detailed summary solution, organized as two tables, is provided below based on the solution.
Table 4.2: Detailed summary with Some Extra Information (1)
Coal
type
1
Weight
of coal
(ton)
1,462.5
Total
Profit($)
Demand
(ton/month)
2,340,00 >=1000
0
2
1,200
1,440,00 >=1200
0
3
1,300
1,820,00 >=1300
0
4
1,500
2,250,00 >=1500
0
Requirement for each kinds of pollution
(ton)
Actual emission after using devices
Pollution generated when producing 1 ton of coal
COx
(ton)
2
Sox
(ton)
9
NOx
(ton)
12
H2S
(ton)
5
PM
(ton)
6
7
1
4
6
8
11
5
3
7
15
5
7
10
4
7
<=2400
0
24000
<=2500
0
17675
<=2300
0
23000
<=3500
0
14328.1
3
<=6500
0
43128.1
3
8
Table 4.3: Detailed summary with Some Extra Information (2)
Device
type
Number Total cost for Pollution decreased by using one device
of
each type
COx(ton) SOx(ton) NOx(ton) H2S(ton)
device
1
0
0
1
6
5.3
2.4
2
2282*
4334800*
4
6
8
6.7
3
0
0
4
7.6
5.1
7.8
*Difference with the output is caused by that value of Y2 is rounded to integer.
PM(ton)
5.2
2.3
4.2
9
4.3 Analysis of the Solution
Analysis the solution, we can find the following information:
1) Maximum profit is $3,514,200, a result of profit of producing and selling coals less expenses
of purchasing and maintaining devices for reducing pollution.
Total profit of coals is the sum of the following:
a. Profit of 1,462.5 ton of coal 1: $2,340,000=1462.5*$1600
b. Profit of 1,200 ton of coal 2: $1,440,000=1200*$1200
c. Profit of 1,300 ton of coal 3: $1,820,000=1300*$1400
d. Profit of 1,500 ton of coal 4: $2,250,000=1500*$1500
Total expense of devices is the sum of the following:
a. Purchase of 2,282 device 2: $3,423,000=2282*$1500
b. Maintenance fee of device 2: $912,800=2282*$400
2) The produced coals are 1462.5, 1200, 1300, and 1500 ton/month for Coal 1, Coal 2, Coal 3,
and Coal 4. This production will fulfill the minimum demand for Coal 2, Coal 3, and Coal 4,
and it will exceed the minimum demand for Coal 1 by 462.5 ton/month.
3) After using that number of Device 2 as given in the solution, emission of COx and NOx will
meet exactly the maximum emission amount as 24000 ton/month and 23000 ton/month.
While emission of SO2 will be 17675 ton/month and it exceeds the maximum requirement by
7325 ton/month. Emission of H2S will be 14328.13 ton/month and it exceeds the maximum
requirement by 20671.88 ton/month. Also, emission of PM will be 43128.13 ton/month and it
exceeds the maximum requirement by 21871.88 ton/month.
10
4.4 Sensitivity Analysis for LP
In this problem, we have seven decision variables and nine constraints, so in total we can
perform sixteen sensitivity analyses. WinQSB is used to perform these sensitivity analyses based
on optimal solution of the LP problem, and the related graphs are generated using Excel.
Figure A2, Appendix A gives sensitivity analysis result given by WinQSB.
For basic variables X1~X4, and Y2, consider realistic ranges around the current values for their
coefficients. For example, coefficient for X1 is 1600, then consider analyzing the effect of
changes for this coefficient in a range of 1200 to 2000, which is in its allowable range [950,
2560.526]. Figure B1~B5 in Appendix B is sensitivity graphs for X1, X2, X3, X4, and Y2.
For non-basic variables Y1 and Y3, there is reduced cost for their coefficient. For coefficient of
Y1, reduced cost is -656.875, and for coefficient of Y3, reduced cost is -735.625. However,
changes in maintenance fee (part of coefficients of Y1~Y3) of one device will change the
coefficients of all three variables, but due to that the coefficient of Y2 will still be in the
allowable range, the bass and bfs will not change, only Z-value will change.
For binding constraints C2, C3, C4, C5, and C7, consider realistic ranges around the current
values for their right-hand side. For example, right-hand side of C2 is 1200, then consider
analyzing the effect of changes for this right-hand side in a range of 830 to 1700, which is in its
allowable range [830, 3379.016]. Figure C1~C5 in Appendix C is sensitivity graphs for C2, C3,
C4, C5, and C7.
For non-binding constraints C1, C6, C8, and C9, there is slack or surplus for their right-hand side.
Slack for C1 is 462.5, for C6 is 7325, for C8 is 20671.88 and for C9 is 21871.88.
Since it is not useful and time efficient to present all 16 sensitivity analysis in the report to the
manager, we can limit the number of sensitivity analysis mentioned in the report, and only
discuss the most useful one among similar variables or constraints. Given the sensitivity analysis
results, the following analysis is chosen.
1) Analysis on coefficient of X4: X4 has the highest value among X1~X4 in the output of
optimal solution and will have more impact in the optimal Z when changing its coefficient.
2) Analysis on coefficient of Y2: Y2 is the only one with positive value among Y1~Y3.
3) Analysis on coefficients of both Y1 and Y3: both of them have value of zero and the reduced
costs can assist in some useful information. However, interpretation for these reduced costs
may be impractical.
4) Analysis on rhs of C3: C3 has the largest shadow price among binding constraints C2~C4
and will have more impact in the optimal Z when changing its rhs.
5) Analysis on rhs of C5: C5 has larger shadow price than C7 and will have more impact in the
optimal Z when changing its rhs.
11
4.5 Report to the Manager
Dear Manager,
Considering the given information about the coals, devices, and the requirements for each kind
of pollutants, the best production and pollution control plan is as follows:
The max profit can be $3,514,200 with a production plan of 1,462.5 ton of Coal 1, 1,200 ton of
Coal 2, 1,300 ton of Coal 3 and 1,500 ton of Coal 4 for each month. To meet the requirements of
pollution emission, we’ll only use Device 2, while no Device 1 and Device 3 is used, and the
number of Device 2 we need is 2,282.
This plan will fulfill demands for all types of coals, and will exceed the demand for Coal 1 by
462.5 ton. All of the pollution requirements will be met. Among which, 17,675 ton of SO2 will
be issued into air, which is 7,325 ton below the requirement. We’ll generate 14,328.13 ton of
H2S, which is 20,671.88 ton less than the maximum generating amount. Also the 43,128.13 ton
of PM will be 21,871.88 ton below the requirement. (Appendix A)
If you would like to use some Device 1, the unit cost of Device 1 should be reduced at least to
$743 from the current total value of $1,400. That can be a combination of reduction in its
purchase price and the monthly maintenance fee. It seems that this is too much of a change,
about 46.9%, and may not be practical. Besides, if you would like to use some Device 3, the unit
cost of Device 3 should be reduced at least to $1,664 from the current total cost of $2,400. The
same problem with Device 1 may happen and cause this change to be also impractical.
(Appendix A)
One alternative to increase the total profit would be to increase the unit profit of Coal 4, on the
basis of its current value $1,500. For each dollar increase in the unit profit of Coal 4, the total
profit will increase by $1,500. Notice that we can only increase the unit profit of Coal 4 to at
most $2,375. In this case the production and pollution control plan won’t change. Since that will
be a 58.3% increase from $1,500 to $2,375 which is not practical, we can just increase the profit
to where we are capable of. (Appendix B, Figure B4)
Besides, the total profit can be increased by decreasing the unit cost of Device 2 from the current
value of $1,900 (purchase cost $1,500+monthly maintenance fee $400). For each dollar of the
decreased cost of Device 2, the total profit can increase by $2,282 and there will not be a change
in the production and pollution control plan as long as it is not lower than $1,347.4. (Appendix B,
Figure B5)
Another alternative to increase the total profit would be reducing the demand for Coal 3, from its
current values of 1,300 ton/month. For each unit of the Coal 3’s demand being reduced, the total
12
profit can increase by $2281.25. But notice that the demand for Coal 3 should not be below
1,105.263 ton/month. And if we reduce the monthly demand for Coal 3, we will continue
producing four kinds of coals and use only Device 2 to help with the control of pollution,
however with different amount of coals produced and Device 2 used. (Appendix C, Figure C2)
Alternatively, you can increase the total profit by increasing the requirement of amount of COx
issued. For each unit of increase in the amount requirement of COx, the total profit can be
increased by $312.5. If we increase the requirement of COx, we will continue to produce four
kinds of coals and use only Device 2 to help with the control of pollutions, however with
different amount of coals produced and Device 2 used. It seems that the requirements on
generating amount of each kind of pollutant are really important rules for us to carry out, so it
may be impractical to change the requirements. (Appendix C, Figure C4)
Please inform of any further questions or feedback. Thank you.
13
5. Goal Programming
5.1 Additional Information for GP
Goal programming is an important technique for decision makers to solve multi objective
problems to achieve solution. Preemptive Goal Programming focuses on completing most
important goals first and afterward succeeding goals.
Goals of our problem include:
Goal 1: Profit of coal production should be at least $6,000,000;
Goal 2: The cost of devices should be at most $3,000,000;
Goal 3: The maintenance fee should be at most $800,000;
Goal 4: Meet the labor hour requirements for three devices.
As for the labor hour requirements, the table below shows detailed information.
Table 5.1: Labor Hours Information
Device
1
2
3
Monthly Required Hours per device
(hrs)
5
3
2
Total Maximum Labor Hours per
Month (hrs)
1500
1300
1200
And the rank of priority of goals is presented below.
Table 5.2: Rank of Priority
Rank
1
Priority
Highest
2
Second Highest
3
Third Highest
4
Least
Goal
Profit of coal should be at
least $6,000,000
The cost of devices should be
at most $3,000,000
The maintenance fee should
be at most $800,000
Meet the labor hour
requirements for three devices
14
5.2 Problem Formulation for GP
The following table lists the GP problem formulation steps. Decision variables and constraints
are the same as linear programming so they are not listed here anymore.
Table 5.3: Formulation for GP
1. Deviational
variables
Description
Equation
SiN = Quantity by which goal i
is not reached.
SiP = Quantity by which goal i is
exceeded.
2. Objective
Function
GP (by order of importance with
1 being most important):
1. Profit of coal production
should be at least $6,000,000
2. The cost of devices should be
at most $3,000,000
3. The maintenance fee should
be at most $800,000
4. Meet the labor hour
requirements for three devices
3. Goals
1. Profit of coal production
should be at least $6,000,000
2. The cost of devices should be
at most $3,000,000
3. The maintenance fee should
be at most $800,000
4. Meet the labor hour
requirements for three devices
4. Sign
Restriction
(Constraints defined in Linear
programming model will also
apply here)
Decision and deviational
variables => 0
Z1=P1S1N
Z2=P2S2P
Z3=P3S3P
Z4=P4S4P+P4S5P+P4S6P
1. 1600X1+100X2+1400X3+1500X4
+S1N-S1P=6,000,000
2. 1000Y1+1500Y2+2000Y3+S2NS2P=3,000,000
3. 400Y1+400Y2+400Y3+S3NS3P=800,000
4. 5Y1+S4N-S4P=1500
5. 3Y2+S5N-S5P=1300
6. 2Y3+S6N-S6P=1200
Xi, Yi, SiN and SiP =>0
15
5.3 Solution for GP and Analysis
The optimal solution given by WinQSB of our GP problem is presented in the table below.
(Appendix D)
Table 5.4: Solution to GP Problem
Variables
Description
Weight of coal 1 produced
Weight of coal 2 produced
Weight of coal 3 produced
Weight of coal 4 produced
Number of device 1 used
Number of device 2 used
Number of device 3 used
Solution
Variables
X1
X2
X3
X4
Y1
Y2
Y3
1,000
1,200
1,300
1,500
0
2,050
0
With this solution, deviations to the targets of goals can be calculated, and the results are showed
below.
Table 5.5: Summary of Goal Fulfillment of GP
Rank
1
Goal
Target
Result
Deviation
Profit of coal
>= 6,000,000
7,110,000
1,110,000
should be at least
$6,000,000
2
The cost of
<=3,000,000
3,075,000
75,000
devices should
be at most
$3,000,000
3
The maintenance <=800,000
820,000
20,000
fee should be at
most $800,000
4
Meet the labor
<=4000
6150
2150
hour
requirements for
three devices
In this solution, it means that SSH will produce 1,000 tons of Coal 1, 1,200 tons of Coal 2, 1,300
tons of Coal 3, and 1,500 tons of Coal 4. In order to reach requirements on pollution emission,
SSH will buy 2,050 Device 2 to reduce pollutant. By doing this, SSH can achieve the first one
out of four goals.
16
5.4 Preemptive GP and Sensitivity Analysis
The goal priorities can be changed to understand effect on fulfillment of requirements. Many
different priority orders have been tried and tables below show the results. (Appendix E)
Table 5.6: Preemptive Goal Programming Results Summary (1)
Priorities
Optimal Solution
Highest
Second Third
Lowest
Highest Highest
G1
G3
G2
G1
G4
G3
X1
X2
X3
X4
Y1
Y2
Y3
G4
1,000
1,200
1,300
1,500
0
2,050
0
G2
G3
1,000
1,200
1,300
1,500
0
1,450
600
G1
G4
G2
1,000
1,200
1,300
1,500
0
1,450
600
G4
G1
G2
G3
1,000
1,200
1,300
1,500
300
433.3
1,541.6
G4
G3
G1
G2
1,000
1,200
1,300
1,500
300
433.3
1,541.6
Table 5.7: Preemptive Goal Programming Results Summary (2)
Priorities
Highest
Second
Highest
Third
Highest
Deviation
Lowest
Z1
Z2
Z3
Z4
G1
G3
G2
G4
0
20,000
75,000
4,850
G1
G4
G2
G3
0
3,050
375,000
20,000
G3
G1
G4
G2
20,000
0
3,050
375,000
G4
G1
G2
G3
18,883.35
0
1,033,333.30
110,000
G4
G3
G1
G2
1,883.3
0
0
1,033,333.30
The table shows that as priorities of goals changes the company produces same amount of coals
as 1000 tons of coal 1, 1200 tons of coal 2, 1300 tons of coal 3, 1500 tons of coal 4 is produced.
In every case goal 1 that is to maximize profit of coal and which should be greater than
$6,000,000 is fulfilled. When the labor hour goal has highest priority, then two goals of
maintenance cost and maximum profit of coal goal are fulfilled but this increases the cost of
devices to $1,033,333.3, which is large amount and may not be practical. In this case company
will have to use all three types of devices thus increase total cost of devices.
17
5.5 Report to the Manager
Dear Manager,
Based on given information, we need to produce 1,000 tons of Coal 1, 1,200 tons of coal2, 1300
tons of coal3, 1500 tons of coal 4 to have profit of $7,110,000. The company uses 2050 number
of devices of device of type 2 to reduce pollution. This results in that out of four requirements,
one requirement is fulfilled. The company has four requirements as follows: main objective of
company is to increase the profit of coal production and it should be greater than $6,000,000.
The company fulfills this requirement with extra profit of $1,110,000, which is good sign. The
next requirement is to reduce the cost of devices which are used for pollution control and this
devices cost should be less than $3,000,000. Company doesn’t fulfill this goal and extra cost of
$75,000 is incurred. The next requirement is that maintenance fee should be less than $800,000
and the company fails to fulfill this requirement with extra of $20,000 fee. The labor hours
requirement for device 1 should not be greater than 1,500 hours, for device 2 should be less than
1,300 and for device 3 hours should not be more than 1,200 hours, as increase in number of
hours will increase the cost of payment for labors. The number of labor hours used for Device 2
is 4,850 hours as Device 1 and Device 3 are not used.
As mentioned earlier, the company uses 2,050 devices of type 2 to reduce pollution. The cost of
Device 2 is greater than that of Device 1, so in order to reduce the cost of devices, the company
should use Device 1. However, when we change our priority order and put the requirement for
labor hour as number one priority, we will end up buying some Device 1 but also great amount
of Device 3. Since Device 3 is more expensive than Device 2, this plan will increase the device
cost. The cases of putting requirement of labor hour in second highest or third highest position
will give us another plan however it will also cost much more than the original plan. So our
original plan is the most practical one. Although it only fulfills the first requirement about coal
profit, combine all costs together it is the most economical one.
As for the fact that only the first requirement is fulfilled and fail with the other three, it may due
to that except from four requirements the plan also need to fulfill constraints about pollution
emission, so a certain amount of devices are required. Since it is not quite possible to change
those requirements on pollution emission, we may have to expand our budget of device costs and
maintenance costs, and be prepared to hire more workers to operate the devices. Or we can try to
find other types of devices which have better effect on reducing pollutants, have lower prices and
costs, or are easier to operate.
Please inform of any further questions or feedback. Thank you.
18
6. Integer Linear Programming
6.1 Motivation for ILP
In the previous discussion, linear programming has been used to solve the SSH company
problem. After applying WinQSB software to solve the problem, optimal solution was obtained
to achieve maximum total profit. However, since number of three devices must be integers in
reality, so in the original formulation, decision variables Y1, Y2, and Y3 are given the constraint
that they must have positive integer values. So the method was to round values of Y1, Y2, and
Y3 to integers, and do the following sensitivity analysis. Doing this would change the value of Zvalue and may have a risk that constraint is no longer satisfied. Due to these considerations,
integer linear programming (ILP) is used to get solution that is more practical.
Because among all seven decision variables, only 3 of them are required to be integer, so this is
actually a mixed integer linear programming problem.
6.2 ILP Problem Solving
Branch and bound is applied to solve this ILP problem. Branch and bound methods find the
optimal solution to an IP by efficiently enumerating the points in a subproblem’s feasible region.
The first subproblem is obviously the original LP problem, and within the solution (see
Appendix A) only Y2 is non-integer, so Y2=2281.25 is branched.
The formulations of subprblems 2 to 5 are as,
a.
b.
c.
d.
Subproblem 2: subproblem 1 formulation + constraint Y2<=2281;
Subproblem 3: subproblem 1 formulation + constraint Y2>=2282;
Subproblem 4: subproblem 2 formulation + constraint Y3=0;
Subproblem 5: subproblem 2 formulation + constraint Y3>=1.
ILP problem is solved by using LINDO 6.1 software. And solving is terminated after solving 5
subproblems. Because this is a max Z-value problem, so the Z-value for subproblem 1 is the
upper bound of all Z-values, so as one goes down the tree, Z-value will decrease. We get two
candidate solutions which are solutions of subproblem 3 and 4. And certainly, there will be more
candidates if we go down the subproblem 5. But Z-values of those subproblems will be smaller
than Z-value of subproblem 5, which is smaller than those of the existing candidates. So the
procedure can be stopped here. Figure 6.1 shows the branch and bound tree, and table 6.1 is a
summary of candidate solutions. (See also Appendix F)
19
Figure 6.1: Branch and bound tree for mixed ILP
Y2<=228
1
Y3=0
Subproblem 4
Z=3515300
X1=1462
X2=1200
X3=1300
X4=1500
Y1=0
Y2=2281
Y3=0
Candidate Solution
Subproblem 2
Z=3515469
X1=1462.42
X2=1200
X3=1300
X4=1500
Y1=0
Y2=2281
Y3=0.21
Subproblem 1
Z=3515625
X1=1462.5
X2=1200
X3=1300
X4=1500
Y1=0
Y2=2281.25
Y3=0
Y3>=1
Y2>=228
2
Subproblem 3
Z=3515210
X1=1462.89
X2=1200.32
X3=1300
X4=1500
Y1=0
Y2=2282
Y3=0
Candidate Solution
Subproblem 5
Z=3514890
X1=1462.14
X2=1200
X3=1300
X4=1500
Y1=0
Y2=2280.07
Y3=1
20
Table 6.1: Summary of candidate solutions
Subproblem 3
Subproblem 4
X1
1462.89
1462
X2
1200.32
1200
X3
1300
1300
X4
1500
1500
Y1
0
0
Y2
2282
2281
Y3
0
0
Z-value
3515210
3515300
Compare their Z-value, subproblem 4 has a larger Z-value, so it is the optimal solution for this
ILP problem.
6.3 Solution Summary as a Table
A detailed summary solution, organized as two tables, is provided below based on the solution.
Max Z-value for ILP is Z=3515300.
Table 6.2: Detailed summary with Some Extra Information (1)
Coal
type
1
Weight
of coal
(ton)
Total
Profit($)
Demand
(ton/month)
Pollution generated when producing 1 ton of coal
COx
(ton)
2
1,462
SOx
(ton)
9
NOx
(ton)
12
H2S
(ton)
5
2,339,20 >=1000
0
2
1,200
1,440,00 >=1200
7
1
4
6
0
3
1,300
1,820,00 >=1300
11
5
3
7
0
4
1,500
2,250,00 >=1500
5
7
10
4
0
Requirement for each kinds of pollution
<=2400 <=2500 <=2300 <=3500
(ton)
0
0
0
0
Actual emission after using devices
24000
17672
22996
14327.3
Table 6.3: Detailed summary with Some Extra Information (2)
PM
(ton)
6
8
15
7
<=6500
0
43125.7
Device
type
Number Total cost for Pollution decreased by using one device
of
each type
COx(ton) SOx(ton) NOx(ton) H2S(ton)
device
PM(ton)
1
2
3
0
2281
0
5.2
2.3
4.2
0
4333900
0
1
4
4
6
6
7.6
5.3
8
5.1
2.4
6.7
7.8
21
6.4 Analysis of the Solution
Analysis the solution (Appendix F), we can find the following information:
1) Maximum profit is $3,515,300, a result of profit of producing and selling coals less expenses
of purchasing and maintaining devices for reducing pollution.
Total profit of coals is the sum of the following:
a. Profit of 1,462 ton of coal 1: $2,339,200=1462*$1600
b. Profit of 1,200 ton of coal 2: $1,440,000=1200*$1200
c. Profit of 1,300 ton of coal 3: $1,820,000=1300*$1400
d. Profit of 1,500 ton of coal 4: $2,250,000=1500*$1500
Total expense of devices is the sum of the following:
a. Purchase of 2,281 device 2: $3,421,500=2281*$1500
b. Maintenance fee of device 2: $912,400=2281*$400
c. The produced coals are 1462, 1200, 1300, and 1500 ton/month for Coal 1, Coal 2,
Coal 3, and Coal 4. This production will fulfill the minimum demand for Coal 2, Coal
3, and Coal 4, and it will exceed the minimum demand for Coal 1 by 462 ton/month.
d. After using that number of Device 2 as given in the solution, emission of COx and NOx will
meet exactly the maximum emission amount as 24000 ton/month and 23000 ton/month.
While emission of SO2 will be 17672 ton/month and it exceeds the maximum requirement by
7328 ton/month. Emission of H2S will be 14327.3 ton/month and it exceeds the maximum
requirement by 20672.7 ton/month. Also, emission of PM will be 43125.7 ton/month and it
exceeds the maximum requirement by 21874.3 ton/month.
22
6.5 Report to the Manager
Dear Manager,
Based on the given information, we can obtain a maximum profit of $3,515,300 after subtracting
cost of devices. To achieve this number, we need to produce 1,462 ton of Coal 1, 1,200 ton of
Coal 2, 1,300 ton of Coal 3, and 1,500 ton of Coal 4. And in order to make pollutants generated
from production of so many coals meet the requirements on air emission, we need to purchase
2,281 Device 2.
This plan can fulfill demands for 4 types of coals and exceeds the minimum demand for Coal 1
by 462 ton. On the other hand, those 2281 Device 2 can help reduce 5 kinds of pollutants
generated in production process, and achieve the requirements on air emission levels. Amount of
COx meets the 24,000 ton requirement. Amount of SOx is 7,328 ton below the 25,000 ton
requirement. Also, NOx emission is 4 ton below the 23,000 ton requirement. Emission of H2S is
14,327.3 ton and is 20,672.7 ton below requirement. Finally, the PM amount is 21,874.3 ton
below the 65,000 ton requirement.
Please inform of any further questions or feedback. Thank you.
23
Appendix A: Solution to the LP Problem
Figure A1: Input of WinQSB
Figure A2: Output of WinQSB (Optimal Solution)
24
Appendix B: Sensitivity Analysis of BVs for LP
Table B1: Summary of Sensitivity Analysis of BVs
BV
Coefficient
X1
X2
X3
X4
Y2
1,600
1,200
1,400
1,500
1,900
Range of Sensitivity Analysis
Max
2,000
1,600
1,800
1,900
2,522
Min
1,200
800
1,000
1,100
1,348
Figure B1: Sensitivity Graph for X1
Figure B1: Sensitivity Graph for Objective Function Coefficient of X1
4300000
4100000
4099200
Value of O.F.
3900000
3700000
3514200
3500000
3300000
3100000
2900000
2929200
2700000
2500000
1000
1200
1400
1600
1800
2000
2200
CBV1: coefficient of X1
25
Figure B2: Sensitivity Graph for X2
Figure B2: Sensitivity Graph for Objective Function Coefficient of X2
4100000
3994200
3900000
Value of O.F.
3700000
3514200
3500000
3300000
3100000
3034200
2900000
2700000
2500000
600
800
1000
1200
1400
1600
1800
CBV2: Coefficient of X2
Figure B3: Sensitivity Graph for X3
Figure B3: Sensitivity Graph for Objective Function Coefficient of X3
4300000
4100000
4034200
Value of O.F.
3900000
3700000
3514200
3500000
3300000
3100000
2994200
2900000
2700000
2500000
800
1000
1200
1400
1600
1800
2000
CBV3: coefficient of X3
26
Figure B4: Sensitivity Graph for X4
Figure B4: Sensitivity Graph for Objective Function Coefficient of X4
4300000
4114200
4100000
3900000
Value of O.F.
3700000
3514200
3500000
3300000
3100000
2914200
2900000
2700000
2500000
900
1100
1300
1500
1700
1900
2100
CBV4: Coefficient of X4
Figure B5: Sensitivity Graph for Y2
Figure B5: Sensitivity Graph for Objective Function Coefficient of Y2
5000000
4773864
4500000
Value of O.F.
4000000
3514200
3500000
3000000
2500000
2094796
2000000
1500000
1200
1400
1600
1800
2000
2200
2400
2600
CBV5: Coefficient of Y2
27
Appendix C: Sensitivity Analysis of Binding Constraint for LP
Table C1: Summary of Sensitivity Analysis for Binding Constriants
Binding Constraints
Right-hand Side
C2
C3
C4
C5
C7
1,200
1,300
1,500
24,000
23,000
Range of Sensitivity Analysis
Min
Max
830
1,700
1,105.263
1,800
1,000
2,000
23,000
25,000
22,000
24,000
Figure C1: Sensitivity Graph for Right-hand Side of C2
Figure C1: Sensitivity Graph for Objective Function r.h.s of Constraint 2
4500000
Value of O.F.
4000000
3999825
3514200
3500000
3000000
2857950
2500000
2000000
650
850
1050
1250
1450
1650
1850
r.h.s of constraint 2
28
Figure C2: Sensitivity Graph for Right-hand Side of C3
Figure C2: Sensitivity Graph for Objective Function r.h.s of Constraint 3
4200000
3958444
3700000
Value of O.F.
3514200
3200000
2700000
2373575
2200000
1700000
900
1000
1100
1200
1300
1400
1500
1600
1700
1800
1900
r.h.s of constraint 3
Figure C3: Sensitivity Graph for Right-hand Side of C4
Figure C3: Sensitivity Graph for Objective Function r.h.s of Constraint 4
4100000
3951700
3900000
Value of O.F.
3700000
3514200
3500000
3300000
3100000
3076700
2900000
2700000
800
1000
1200
1400
1600
1800
2000
2200
r.h.s of constraint 4
29
Figure C4: Sensitivity Graph for Right-hand Side of C5
Figure C4: Sensitivity Graph for Objective Function r.h.s of Constraint 5
3900000
3826700
3800000
Value of O.F.
3700000
3600000
3514200
3500000
3400000
3300000
3200000
3100000
22500
3201700
23000
23500
24000
24500
25000
25500
r.h.s of constraint 5
Figure C5: Sensitivity Graph for Right-hand Side of C7
Figure C5: Sensitivity Graph for Objective Function r.h.s of Constraint 7
3620000
3600000
3595450
Value of O.F.
3580000
3560000
3540000
3520000
3514200
3500000
3480000
3460000
3440000
3420000
21500
3432950
22000
22500
23000
23500
24000
24500
r.h.s of constraint 7
30
Appendix D: Solution to the GP Problem
Figure D1: Input of Goal Programming
31
Figure D2: Output of Goal Programming
32
Appendix E: Preemptive Goal Programming
Priority Order: G1>>G3>>G2>>G4
Figure E1: Input of G1>>G3>>G2>>G4 GP
Figure E2: Output of G1>>G3>>G2>>G4 GP
33
Priority Order: G1>>G4>>G3>>G2
Figure E3: Input of G1>>G4>>G3>>G2 GP
Figure E4: Output of G1>>G4>>G3>>G2 GP
34
Priority Order: G3>>G1>>G4>>G2
Figure E5: Input of G3>>G1>>G4>>G2 GP
Figure E6: Output of G3>>G1>>G4>>G2 GP
35
Priority Order: G4>>G1>>G2>>G3
Figure E7: Input of G4>>G1>>G2>>G3 GP
Figure E8: Output of G4>>G1>>G2>>G3 GP
36
Priority Order: G4>>G3>>G1>>G2
Figure E9: Input of G4>>G3>>G1>>G2 GP
Figure E10: Output of G4>>G3>>G1>>G2 GP
37
Appendix F: Solution to the ILP Problem
*Solution to subproblem 1 is in Appendix A
Figure F1: Solution to Subproblems of ILP
Subproblem 2
Input
Output
38
Subproblem 3
Input
Output
39
Subproblem 4
Input
Output
40
Subproblem 5
Input
Output
41
Figure F2: Sensitivity Analysis of Subproblem 4
Sensitivity analysis provided by LINDO for subproblem 4
42
Appendix G: Flow of Information
Figure G1: Information Flow Diagram
S
Demand
Profit
Purchase
Price
Maintenance
cost
Coal of type
(1, 2, 3, 4)
Devices
available (1, 2,
3)
Max limit
Amount of
Pollutants
produced
Amounts of
Pollutants reduced
(1, 2, 3, 4)
(1, 2, 3, 4)
Given Information
LP Model
Solution from LP
Total Profit
Sensitivity
Info
Amount of
coal produced
Number of
devices used
Report
43
Appendix H: Comparison of WinQSB, LINDO and Excel Solver
In this project, our team has applied WinQSB to solve linear programming and goal
programming problems, and LINDO 6.1 to solve integer linear programming problem. But when
solving the linear programming problem, we also tried the Excel Solver to solve it. We have
concluded our user experience and present here in the table below.
Software
WinQSB
LINDO 6.1
Excel Solver
LP
GP
ILP
Input in programming
style;
Output includes
solution and
sensitivity analysis;
Not able to draw
sensitivity graph
Define everything in
spreadsheet first;
Difficult to solve;
Need to do many
times of iterations
Same as LP
44
Team Evaluation Sheet
Date: Dec. 2, 2014
Course: ISE 530: Optimization Methods for Analytics
Project Name: SSH Company: Coal Production Strategy
Partner Name
% of the case/research
performed by this partner (out
of 100%)
Signature
Comments:
45