Integer Programming, Goal Programming, and Nonlinear Programming

Chapter 11
Integer Programming,
Goal Programming,
and Nonlinear
Programming.
Prepared By: Salah A. Skaik

In this Chapter:
• Integer Programming.
• Goal Programming.
• Nonlinear Programming.

Integer Programming
 IP is the extension of LP that solves
problems requiring Integer Solutions.
Ex. (Airline)
 There are two ways to solve IP
Problems:
1- Graphically.
2- The Branch & Bound Method.

Goal Programming
 GP is the extension of LP that permits
Multiple Objectives to be stated.
Ex. (Max profit & Max market share).

Nonlinear Programming
 NLP is the case in which Objectives
or Constraints are Nonlinear.
Ex. (Maximizing profit =
25X1 - 0.4X²1 + 30X2 – 0.5X²2 ).
Ex. (12X1 - 0.6X²1 ≥ 3,500 ).

Integer Programming
• There are three types of Integer
Programs:
1- Pure Integer Programs.
2- Mixed-Integer Programs.
3- Zero-One Integer Programming.
(special cases).

Example of Integer Programming
Harrison Electric Company

Chandeliers
2 hrs to
wire
6 hrs to
assembly
Ceiling Fans
3 hrs to
wire
5 hrs to
assembly

• 12 hrs of wiring
• 30 hrs of assembly
Production
Capability
• 7$ each chandelier
• 6$ each fanNet Profit

Let:
X1 = Number of chandeliers produced.
X2 = Number of ceiling fans produced.

Objective Function:
Maximize Profit = $7 X1 + $6 X2
Subject to:
2X1 + 3X2 ≤ 12 (wiring hours)
6X1 + 5X2 ≤ 30 (assembly hours)

6 –
5 –
4 –
3 –
2 –
1 –
0 –
| | | | | | |
1 2 3 4 5 6 X1
X2
+
++
++++
+
6X1 + 5X2 ≤ 30
2X1 + 3X2 ≤ 12
+ = Possible Integer Solution
Optimal LP Solution
(X1 =3.75, X2 = 1.5,
Profit = $35.25)

 In case the company doesn’t produce a
fraction of the product:
1- Rounding Off.
2- Enumeration Method.

 Rounding Off has two problems:
1- Integer solution may not be in feasible
region. (X1=4, X2=2). (unpractical solution)
2- May not be the optimal feasible Integer
solution. (X1=4, X2=1).

Enumeration Method

Optimal solution to
integer programming
problem
Solution if rounding
is used
CHANDELIERS (X1) CEILING FANS (X2) PROFIT ($7X1 + $6X2)
0 0 $0
1 0 7
2 0 14
3 0 21
4 0 28
5 0 35
0 1 6
1 1 13
2 1 20
3 1 27
4 1 34
0 2 12
1 2 19
2 2 26
3 2 33
0 3 18
1 3 25
0 4 24

An integer solution can never be better
than the LP solution and is usually a lesser
solution.

Six Steps in Solving IP
Maximization Problems by
Branch and Bound
Branch-and-Bound Method

Step(1):
Solve the original problem using LP. If the
answer satisfies the integer constraints, we
are done. If not, this value provides an
initial upper bound.

Step(2):
Find any feasible solution that meets the
integer constraints for use as a lower
bound. Usually, rounding down each
variable will accomplish this.

Step(3):
Branch on one variable from step 1 that does
not have an integer value. Split the
problem into two sub-problems based on
integer values that are immediately above
or below the non-Integer value.

Step(4):
Create nodes at the top of these new
branches by solving the new problem.

Step(5-a):
If a branch yields a solution to the LP
problem that is not feasible, terminate
the branch.

Step(5-b):
If a branch yields a solution to the LP
problem that is feasible, but not an integer
solution, go to step 6.

Step(5-c):
If the branch yields a feasible integer solution,
examine the value of the objective function. If
this value equals the upper bound, an optimal
solution has been reached. If it not equal to
the upper bound, but exceeds the lower
bound, set it as the new lower bound and go
to step 6. finally, if it’s less than the lower
bound terminate this branch.

Step(6):
Examine both branches again and set the
upper bound equal to the maximum value
of the objective function at all final nodes. If
the upper bound equals the lower bound,
stop. If not, go back to step 3.

NOTE:
Minimization problems involved reversing the
roles of the upper and lower bounds.

Harrison Electric Company Revisited
 Recall that the Harrison Electric Company’s integer
programming formulation is
Maximize profit = $7X1 + $6X2
subject to 2X1 + 3X2 ≤ 12
6X1 + 5X2 ≤ 30
where
X1 = number of chandeliers produced
X2 = number of ceiling fans produced
 And the optimal non-integer solution is
X1 = 3.75 chandeliers, X2 = 1.5 ceiling fans
profit = $35.25

 Since X1 and X2 are not integers, this solution is not
valid.
 The profit value of $35.25 will provide the initial upper
bound.
 We can round down to X1 = 3, X2 = 1, profit = $27,
which provides a feasible lower bound.
 The problem is now divided into two sub-problems.

6X1 + 5X2 ≤ 30
X1 ≥ 4
Subproblem A
6X1 + 5X2 ≤ 30
X1 ≤ 3
Subproblem B

 If you solve both sub-problems graphically
[X1 = 4, X2 = 1.2, profit = $35.20]
Sub-problem A’s optimal
solution:
Sub-problem B’s optimal
solution: [X1 = 3, X2 = 2, profit = $33.00]
 We have completed steps 1 to 4 of the branch-and-bound method.

 Harrison Electric’s first branching:
subproblems A and B
Subproblem A
Next Branch (C)
Next Branch (D)
Upper Bound = $35.25
Lower Bound = $27.00 (From
Rounding Down)
X1 = 4
X2 = 1.2
P = 35.20
X1 = 3
X2 = 2
P = 33.00
StopThis Branch
Solution Is Integer, Feasible
Provides New Lower Bound of $33.00
X1 = 3.75
X2 = 1.5
P = 35.25
Infeasible (Noninteger) Solution
Upper Bound = $35.20
Lower Bound = $33.00
Subproblem B

6X1 + 5X2 ≤ 30
X1 ≥ 4
X2 ≥ 2
Subproblem C
6X1 + 5X2 ≤ 30
X1 ≥ 4
X2 ≤ 1
Subproblem D
 Subproblem A has branched into two new subproblems, C
and D.

 Subproblem C has no feasible solution because the all the
constraints can not be satisfied
 We terminate this branch and do not consider this
solution
 Subproblem D’s optimal solution is X1 = 4.17, X2 = 1,
profit = $35.16
 This noninteger solution yields a new upper bound of
$35.16

6X1 + 5X2 ≤ 30
X1 ≥ 4
X1 ≤ 4
X2 ≤ 1
Subproblem E
6X1 + 5X2 ≤ 30
X1 ≥ 4
X1 ≥ 5
X2 ≤ 1
Subproblem D
 Finally we create subproblems E and F
Optimal solution to E:
X1 = 4, X2 = 1, profit = $34
Optimal solution to F:
X1 = 5, X2 = 0, profit = $35

Subproblem F
X1 = 5
X2 = 0
P = 35.00
Subproblem E
X1 = 4
X2 = 1
P = 34.00
 Harrison Electric’s full branch and bound solution
Feasible, Integer
Solution
Optimal
Solution
Upper Bound
= $35.25
Lower Bound
= $27.00
Subproblem C
No Feasible
Solution
Region
Subproblem D
X1 = 4.17
X2 = 1
P = 35.16
Subproblem A
X1 = 4
X2 = 1.2
P = 35.20
Subproblem B
X1 = 3
X2 = 2
P = 33.00
X1 = 3.75
X2 = 1.5
P = 35.25

Using Software To Solve Harrison Integer
Programming Problem
 QM forWindows input screen with Harrison Electric data

Programming Problem
 QM forWindows solution screen for Harrison Electric
data

Programming Problem
 QM forWindows iteration results screen for Harrison
Electric data

Programming Problem
 Using Excel’s Solver to formulate Harrison’s integer
programming model

Programming Problem
 Integer variables are specified with a drop-down menu in
Solver

Programming Problem
 Excel solution to the Harrison Electric integer
programming model

Mixed-Integer Programming Problem
There are many situations in which some of
the variables are restricted to be integers
and some are not.

 Bagwell wants to maximize profit
 We let X = number of 50-pound bags of xyline
 We let Y = number of pounds of hexall
 This is a mixed-integer programming problem as Y is not
required to be an integer.
AMOUNT PER 50-POUND
BAG OF XYLINE (LB)
AMOUNT PER POUND
OF HEXALL (LB)
AMOUNT OF
INGREDIENTS
AVAILABLE
30 0.5 2,000 lb–ingredient A
18 0.4 800 lb–ingredient B
2 0.1 200 lb–ingredient C

 The model is
Maximize profit = $85X + $1.50Y
subject to 30X + 0.5Y ≤ 2,000
18X + 0.4Y ≤ 800
2X + 0.1Y ≤ 200
X, Y ≥ 0 and X integer

 Using QM forWindows and Excel to solve Bagwell’s IP
model

 Excel formulation of Bagwell’s IP problem with Solver

 Excel solution to the Bagwell Chemical problem

ModelingWith 0-1 (Binary)Variables
 We can demonstrate how 0-1 variables can be
used to model several diverse situations.
 Typically a 0-1 variable is assigned a value of 0 if a
certain condition is not met and a 1 if the
condition is met.
 This is also called a binary variable

Capital Budgeting Example
 A common capital budgeting problem is selecting from a set of
possible projects when budget limitations make it impossible to
select them all
 A 0-1 variable is defined for each project
 Quemo Chemical Company is considering three possible
improvement projects for its plant
◦ A new catalytic converter
◦ A new software program for controlling operations
◦ Expanding the storage warehouse
 It can not do them all
 They want to maximize net present value of projects
undertaken

 The basic model is
Maximize net present value of projects undertaken
subject to Total funds used in year 1 ≤ $20,000
Total funds used in year 2 ≤ $16,000
 Quemo Chemical Company information
PROJECT NET PRESENT VALUE YEAR 1 YEAR 2
Catalytic Converter $25,000 $8,000 $7,000
Software $18,000 $6,000 $4,000
Warehouse expansion $32,000 $12,000 $8,000
Available funds $20,000 $16,000

 The mathematical statement of the integer programming
problem becomes
Maximize NPV = 25,000X1 + 18,000X2 + 32,000X3
subject to 8,000X1 + 6,000X2 + 12,000X3 ≤ 20,000
7,000X1 + 4,000X2 + 8,000X3 ≤ 16,000
X1, X2, X3 = 0 or 1
 The decision variables are
X1 = 1 if catalytic converter project is funded
0 otherwise
X2 = 1 if software project is funded
0 otherwise
X3 = 1 if warehouse expansion project is funded
0 otherwise

 Solved with computer software, the optimal
solution is X1 = 1, X2 = 0, and X3 = 1 with an
objective function value of 57,000
 This means that Quemo Chemical should fund
the catalytic converter and warehouse expansion
projects only
 The net present value of these investments will
be $57,000

Limiting the Number of Alternatives Selected
 One common use of 0-1 variables involves limiting the
number of projects or items that are selected from a
group
 Suppose Quemo Chemical is required to select no more
than two of the three projects regardless of the funds
available
 This would require adding a constraint
X1 + X2 + X3 ≤ 2
 If they had to fund exactly two projects the constraint
would be
X1 + X2 + X3 = 2

Dependent Selections
 At times the selection of one project depends on the
selection of another project
 Suppose Quemo’s catalytic converter could only be
purchased if the software was purchased
 The following constrain would force this to occur
X1 ≤ X2 or X1 – X2 ≤ 0
 If we wished for the catalytic converter and software
projects to either both be selected or both not be
selected, the constraint would be
X1 = X2 or X1 – X2 = 0

Fixed-Charge Problem Example
 Often businesses are faced with decisions involving a fixed
charge that will affect the cost of future operations
 Sitka Manufacturing is planning to build at least one new
plant and three cities are being considered in
◦ Baytown,Texas
◦ Lake Charles, Louisiana
◦ Mobile,Alabama
 Once the plant or plants are built, the company want to
have capacity to produce at least 38,000 units each year

 Fixed and variable costs for Sitka Manufacturing
SITE
ANNUAL
FIXED COST
VARIABLE COST
PER UNIT
ANNUAL
CAPACITY
Baytown, TX $340,000 $32 21,000
Lake Charles, LA $270,000 $33 20,000
Mobile, AL $290,000 $30 19,000

 We can define the decision variables as
X1 = 1 if factory is built in Baytown
0 otherwise
X2 = 1 factory is built in Lake Charles
0 otherwise
X3 = 1 if factory is built in Mobile
0 otherwise
X4 = number of units produced at Baytown plant
X5 = number of units produced at Lake Charles plant
X6 = number of units produced at Mobile plant

 The integer programming formulation becomes
Minimize cost = 340,000X1 + 270,000X2 + 290,000X3
+ 32X4 + 33X5 + 30X6
subject to X4 + X5 + X6 ≥ 38,000
X4 ≤ 21,000X1
X5 ≤ 20,000X2
X6 ≤ 19,000X3
X1, X2, X3 = 0 or 1;
X4, X5, X6 ≥ 0 and integer
 The optimal solution is
X1 = 0, X2 = 1, X3 = 1, X4 = 0, X5 = 19,000, X6 = 19,000
Objective function value = $1,757,000

Financial Investment Example
 Numerous financial applications exist with 0-1 variables
 Simkin, Simkin, and Steinberg specialize in recommending
oil stock portfolios for wealthy clients
 One client has the following specifications
◦ At least twoTexas firms must be in the portfolio
◦ No more than one investment can be made in a foreign oil
company
◦ One of the two California oil stocks must be purchased
 The client has $3 million to invest and wants to buy large
blocks of shares

 Oil investment opportunities
STOCK COMPANY NAME
EXPECTED ANNUAL
RETURN ($1,000s)
COST FOR BLOCK
OF SHARES ($1,000s)
1 Trans-Texas Oil 50 480
2 British Petroleum 80 540
3 Dutch Shell 90 680
4 Houston Drilling 120 1,000
5 Texas Petroleum 110 700
6 San Diego Oil 40 510
7 California Petro 75 900

 Model formulation
Maximize return = 50X1 + 80X2 + 90X3 + 120X4 + 110X5 + 40X6 + 75X7
subject to
X1 + X4 + X5 ≥ 2 (Texas constraint)
X2+ X3 ≤ 1 (foreign oil constraint)
X6 + X7 = 1 (California constraint)
480X1 + 540X2 + 680X3 + 1,000X4 + 700X5
+ 510X6 + 900X7 ≤ 3,000 ($3 million limit)
All variables must be 0 or 1

Using Excel to Solve the Simkin Example
 Solver input for Simkin’s 0-1 variables

 Complete Solver input for Simkin’s 0-1 integer
programming problem

 Excel solution to Simkin’s 0-1 integer programming
problem

Solved Problem 11-1
 Consider the 0-1 integer programming problem that follows:
Maximize = 50X1 + 45X2 + 48X3
Subject to: 19X1 + 27X2 + 34X3 ≤ 80
22X1 + 13X2 + 12X3 ≤ 40
X1, X2, X3 must be either 0 or 1

Solved Problem 11-1
 Additional constraints:
1- No more than two of the three variables can take on a
value equal to 1 in the solution.
2- Make sure that if X1 = 1, then X2 = 1 also.

Solved Problem 11-1
 The model is
Maximize = 50X1 + 45X2 + 48X3
Subject to: 19X1 + 27X2 + 34X3 ≤ 80
22X1 + 13X2 + 12X3 ≤ 40
X1 + X2 + X3 ≤ 2
X1 - X2 ≤ 0
X1, X2, X3 must be either 0 or 1

Integer Programming, Goal Programming, and Nonlinear Programming

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