Graduate Program in
Business Information
Systems
Integer and Goal Programming
Aslı Sencer
Integer Program

Structurally like the LP models where the
solutions must be integers!
Ex:



Production order quantities
Number of items sent from plants to warehouses
Binary decisions



produce or do not produce
Assign staff i to job j or not
Locate facility i in location j or not
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Ex:Redwood Furniture Model
(revisited)
Max
6 XT  8 XC
subject to 30 X T  20 X C  310 (wood)
5 X T  10 X C  113 (labor)
X T ,X C  0
Optimal solution:
X T  4.2
*
X C  9 .2
*
P *  98.8
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The Branch and Bound Method

An effective search procedure that involves solving
a succession of carefully formulated LP’s.

If the optimal solution of the parent program
includes fractions, then two descendent LP’s are
developed and solved to force the noninteger
variable to be an integer.

The optimal objective function of the parent program
constitues a bound on the optimal objective function
value of the descendent LP’s.
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Graphical Solution
Xc
15
Problem 2
Problem1
11
Problem 3
Xt
4
5
10
22
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Branch and Bound Solution
Problem 2
XT  4
Problem 1
XC  9
Xt=4
Xc=9
P=96
Xt=4
Xc=9.3
P=98.4
Xt=4.2
Xc=9.2
P=98.8
OPTIMAL SOLUTION
X C  10
XT  5
Problem 3
Xt=5
Xc=8
P=94
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Problem 4
Problem 5
Xt=2.6
Xc=10
P=95.6
6
When do we round off the LP?

For problems having small number of
variables, IP is solved in a short time.

Bigger problems require a lot of time.

If no integer solution has been found, a
rounded LP may be used.

Which solution to be used is judgemental.
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Solving Integer Programs with
a Spreadsheet

When using Excel Spreadsheet, it is sufficient
to add a new constraint and define the
decision variables as integers.

You can also use WinQSB or QuickQuant to
solve LP and IP problems.
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Ex: Maui Miser Car Rentals








Budget $500,000.
Three types of cars: Economy vans, compact cars,
large cars
Income/vehicle: $15,000, $7,600, $10,600
Costs/vehicle: $25,000, $15,000, $21,000
At least 25 vehicles
At least 5 vans, at least 5 large cars, at least 12
passenger cars
# large cars can not exceed the #compacts
Average daily rental rate should be less than $39
where Daily rental prices/vehicle: $50, $30, $40
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Ex: Formulation
X V :# economy vans to purchase
X C :# compact cars to purchase
X L :# large cars to purchase
Max Income  15,000 X V  7,600 X C  10,600 X L
subject to 25,000 X V  15,000 X C  21,000 X L  500,000
X V  X C  X L  25
XV
5
XL  5
X C  X L  12
 XC  X L  0
11 X V  9 X C  X L  0
X V , X C , X L are integers
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Optimal Solution
LP Solution
IP Solution
X V  8.92
XV  8
X C  11.46
X C  13
XL  5
XL  5
P  $273,953.81
P  $271,800
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Goal Programming
Maximize profit
OR
Minimize cost
Single Goal
Sometimes we have Multiple Goals.
Minimize Cost AND Maximize Customer Satisfaction
Maximize Return AND Minimize Risk
What is the optimal balance in attaining the conflicting goals?
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Basic Idea

It is impossible to meet all goals
simultaneously.

The focus aims at achieving certain targets
for each goal.

The overall objective is to find the solution
that collectively minimizes deviations from
these targets.
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Ex:Goal Programming in
Redwood Furniture Co.



Objectives
Maximize Profit
Maximize Revenue to maintain marketshare growth
Maximize training time to increase work-force productivity
Target
Table
Chair
Profit/unit
$6
$8
$90
Revenue/unit
$50
$25
$450
Train. time/unit
1hr.
3hr.
30hr.
Each LP has a different optimal solution.
An idea is to set targets for the goals.
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LP with Goal Targets
Idea 1: We can treat targets as the lower bounds on the goals
Goal 1 : 6 X T  8 X C  90 (Profit target  90)
Goal 2 : 50 X T  25 X C  450
Goal 3 : X T  3 X C  30
(Revenue target  450)
(Training target  30)
Warning: When these constraints are added to the
Redwood Product Mix Formulation there may be no
feasible solution satisfying the minimum requirements of
all goals!
What should we do then? Any comments?
Answer: Try to lower the targets
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Goal Deviations
Idea 2: We can allow some goals to be partially met!


Compared to LP, goal programming is flexible.
Optimal balance between goals can be achieved in
several ways.
Goal Deviation: Amount of deviation from the goal
target is a variable to be minimized.
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Defining Goal Deviations
Goal 1 : 6 X T  8 X C  90 ( profit )
Yp: The amount by which
profit deviates from the
target level


Let YP  (6 X T  8 X C )  90,
then (6 X T  8 X C )  YP  90


Here , YP  YP  YP , YP , YP  0


When Yp is positive,
YP  YP , YP  0
When Yp is negative
YP  YP , YP  0

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Goal Deviation Constraints


6 X T  8 X C  (YP  YP )  90


50 X T  25 X C  (YR  YR )  450


X T  3 X C  (YT  YT )  30
Regular Constrains:
30 X T  20 X C  300
5 X T  10 X C  110
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Goal Programming Objective





Minimize C  0 XT  0 X C  0YP  1YP  0YR  2YR  0YT  0.5YT

It costs $1 to fall a dollar below the profit target. However exceeding
the profit target is desirable.



6 X T  8 X C  (YP  YP )  90  YP  1

89


Falling below revenue target lowers market share, lost earnings is $2
for a dollar below revenue target.
Training time upto 30hrs enhances future efficiency. It costs $0.50
per hour obtained. Beyond 30hrs. training effect is negligable.
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
Ex: Optimal Solution


XT  6
YP  0
XC  6
YP  6 YR  0

YR  0


YT  0

YT  6
C 9
Profit and Training levels are $6 and 6hrs. below their targets.
Revenue target is met.
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Ex: Maui Miser Case (revisited)
Achieve annual rental income of at least
$300,000 from fleet addition
 Spend as close to as possible to $500,000
 Do not exceed average daily rental of $39 for
the vehicles in the fleet addition.
Penalty Costs:
$1 for each dollar violation of goals 1 and 2.
$10,000 for each dollar violation of goal 3.

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Ex: Maui Miser Case
GP Formulation



Min C  Y1  Y2  Y2  10,000Y3



15,000 X V  7,600 X C  10,600 X L  Y1  Y1  300,000




25,000 X V  15,000 X C  21,000 X L  Y2  Y2  500,000
11 X V  9 X C  X L  Y3  Y3  0
X V  X C  X L  25
XV
5
XL  5
X C  X L  12
 XC  X L  0
all variables are positive
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Ex: Maui Miser Case
GP Optimal Solution
GP Solution
X V  8.923
X C  11.462
XL  5

Y1  26,046.154
C  $26,046.154
IGP Solution
XV  8
X C  13
XL  5

Y1  28,200

Y3  24
C  $28,200
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