Graduate Program in
Business Information Systems
LINEAR PROGRAMMING AND
APPLICATIONS
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OPERATIONS RESEARCH
2
 What is Operations Research?
 Collection of techniques used to
 allocate the scarce resources
 in the “best” –OPTIMAL – way!
 Best of what?
 We need an “objective” to be minimized or maximized!
 Profit, Cost, Utility, Delay, Distance, Flow, etc.
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Some applications
3
 Resource allocation
 Production and inventory planning
 Capacity Planning
 Workers and machine scheduling
 Investment planning
 Formulating marketing and military
strategies
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Some news about OR:
4
 A Wall Street Journal Article lists the use of
LP as one of the greatest technological
innovations of the past 1000 years.
 1975 Nobel Prize for economics: T.C.
Koopmans and L.V. Kantoprovich for their
contributions in the field
 1992 Nobel Prize: Harry Markowitz for his LP
based research.
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Basic Optimization Techniques
5
 Linear Programming
 Integer and Goal Programming
 Transportation, Assignment Models
 Network Models
 Nonlinear Programming
 Stochastic Programming
 Also Simulation
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LINEAR PROGRAMMING
6
 Most successful of all modern quantitative methods.
 Program here is not a computer code! It is a plan
that efficiently allocate limited resources to achive a
goal.
 Involves linear relationships, i.e. relations are in the
form of lines, planes!
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Basic LP Models: Product Mix
7
 Redwood Furniture Co.
Unit Requirements
Table
30
Chair
20
Amount
Available in
a Period
300
Labor (hrs)
5
10
110
Unit profit
$6
$8
Resource
Wood (ft)
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What is the optimal plan to max. Profit?
8
 Option 1: Allocate all resources to the more
profitable item.
Total quantity, profit? Any resource left?
 Option 2: Is it more profitable to produce
less chairs and more tables?
Linear Programming
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Formulating a Linear Problem
9
 Define variables:
Xt:
number of tables produced in a period
X c:
number of chairs produced in a period
 Define constraints:
30 X t  20 X c  300
( wood)
5 X t  10 X c  110 ( labor)
X t , X c  0 ( nonnegativ ity )
 Define Objective Function
Maximize
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Profit  6X t  8 X c
How is an LP solved?
10
 Graphical Method: Applicable to a maximum of two
decision variables.
 Simplex Method: Applicable to all LP.
Takes long to implement manually.
 Use softwares based on simplex and other
techniques.
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Graphical Solution
11
Xc
15
11
Optimal Solution:
Xt=4 tables, Xc=9 Chairs
Profit*=$96
Constraint 1
(4,9)
Constraint 2
Xt
10
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22
Basic LP Models: Feed Mix
12
 Two types of seeds are mixed to formulate
the wheat of wild birdseed.
Proportional Content
Nutritional
Item
Buckwheat
Sunflower
wheat
Total
Requirement
Fat
.04
.06
≥480 lb
Protein
.12
.10
≥1200 lb
Roughage
.10
.15
≤1500
Cost per lb
$.18
$.10
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LP Formulation
13
X b : Amount
of buckweat
(lb) in mixture
X s : Amount
of sunflower
(lb) in mixture
Minimize
Subject
to
Cost  0.18X
b
 0 . 10 X s
0.04X
b
 0 . 06 X s  480 (fat)
0.12X
b
 0 . 10 X s  1200 (protein)
0.10X
b
 0 . 15 X s  1500 (roughage)
X
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b
, X s  0 (nonnegati
vity)
Graphical Solution to Feed Mix Problem
14
Xs
Optimal Solution:
Xb*=3750 lb, Xs*=7500 lb
Cost*=$1425
(3750,7500)
(10000,0)
Fat
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Protein
(15000,0)
Xb
Roughage
Types of Feasible Regions
15
Bounded Feasible
Region
Unbounded Feasible
Region
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Cases in an LP: Infeasible Solution
16
Do all LP has an optimal Solution?
No feasible region
If an LP has no feasible region, then the solution is INFEASIBLE!
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Cases in an LP: Multiple Optima
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P  1 0 X 1  12 X 2
Maximize
Subject
Optimal Solutions:
Point(1): X1*=4 2/7, X2*=6 3/7
Point(2): X1*=6 6/7, X2*=4 2/7
P*=$120

X1 
4




*
X2 
6

*
to
2
 6
6

7  (1   )  7 
 2
3

4 
7
 7
5X 1  6 X 2  60
8X 1  4 X 2  72
3X 1  5 X 2  45
X
Infinite number of optimal
solutions exist in the form
(1)
(2)
0   1
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1
,X2  0
Cases in an LP: Unbounded Optimal Solution
18
Optimal Solution:
X1*=, X2*= 
P*= 

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Solving Linear Programs with a Spreadsheet
19
 Write out the formulation table
 Put the formulation table into a spreadsheet
 Use Excel’s Solver to obtain a solution
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Solution in the Excel Solver
20
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Applications of LP:Transportation Models
21
 Sporting goods company
Capacity Plants
Warehouses
Demand
Frankfurt
150
Seoul
NY
100
Tel Aviv
Phoenx
200
Yokohama
150
100
Juarez
300
200
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LP:Transportation Models (cont’d.)
22
What are the optimal shipping quantities from the
plants to the warehouses, if the demand has to be met
by limited capacities while the shipping cost is minimized?
Shipping Costs per pair of skis
Destination
From Plant
Frankfurt
NY
Phoenix
Yokohama
Juarez
$19
$7
$3
$21
Seoul
15
21
18
6
Tel Aviv
11
14
15
22
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LP:Transportation Models (cont’d.)
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Xij: Number of units shipped from plant i to warehouse j.
i=1,2,3 and j=1,2,3,4.
Minimize shipping costs=19X11+7X12+3X13+21X14
+15X21+21X22+18X23+6X24
+11X31+14X32+15X33+22X34
From
Plant
Destination
Frankfurt
NY
Phoenix
Yokohama
Capacity
Juarez
X11
X12
X13
X14
100
Seoul
X21
X22
X23
X24
300
Tel Aviv
X31
X32
X33
X34
200
Demand
150
100
200
150
600
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LP:Transportation Models (cont’d.)
24
subject to
#shipped from a plant can not exceed the capacity:
X11+X12+X13+X14≤100 (Juarez Plant)
X21+X22+X23+X24≤300 (Seoul Plant)
X31+X32+X33+X34≤200 (Tel Aviv Plant)
#shipped to a warehouse can not be less than the demand:
X11+X21+X31≥150 (Frankfurt)
X12+X22+X32≥100 (NY)
X13+X23+X33≥200 (Phoenix)
X14+X24+X34≥150 (Yokohama)
Nonnegativity
Xij ≥0 for all i,j.
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LP:Transportation Models (cont’d.)
25
Optimal Solution: Optimal cost=$6,250
Capacity
100
Warehouses
Plants
Juarez
50
Demand
Frankfurt
150
NY
100
Phoenx
200
Yokohama
150
100
300
Seoul
100
100
200
Tel Aviv
100
150
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LP: Marketing Applications
26
 How to allocate advertising budget between mediums
such as TV, radio, billboard or magazines?
Ex: Real Reels Co. Allocated ad. Budget=$100,000
Playboy
True
Esquire
10 million
6 million
4 million
Significant
Buyers
10%
15%
7%
Cost per ad
$10,000
$5,000
$6,000
1,000,000
900,000
280,000
Readers
Exposures per
ad
•No more than 5 ads in True and at least two ads in Playboy and Esquire
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LP: Marketing Applications (cont’d.)
27
X
p
:# adds in Playboy
X T :# adds in True
X E :# adds in Esquire
Max Total Exposure
10 , 000 X
p
 X
p
 0 . 9 X T  0 . 28 X E
 5 , 000 X T  6 , 000 X E  100 , 000
( budget )
XT  5
XP  2
XE  2
XT , X P, X E  0
Optimal
Solution
Not integer?
: X p  6 . 3 ads, X T  5 ads, X E  2 ads.
Optimal
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Exposure
 11.36 million
LP: Assignment Models
28
 Assignment of a set of workers to a set of
jobs
Time required to complete one job
Individual
Ann
Drilling
5min
Grinding
10min
Lathe
10min
Bud
10
5
15
Chuck
15
15
10
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LP: Assignment Models (cont’d.)
X ij
29
to job j
1, if worker i is assigned

 0 , if worker i is not assigned to job j
i , j  1, 2 , 3
Optimal
Solution
:
X X
 1 , X 10
 1 ,X
X 1
Min Total Time of Jobs  5 X 11  10
12
13
11
22
33
Total Time  20 min .
 10 X 21  5 X 22  15 X 23
 15 X 31  15 X 32  10 X 33
A worker can be assigned to a single job
X 11  X 12  X 13  1
( Ann )
X 21  X 22  X 23  1
( Budd )
X 31  X 32  X 33  1
( Chuck )
Each job is performed
by a single worker
X 11  X 21  X 31  1
( Drilling )
X 12  X 22  X 32  1
( Grinding )
X 13  X 23  X 33  1
( Lathe )
X ij  0
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LP: Diet Problem
30
 How much to use of each ingredient so that the
nutritional requirements are met in the cheapest
way?
 Ex: Feed Mix problem given at the beginning of the
lecture
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LP:Labor Planning
31
 Addresses staffing needs over a specific time period.
Hong Kong Bank of Commerce:
 12 Full time workers available, but may fire some.
 Use part time workers who has to work for 4 consequtive
hours in a day.
 Luch time is one hour between 11a.m. and 1p.m. shared by
full time workers.
 Total part time hours is less than 50% of the day’s total
requirement.
 Part-timers earn $4/hr (=$16/day) and full timers earn
$50/day.
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LP:Labor Planning (Cont’d.)
32
Time Period
9a.m.-10a.m.
Minimum labor required
10
10a.m.-11a.m.
11a.m.-noon
Noon-1p.m.
1p.m.-2p.m.
12
14
16
18
2p.m.-3p.m.
3p.m.-4p.m.
4p.m.-5p.m.
17
15
10
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LP:Labor Planning (cont’d.)
33
F : # Full time tellers per day
Pi : # Part time
tellers who start work
Min Daily Personnel
Cost  $ 50 F  $ 16  Pi
F  P1
 10
F  P1  P2
 12
0 . 5 F  P1  P2  P3
 14
0 . 5 F  P1  P2  P3  P4
 16
F
 P2  P3  P4  P5  18
F
 P3  P4  P5  17
F
 P4  P5  15
F
 P5  10
F  12
4  Pi  0 . 5 (10  12  14  ...  10 )
F , Pi  0
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at time slot i , i  1, 2 ..., 5 .