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How to optimally allocate scarce resources!
1
An Introduction to Linear Programming
(LP)
Please hold your applause until the end.
What is a Linear Programming
2
A linear program (LP) is an optimization problem consisting
of a function to be maximized or minimized subject to one or
more limitations (called constraints) on the variables of the
function.
Both the optimization function (called the objective function)
and the constraints have only linear relationships.
George Dantzig
Inventor of linear programming and the simplex algorithm
3
In 1947 Dantzig made the contribution to
mathematics for which he is most famous, the
simplex method of optimisation. It grew out of
his work with the U.S. Air Force known as
"programming", a military term that, at that
time, referred to plans or schedules for
training, logistical supply or deployment of
men. Dantzig mechanised the planning process
by introducing "programming in a linear
structure" Born: 8 Nov 1914
Died: 13 May 2005
The Road Ahead
4
Example LP
The Graphical Solution
LP Defined
Application and Examples
Solving a LP
The Computer Way
The Algebraic Way
The Simplex Way
Our Very First Example
5
The Opti Mize Company manufactures two products that compete
for the same (limited) resources. Relevant information is:
Product A B Available resources
Labor-hrs/unit 1 2 20 hrs/day
Machine hrs/unit 2 2 30 hrs/day
Cost/unit $6 $20 $180/day
Profit/unit $5 $15
The Model
6
Let X = number of units of product A to manufacture
Y = number of units of product B to manufacture
Max Profit = z = 5 X + 15 Y
subject to:
X + 2 Y <= 20 (labor-hours)
2 X + 2 Y <= 30 (machine hours)
6 X + 20Y <= 180 ($ - budget)
X >= 0, Y >= 0
The Graphical Solution
7
X
Y
5 10 15 25 20 30
5
10
20
15
X + 2 Y = 20
The Graphical Solution
8
X
Y
5 10 15 25 20 30
5
10
20
15
X + 2 Y = 20
9
6X + 20Y = 180
The Graphical Solution
9
X
Y
5 10 15 25 20 30
5
10
20
15
X + 2 Y = 20
2X + 2Y = 30
9
6X + 20Y = 180
The Graphical Solution
10
X
Y
5 10 15 25 20 30
5
10
20
15
X + 2 Y = 20
2X + 2Y = 30
9
6X + 20Y = 180
The feasible region
The Graphical Solution (continued)
11
X
Y
5 10 15 25 20 30
10
20
15
9 Z = 5X + 15Y = 30
2
6
The Graphical Solution (continued)
12
X
Y
5 10 15 25 20 30
10
20
15
9 Z = 5X + 15Y = 30
2
6
4
Z = 5X + 15Y = 60
12
The Graphical Solution (continued)
13
X
Y
5 10 15 25 20 30
10
20
15
9 Z = 5X + 15Y = 30
2
6
4
Z = 5X + 15Y = 60
12
(5, 7.5)
Z = 5 (5) + 15 (7.5) = 137.5
The Graphical Solution
Alternate Approach
14
X
Y
5 10 15 25 20 30
5
10
20
15
9
(x = 5, y = 7.5; z = 137.5)
(x = 10, y = 5; z = 125)
Z = 5X + 15Y
(x = 15, y = 0; z = 75) (x = 0, y = 9; z = 135)
(x = 0, y = 0; z = 0)
15
This is powerful stuff!
Can we see another
example followed by a
description of the
general model?
Sure can ……..
A Minimization Problem
16
The Classical Diet Problem
The Problem
17
Mr. U. R. Fatte has been placed on a diet by his Doctor (Dr. Ima
Quack) consisting of the foods shown below. The doctor warned
him to insure proper consumption of nutrients to sustain life.
Relevant information is:
Nutrients Beer Pizza Monthly Requirement
Vitamin A 2 mg/oz 3 mg/oz 3500 mg (min rqmt)
Protein 6 mg/oz 2 mg/oz 7000 mg (min rqmt)
Fat 4g/oz 2g/oz 8000 grams (max rqmt)
cost/oz $0.50 $0.20
The Mathematical Model
18
Let X = ounces of beer consumed per week
Y = ounces of pizza consumed per week
Min cost = z = 0.5X + 0.2Y
subject to:
2X + 3Y >= 3500
6X + 2Y >= 7000
4X + 2Y <= 8000
X, Y >= 0
The Boundary Equations
19
X
Y
1000
3000
2000
3000 2000
1000
4000 6x + 2y = 7,000
2x + 3y = 3,500
4x + 2y = 8,000
2X + 3Y >= 3500
6X + 2Y >= 7000
4X + 2Y <= 8000
Feasible Region
20
X
Y
1000
3000
2000
3000 2000
1000
4000
Feasible Region with Objective Function
21
X
Y
1000
3000
2000
3000 2000
1000
4000 z = .5X + .2Y
z = .5X + .2Y = $800
(0,4000) and (1600,0)
Optimal Solution
22
X
Y
1000
3000
2000
3000 2000
1000
4000
6 2 7,000
2 3 3,500
6 2 7,000
6 9 10,500
7 3,500 * 500
7,000 2* 1,000
6
* .5 .2 $600
x y
x y
x y
x y
y or y
yx
z x y
Optimal Solution – alternate approach
23
X
Y
1000
3000
2000
3000 2000
1000
4000
(x = 0, y = 4000, z = $800)
(x = 0, y = 3500, z = $700)
(x = 1000, y = 500, z = $600)
(x =2000, y = 0, z = $1,000)
(x = 1,750, y = 0, z = $875)
z = .5X + .2Y
Surgeon General’s Warning
24
The minimum cost diet found in solving this problem is illustrative only. One should not attempt to
follow this diet for any prolonged period of time.
The General Diet Problem
25
Minimize the "cost of the menu" subject to the nutrition requirements: eat enough but not too much of Vitamin A . . eat enough but not too much Vitamin C
Eat at least a certain minimum number of servings of beef but not more than the maximum number of servings of beef you want. . .
Eat at least a certain minimum number of servings of carrots but not more than the maximum number of servings of carrots you want…
Special student exercise: solve the diet problem found on
The supplement homework list of the course website!
The General LP Model
26
Max or Min z = c1 x1 + c2x2 + . . . + cnxn
subject to:
A11x1 + A12x2 + . . . + A1nxn <= b1
A21 x1 + A22x2 + . . . + A2nxn <= b2
.
.
Am1x1 + Am2x2 + . . . + Amnxn <= bm
x1, x2, . . . xn >= 0
Objective
Function
Constraints
xj = decision variables or activity levels
cj = profit or cost coefficient
Aij = technology coefficient
bi = resource capacities (right hand side values)
Assumptions
27
Deterministic
all input data (parameters) are known and constant
no statistical uncertainty
Linear
cost or profit is additive and proportional to the
activity levels
output or resources consumed are additive and
proportional to the activity levels
Non-integer
variables (activity levels) are continuous
Solution Procedures
28
Graphical
two or three variables only
Algebraic
solve systems of equations for corner points
Simplex algorithm
numerical, iterative approach
Ellipsoid
theoretical importance more than applied
29
Can we see
more examples
of this LP thing?
Please!!!!!
Let’s do it.
A Blending Problem
30
The B. A. Nutt Company sells mixed nuts of two quality levels.
The expensive mix should not contain more than 25% peanuts
nor less than 40% cashews. The cheap mix should not have more
than 60% peanuts and no less than 20% cashews. Cashews cost 50
cents a pound and peanuts cost 20 cents a pound. The expensive
mix sells for 80 cents a pound and the cheap mix for 40 cents a
pound. What should the blend of each mix be in order to maximize
profit. The company has $100 a day with which to purchase
nuts.
The Model
31
Let x1 = pounds of cashews in expensive mix
x2 = pounds of peanuts in expensive mix
y1 = pounds of cashews in cheap mix
y2 = pounds of peanuts in cheap mix
Max z = .80 (x1 + x2) + .40 (y1 + y2) - .5(x1+y1) - .2(x2 + y2)
subject to:
.5 (x1 + y1) +.2(x2 + y2 ) <= 100
x2 / (x1+x2) <= .25 y2/(y1 + y2) <= .6
x1 / (x1+x2) >= .40 y1/(y1 + y2) >= .20
Re-formulate the Model
32
Max z = = .3x1 + .6x2 -.1y1 + .2y2
subject to:
.5x1 +.2x2 +.5y1 +.2 y2 <= 100
-.25x1 + .75x2 <= 0
-.60y1 + .40y2 <= 0
.60x1 - .40x2 >= 0
.80y1 - .20y2 >= 0
A Marketing Example
33
The I. B. Adman Advertising Company is planning a large media blitz
covering television, radio, and magazines to sell management science
to the public. The company’s objective is to reach as many people as
possible. Results of a market survey show:
Television
Day time Prime Time Radio Magazines
cost per unit $40,000 75,000 30,000 15,000
# people 400,000 900,000 500,000 200,000
# business 300,000 400,000 200,000 100,000
The company has a budget of $800,000 to spend on the campaign. It
requires at least two million exposures among the business community.
Television must be limited to $500,000, and at least 3 units of day time and
2 units of prime time must be purchased. Advertising units on both radio and
magazines should be between 5 and 10.
The Model
34
Let x1, x2, x3, and x4 be the number of advertising units
bought in daytime TV, primetime TV, radio and magazines.
Max z = 400x1 + 900x2 + 500x3 + 200x4 (in thousands)
subject to:
40,000x1 + 75,000x2 + 30,000x3 + 15,000x4 <= 800,000
300,000x1 + 400,000x2 + 200,000x3 + 100,000x4 >= 2,000,000
40,000x1 + 75,000x2 <= 500,000
x1 >= 3; x2 >= 2; 5 <= x3 <= 10; 5 <= x4 <= 10
35
The Transportation Problem
A company manufactures a single product at
each of m factories. Factory i has a capacity of
Si per month. There are n warehouses receiving
this product. The demand at warehouse j is Dj.
It cost factory i, cij dollars to ship one unit to
warehouse j. How many units should each factory
send to each warehouse in order to minimize the
total transportation costs?
This is a really neat problem.
36
The LP Formulation
Min z c x
subject to
x S i m
x D j n
x
ij ij
j
n
i
m
ij
j
n
i
ij j
i
m
ij
11
1
1
1 2
1 2
0
:
, ,...,
, ,...,
Let xij = the number of units sent from factory i to warehouse j
Job-Training
37
The Never-Say-Die Life Insurance Company hires and
trains a large number of salespersons each month to replace
those who have departed. Trained salespersons must be used
to train new salespersons. Training takes one month and there
is a 20 percent attrition rate by the end of the month. While a
salesperson is training a new employee, that person cannot be
used in the field selling insurance. The monthly demand for
experienced salespeople is:
Month Demand (in the field)
January 100
February 150
March 200
April 225
May 175
Trainees receive $400 per month while it costs the company $850 per
month for an experienced salesperson. Ten percent of all experienced
salespeople will leave the company by the end of each month.
The Model
38
Let xi = the number of trainees during month i
yi = the number of experienced salespeople in month i
Min z = 1250 (x1 + x2 + x3 + x4 + x5)
subj to: y1 - x1 >= 100
y2 - x2 >= 150
y3 - x3 >= 200
y4 - x4 >= 225
y5 - x5 >= 175
yi = .9yi-1 + .8xi-1 for i = 1, 2, 3, 4, 5
example: yfeb = .9 yjan + .8xjan
A Large Department Store Somewhere in
the Midwest
Problem Set 2.3F 6
Store operates 7 days a week
Min number of salespersons required is 12 for Mon., 18 for Tues.,
20 for Wed., 28 for Thurs., 32 for Fri., 40 for Sat. and Sun.
Each salesperson works 5 days a week with 2 consecutive days off
What is the minimum number of salespersons and how should
their days off be allocated?
The Formulation
Let xi = the number of salesperson that start their week on day i (i =
M, T, W, Th, F, Sa, Su)
. . 12
18
20
28
32
40
40
M T W TH F Sa Su
M TH F Sa Su
M T F Sa Su
M T W Sa Su
M T W Th Su
M T W Th F
T W Th F Sa
W Th F Sa Su
Min z x x x x x x x
s t x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
x x x x x
Monday
Tuesday
Wednesday
Thursday
Friday
Saturday
Sunday
Urban Renewal
41
ORsville has severe budget shortage City council votes to improve the tax base
condemning inner-city housing replacing with new development
Up to 300 substandard houses may be demolished each house occupies .25 acre lot and cost $2,000 to demolish
Lot sizes for single, double, triple, and quadruple family unit are .18, .28, .4, and .5 acre respectively
15% of available acreage need for utilities, streets, green space, etc. Triple and quadruple units must be at least 25% of total, single at least
20%, and double at least 10% Tax levied per single, double, triple, and quadruple is $1,000, $1,900,
$2,700, and $3,400 respectively Construction costs per single, double, triple, and quadruple is $50,000,
$70,000, $130,000, and $160,000 respectively Max of $15 million available for construction
The Decision Variables
42
x1 = number of units of single-family homes
x2 = number of units of double-family homes
x3 = number of units of triple-family homes
x4 = number of units of quadruple-family homes
x5 = number of old homes to be demolished
A city council member tosses
a coin to see if a particular
house is to be demolished.
The Model
43
1 2 3 4
1 2 3 4 5
5
1 1 2 3 4
2 1 2 3 4
3 4 1 2 3 4
1 2 3 4 5
1 2 3 4 5
1000 1900 2700 3400
:
.18 .28 .4 .5 .25 .85
300
.2
.1
.25
50 70 130 160 2 15,000
, , , , 0
Max z x x x x
subject to
x x x x x
x
x x x x x
x x x x x
x x x x x x
x x x x x
x x x x x
tax collection
available acreage
max nbr to demolish
minimum nbr of each home type
budget in $1,000
The Model’s Solution
44
x1 = 35.83
x2 = 98.53
x3 = 44.79
x4 = 0
x5 = 244.49
z = $343,965
I will not live in a house that is only .83
complete!
Is it okay if we work more of these on our
own?
Gee, this LP stuff can solve just about any
problem.
Computer Solutions
46
LINDO
LINGO
Excel Solver
What’s Best (Excel add-on)
AMPL
TORA
Can you show us how
the computer solves
these LP problems?
More LP
The Simplex Algorithm
47
Next
Don”t touch that drop card!
The fun never stops!
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