02_Intro to LP.pdf

Deterministic Operations Research
OR 6205
Module 2: Introduction
to Linear Programming

Objectives
! Introduce modeling components in Linear
Programs
! Solve smaller-size linear programs manually by
graphic methods
! Formulate and solve LP in Excel
! Introduce Lingo and Python for optimization
2

2
What Is a Linear Programming
Problem?
„ Linear Programming (LP) is a tool for solving
optimization problems.
„ Linear programming problems involve
important terms that are used to describe
linear programming.

3
E 1:
Woodcarving
„ , .,
and trains.
† Each soldier built:
„ Sell for $27 and uses $19 worth of raw materials.
„ /
$14.
„ Requires 2 hours of finishing labor.
„ Requires 1 hour of carpentry labor.
† Each train built:
„ Sell for $21 and used $9 worth of raw materials.
„ /
by $10.
„ Requires 1 hour of finishing labor.
„ Requires 1 hour of carpentry labor.
0

4
Ex. 1 - continued
„ Each week Giapetto can obtain:
† All needed raw material.
† Only 100 finishing hours.
† Only 80 carpentry hours.
„ Demand for the trains is unlimited.
„ At most 40 soldiers are bought each week.
„ Giapetto wants to maximize weekly profit
(revenues costs).
„ F
situation that can be used maximize weekly
profit.

5
Example 1: Solution
„ The Giapetto solution model incorporates the
characteristics shared by all linear
programming problems.
† Decision variables should completely describe the
decisions to be made.
„ x1 = number of soldiers produced each week
„ x2 = number of trains produced each week
† The decision maker wants to maximize (usually
revenue or profit) or minimize (usually costs) some
function of the decision variables. This function to
maximized or minimized is called the objective
function.
„ For the Giapetto problem, fixed costs do not depend
upon the the values of x1 or x2.

6
Ex. 1 - Solution continued
„
terms of the decision variables x1 and x2:
Weekly profit =
weekly revenue weekly raw material costs the
weekly variable costs = 3x1 + 2x2
„ , x1 and x2
to maximize weekly profit. The variable z
denotes the objective function value of any LP.
„
Maximize z = 3x1 + 2x2
„ The coefficient of an objective function variable
is called an objective function coefficient.

7
„ As x1 and x2 ,
function grows larger.
„ For Giapetto, the values of x1 and x2 are
limited by the following three restrictions
(often called constraints):
† Each week, no more than 100 hours of finishing time
may be used. (2 x1 + x2 100)
† Each week, no more than 80 hours of carpentry time
may be used. (x1 + x2 80)
† Because of limited demand, at most 40 soldiers
should be produced. (x1 40)

8
„ The coefficients of the decision variables in the
constraints are called the technological coefficients.
The number on the right-hand side of each constraint is
-hand side (or rhs).
„ To complete the formulation of a linear programming
problem, the following question must be answered for
each decision variable.
† Can the decision variable only assume nonnegative values,
or is the decision variable allowed to assume both positive
and negative values?
„ If the decision variable can assume only nonnegative
values, the sign restriction xi 0 .
„ If the variable can assume both positive and negative
values, the decision variable xi is unrestricted in sign
(often abbreviated urs).

9
„ For the Giapetto problem model, combining the
sign restrictions x1 0 x2 0
objective function and constraints yields the
following optimization model:
Max z = 3x1 + 2x2 (objective function)
Subject to (s.t.)
2 x1 + x2 ≤ 100 (finishing constraint)
x1 + x2 ≤ 80 (carpentry constraint)
x1 ≤ 40 (constraint on demand for soldiers)
x1 ≥ 0 (sign restriction)
x2 ≥ 0 (sign restriction)

10
„ A function f(x1, x2, , xn of x1, x2, , xn is a
linear function if and only if for some set of
constants, c1, c2, , cn, f(x1, x2, , xn) = c1x1
+ c2x2 + + cnxn.
„ For any linear function f(x1, x2, , xn) and any
number b, the inequalities f(x1, x2, , xn) b
and f(x1, x2, , xn) linear
inequalities.

11
„ A linear programming problem (LP) is an
optimization problem for which we do the
following:
† Attempt to maximize (or minimize) a linear function
(called the objective function) of the decision
variables.
† The values of the decision variables must satisfy a set
of constraints. Each constraint must be a linear
equation or inequality.
† A sign restriction is associated with each variable.
For any variable xi, the sign restriction specifies
either that xi must be nonnegative (xi 0) xi
may be unrestricted in sign.

12
„ The fact that the objective function for an LP
must be a linear function of the decision
variables has two implications:
1. The contribution of the objective function from each
decision variable is proportional to the value of the
decision variable. For example, the contribution to
the objective function for 4 soldiers is exactly fours
times the contribution of 1 soldier.
2. The contribution to the objective function for any
variable is independent of the other decision
variables. For example, no matter what the value of
x2, the manufacture of x1 soldiers will always
contribute 3x1 dollars to the objective function.

13
„ Analogously, the fact that each LP constraint
must be a linear inequality or linear equation
has two implications:
1. The contribution of each variable to the left-hand side
of each constraint is proportional to the value of the
variable. For example, it takes exactly 3 times as
many finishing hours to manufacture 3 soldiers as it
does 1 soldier.
2. The contribution of a variable to the left-hand side of
each constraint is independent of the values of the
variable. For example, no matter what the value of
x1, the manufacture of x2 trains uses x2 finishing
hours and x2 carpentry hours

14
„ The first item in each list is called the
Proportionality Assumption of Linear
Programming.
„ The second item in each list is called the
Additivity Assumption of Linear
Programming.
„ The divisibility assumption requires that
each decision variable be permitted to assume
fractional values.

15
„ The certainty assumption is that each
parameter (objective function coefficients,
right-hand side, and technological coefficients)
are known with certainty.
„ The feasible region of an LP is the set of all
sign restrictions.
„ For a maximization problem, an optimal
solution to an LP is a point in the feasible
region with the largest objective function
value.
† Similarly, for a minimization problem, an optimal
solution is a point in the feasible region with the
smallest objective function value.

16
The Graphical Solution to a Two-
Variable LP Problem
„ Any LP with only two variables can be solved
graphically.
† The variables are always labeled x1 and x2 and the
coordinate axes the x1 and x2 axes.
Satisfies 2x1 + 3x2 ≥ 6
Satisfies 2x1 + 3x2 ≤ 6 X1
1 2 3 4
1
2
3
4
X2
-1
-1

17
„ Since the Giapetto LP has two variables, it may
be solved graphically.
„ The feasible region is the set of all points
satisfying the constraints
2 x1 + x2 100 ( )
x1 + x2 80 ( )
x1 40 ( )
x1 0 ( )
x2 0 ( )

18
„ The set of points satisfying the Giapetto LP is
bounded by the five sided polygon DGFEH.
Any point on or in the interior of this polygon
(the shade area) is in the feasible region.
X1
X2
10 20 40 50 60 80
20
40
60
80
100
finishing constraint
carpentry constraint
demand constraint
z = 60
z = 100
z = 180
Feasible Region
G
A
B
C
D
E
F
H

19
„ Having identified the feasible region for the
Giapetto LP, a search can begin for the optimal
solution which will be the point in the feasible
region with the largest z-value.
„ To find the optimal solution, graph a line on
which the points have the same z-value. In a
max problem, such a line is called an isoprofit
line while in a min problem, this is called the
isocost line. The figure shows the isoprofit
lines for z = 60, z = 100, and z = 180

20
„ A constraint is binding if the left-hand and
right-hand side of the constraint are equal
when the optimal values of the decision
variables are substituted into the constraint.
† In the Giapetto LP, the finishing and carpentry
constraints are binding.
„ A constraint is nonbinding if the left-hand side
and the right-hand side of the constraint are
unequal when the optimal values of the
decision variables are substituted into the
constraint.
† In the Giapetto LP, the demand constraint for
wooden soldiers is nonbinding since at the optimal
solution (x1 = 20), x1 < 40.

21
„ A set of points S is a convex set if the line
segment jointing any two pairs of points in S is
wholly contained in S.
„ For any convex set S, a point p in S is an
extreme point if each line segment that lines
completely in S and contains the point P has P
as an endpoint of the line segment.
„ Extreme points are sometimes called corner
points, because if the set S is a polygon, the
extreme points will be the vertices, or corners,
of the polygon.
† The feasible region for the Giapetto LP will be a
convex set.

22
Example 2 : Dorian Auto
„ Dorian Auto manufactures luxury cars and
trucks.
„ The company believes that its most likely
customers are high-income women and men.
„ To reach these groups, Dorian Auto has
embarked on an ambitious TV advertising
campaign and will purchase 1-mimute
commercial spots on two type of programs:
comedy shows and football games.

23
Ex. 2: continued
„ Each comedy commercial is seen by 7 million
high income women and 2 million high-income
men and costs $50,000.
„ Each football game is seen by 2 million high-
income women and 12 million high-income
men and costs $100,000.
„ Dorian Auto would like for commercials to be
seen by at least 28 million high-income women
and 24 million high-income men.
„ Use LP to determine how Dorian Auto can meet
its advertising requirements at minimum cost.

24
Example 2: Solution
„ Dorian must decide how many comedy and football ads
should be purchased, so the decision variables are
† x1 = number of 1-minute comedy ads
† x2 = number of 1-minute football ads
„ Dorian wants to minimize total advertising cost.
„ D
min z = 50 x1 + 100x2
„ Dorian faces the following the constraints
† Commercials must reach at least 28 million high-income
women. (7x1 + 2x2 28)
† Commercials must reach at least 24 million high-income
men. (2x1 + 12x2 24)
† The sign restrictions are necessary, so x1, x2 0.

25
Ex. 2 Solution continued
„ Like the Giapetto LP, The Dorian LP has a
convex feasible region.
„ The feasible region for the Dorian problem,
however, contains points for which the value of
at least one variable can assume arbitrarily
large values.
„ Such a feasible region is called an unbounded
feasible region.

26
„ To solve this LP graphically begin by graphing
the feasible region.
X1
X2
2
4
6
8
10
12
14
2 4 6 8 10 12 14
z = 600
z = 320
A C
D
E
B
Feasible
Region
(unbounded)
High-income women constraint
High-income men constraint

27
„ Since Dorian wants to minimize total
advertising costs, the optimal solution to the
problem is the point in the feasible region with
the smallest z value.
„ An isocost line with the smallest z value passes
through point E and is the optimal solution at
x1 = 3.4 and x2 = 1.4.
„ Both the high-income women and high-income
men constraints are satisfied, both constraints
are binding.

28
 Does the Dorian model meet the four
assumptions of linear programming in reality?
 The Proportionality Assumption is violated because at
a certain point advertising yields diminishing returns.
 Even though the Additivity Assumption was used in
writing: (Total viewers) = (Comedy viewer ads) +
(Football ad viewers) many of the same people might
view both ads, double-counting of such people would
occur thereby violating the assumption.
 The Divisibility Assumption is violated if only 1-
minute commercials are available. Dorian is unable to
purchase 3.6 comedy and 1.4 football commercials.
 The Certainty assumption is also violated because
there is no way to know with certainty how many
viewers are added by each type of commercial.

29
Special Cases
„ The Giapetto and Dorian LPs each had a unique
optimal solution.
„ Some types of LPs do not have unique
solutions.
† Some LPs have an infinite number of solutions
(alternative or multiple optimal solutions).
† Some LPs have no feasible solutions (infeasible LPs).
† Some LPs are unbounded: There are points in the
feasible region with arbitrarily large (in a
maximization problem) z-values.
„ The technique of goal programming is often
used to choose among alternative optimal
solutions.

30
„
empty, resulting in an infeasible LP.
„ Because the optimal solution to an LP is the
best point in the feasible region, an infeasible
LP has no optimal solution.
„ For a max problem, an unbounded LP occurs if
it is possible to find points in the feasible
region with arbitrarily large z-values, which
corresponds to a decision maker earning
arbitrarily large revenues or profits.

31
„ For a minimization problem, an LP is unbounded if there
are points in the feasible region with arbitrarily small z-
values.
„ Every LP with two variables must fall into one of the
following four cases.
† The LP has a unique optimal solution.
† The LP has alternative or multiple optimal solutions: Two or
more extreme points are optimal, and the LP will have an
infinite number of optimal solutions.
† The LP is infeasible: The feasible region contains no points.
† The LP unbounded: There are points in the feasible region
with arbitrarily large z-values (max problem) or arbitrarily
small z-values (min problem).

Example: Solve problem by graphic method

Lesson 2: Formulating and
solving LP models on a
spreadsheet

Capacitated Plant Location Model
qDecide in which regions to locate facilities and
how to allocate plant capacity across regions
qTrade-off reduced transportation times and
tariffs from more regional facilities vs. reduced
facilities cost for fewer centralized facilities

Potential
Sources Destinations
Limited production capacity Demand to be satisfied
K1
K2
K3
K4
D1
D2
D3

qInput data:
qm customer sites, with annual demand Dj
qn possible plant location sites, with annual capacity Ki
qfi = annualized fixed cost of opening facility i
qcij = cost to produce at i and ship to j
qDecision variables:
qyi = 1 if a plant is located at site i, 0 otherwise
qxij = quantity shipped from plant site i to customer j

Example:
qThere are two potential plant locations, in
Boston and Los Angeles
qAnnualized cost of opening plant in Boston is
$150K, in Los Angeles is $100K
qThere are three customers, located in
Albany, Chicago and San Diego

qLet XBA, XBC, XBS, XLA, XLC, XLS be amounts shipped
from each plant to each customer
qLet YB and YL indicate whether or not open plant
in Boston and Los Angeles
qFind these values to minimize total cost:

qMust satisfy all demand at each customer:

qTotal shipments out of plant must not exceed
available capacity at plant:

qShipment amounts cannot be negative:
qPlant location decision variables must be
either 0 or 1:

qHint for building these models in solver:
qThese models must be linear programs
qDo not use “if” function in Excel
• Example: If we build a plant at i then the capacity is Ki, otherwise
the capacity is 0
qSolver may not work with these “if” statements
• Instead should use binary variables:
– Capacity at i = Ki yi, where yi = 1 if build plant,
yi = 0 if do not build plant

qA more complex example:
qSunOil is a manufacturer of petrochemical products
with worldwide sales
qDivides the world into five regions:
• N. America, S. America, Europe, Asia and Africa
qMust satisfy all demand in each region

• Need to decide where to locate facilities, how
many facilities to have, and how large facilities
should be (capacity)
• Could potentially locate a facility in each region,
regions have different production, transportation
and facility costs
• Facilities may be small (capacity of 10 million units)
or large (capacity of 20 million units)

qUsing solver in Excel to solve SunOil’s
capacitated plant location model:
qDownload the capacitated plant location
spreadsheet from the Canvas
qContains all required input data:
• Unit production and transportation costs, fixed costs,
capacities and demands

qStep 1:
qCreate a table for the decision variables
• Shipment quantities (in 1,000,000s) from each supply
region to each demand region
– Same format as production & transportation cost table
• Two sets of binary variables
– 1= open a small facility at each location, 0 = do not open
– 1 = open a large facility at each location, 0 = do not open

qStep 2:
qCalculate quantities need for the constraints
• Quantity shipped into demand region must equal demand in
that region
– Excess demand = Demand in region – Shipments into region = 0
• Quantity shipped out of supply region must be less than or
equal to capacity in that region
– Excess capacity = Actual capacity from plants in region
– Shipments out of region >=0

qStep 3:
qCalculate the total costs (use sumproduct
function)
• Production and transportation costs
• Costs from opening small plants
• Costs from opening large plants
• Add these three to get total costs

qStep 4:
qSet up model in Solver
• Minimize total costs
• By changing the decision variables
• Subject to the constraints:
– Excess demand = 0
– Excess capacity >= 0
– Shipment quantities must be non-negative
– Plant size decision variables must be binary

qOften, we are working within an existing
supply chain
qHave set of facilities with given capacities
qConsidering adding new facilities
qHow can we use and / or modify basic model to
accommodate this?

capacitated plant location model, now assume
that we already have a large capacity plant in
North America and that we plan to continue
to operate this plant
qAdd the appropriate constraint and resolve
qHow does the solution change?

qSuppose we would like to limit # of facilities
qMotivation?
qForm of additional constraint?
qLet k = maximum number of facilities
qSensitivity analysis on number of facilities can
be useful for decision-maker

capacitated plant location model, with
additional constraint that can open at most
one large plant
qProblems with solution?

capacitated plant location model, with
additional constraint that can open at most
one large plant and can open at most one
plant at each location

Plant Location Model with Single
Sourcing
qCould require each market to be served by just a single
plant
qMotivation for this constraint?
qyi = 1 if plant is located at site i, 0 otherwise
qxij = 1 if market j supplied by factory i, 0 otherwise
qIf xij = 1, ship quantity Dj from factory i to market j

qSuppose we would like the option of not
satisfying all demands
qWhat do we lose if we do not satisfy some
demands?
qHow do we incorporate this loss in model?
qOther modifications to model required?

Lesson 3: Formulating and
solving LP models using
Lingo and/or AMPL

Overview
qGraphical approach is intuitive but limited to very
small size problems (2 variables and a few
constraints)
qScale up to solve any size of problem
qExplicit mode is useful for small size problems where the
user enters the equations as they are read, combining the
coefficients of the variables with the variables themselves
qFor medium & large size problems, a modeling language
aids the user by entering a general form of each type of
constraint. Data of the model parameters are separated
from the equations & they may even be located in a
separate database or spreadsheet file.
5

LINGO
qGo to www.lindo.com
qDownloads
qDownload LINGO
2

LINGO in Explicit Mode
Simple example:
Max: Z= 20*x +31*y;
2*x+5*y <= 16;
4*x -3*y = 6;
And x >=0, y>= 0.
3

LINGO
7
model
Optimal
Results:

Importing and exporting
Spreadsheet data in LINGO
8

02_Intro to LP.pdf

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