Chapter 2
Linear programming
1
Introduction
• Many management decisions involve trying to make the most
effective use of an organization’s resources.
• Resources typically include machinery, labor, money, time,
warehouse space, or raw materials.
• Resources may be used to produce products (such as
machinery, furniture, food, or clothing) or services (such as
schedules for shipping and production, advertising policies,
or investment decisions).
• Linear programming (LP) is a widely used mathematical
technique designed to help managers in planning and
decision making relative to resource allocation.
• Despite the name, linear programming, and the more general
category of techniques called “mathematical programming”,
have very little to do with computer programming.
• In the world of Operations Research, programming refers to
modeling and solving a problem mathematically.
• Computer programming has, however, played an important
role in the advancement and use of LP to solve real-life LP
problems
2
Linear Programming Model
Most of the deterministic OR models can be formulated as
mathematical programs.
"Program," in this context, has to do with a “plan,” not a
computer program.
General form of Linear programming model
Maximize / Minimize z = f(x1, x2 ,…, xn)
Subject to
gi(x1, x2 , …, xn)
{}


=
bi
i =1,…,m
xj ≥ 0, j = 1,…,n
3
Model Components
• xj are called decision variables. These are things
that you control and you want to determine its
values
• gi(x1, x2 ,…, xn)
{}


=
bi are called structural
(or functional or technological) constraints
• xj ≥ 0 are nonnegativity constraints
• f(x1, x2 ,…, xn) is the objective function
4
Example: Giapetto woodcarving Inc.,
• Giapetto Woodcarving, Inc., manufactures two types of
wooden toys: soldiers and trains. A soldier sells for
$27 and uses $10 worth of raw materials. Each soldier
that is manufactured increases Giapetto’s variable
labor and overhead cost by $14. A train sells for $21
and uses $9 worth of raw materials. Each train built
increases Giapetto’s variable labor and overhead cost
by $10. The manufacture of wooden soldiers and
trains requires two types of skilled labor: carpentry
and finishing. A soldier requires 2 hours of finishing
labor and 1 hour of carpentry labor. A train requires 1
hour of finishing and 1 hour of carpentry labor. Each
week, Giapetto can obtain all the needed raw material
but only 100 finishing hours and 80 carpentry hours.
Demand for trains is unlimited, but at most 40
soldiers are bought each week. Giapetto wants to
maximize weekly profit. Formulate a linear
programming model of Giapetto’s situation that can be
5
used to maximize Giapetto’s weekly profit
Solution: Giapetto woodcarving Inc.,
• Step 1: Model formulation
1. Decision variables: we begin by finding the
decision variables. In any LP, the decision
variables should completely describe the
decisions to be made. Clearly, Giapetto must
decide how many soldiers and trains should
be manufactured each week. With this in
mind, we define:
X1 = number of soldiers produced each
week
X2 = number of trains produced each
week
6
Solution: Giapetto woodcarving Inc.,
2. Objective function: in any LP, the decision
maker wants to maximize (usually revenue or
profit) or minimize (usually costs) some
function of the decision variables. The
function to be maximized or minimized is
called the objective function. For the Giapetto
problem, we will maximize the net profit
(weekly revenues – raw materials cost – labor
and overhead costs).
Weekly revenues and costs can be expressed in
terms of the decision variables, X1 and X2 as
following:
7
Solution: Giapetto woodcarving Inc.,
• Weekly revenues = weekly revenues from
soldiers + weekly revenues from trains
= 27 X1 + 21 X2
Also,
Weekly raw materials costs = 10 X1 + 9 X2
Other weekly variable costs = 14 X1 + 10 X2
Therefore, the Giapetto wants to maximize:
(27 X1 + 21 X2) – (10 X1 + 9 X2) – (14 X1 + 10 X2) =
3 X1 + 2 X2
Hence, the objective function is:
Maximize Z = 3 X1 + 2 X2
8
Solution: Giapetto woodcarving Inc.,
3. Constraints: as X1 and X2 increase,
Giapetto’s objective function grows larger.
This means that if Giapetto were free to
choose any values of X1 and X2, the company
could make an arbitrarily large profit by
choosing X1 and X2 to be very large.
Unfortunately, the values of X1 and X2 are
limited by the following three restrictions
(often called constraints):
Constraint 1: each week, no more than 100
hours of finishing time may be used.
Constraint 2: each week, no more than 80
hours of carpentry time may be used.
Constraint 3: because of limited demand, at
most 40 soldiers should be produced.
9
Solution: Giapetto woodcarving Inc.,
• The three constraints can be expressed in
terms of the decision variables X1 and X2 as
follows:
Constraint 1: 2 X1 + X2  100
Constraint 2: X1 + X2  80
Constraint 3: X1  40
Note:
The coefficients of the decision variables in the
constraints are called technological
coefficients. This is because its often reflect
the technology used to produce different
products. The number on the right-hand side
of each constraint is called Right-Hand Side
(RHS). The RHS often represents the quantity
of a resource that is available.
10
Solution: Giapetto woodcarving Inc.,
• Sign restrictions: to complete the formulation
of the LP problem, the following question must
be answered for each decision variable: can
the decision variable only assume nonnegative
values, or it is allowed to assume both
negative and positive values?
If a decision variable Xi can only assume a
nonnegative values, we add the sign
restriction (called nonnegativity constraints)
Xi  0.
If a variable Xi can assume both positive and
negative values (or zero), we say that Xi is
unrestricted in sign (urs).
In our example the two variables are restricted
in sign, i.e., X1  0 and X2  0
11
Solution: Giapetto woodcarving Inc.,
• Combining the nonnegativity
constraints with the objective function
and the structural constraints yield the
following optimization model (usually
called LP model):
Max Z = 3 X1 + 2 X2 (objective function)
subject to (st)
2 X1 + X2  100
(finishing constraint)
X1 + X2  80
(carpentry constraint)
X1  40
(soldier demand constraint)
X1  0 and X2  0 (nonnegativity constraint)
The optimal solution to this problem is :
12
X1 = 20, and X2 = 60, Z = 180
What is Linear programming
problem (LP)?
• LP is an optimization problem for which we do the
following:
1. We attempt to maximize (or minimize) a linear
function of the decision variables. The function that is
to be maximized or minimized is called objective
function.
2. The values of decision variables must satisfy a set of
constraints. Each constraint must be a linear
equation or linear inequality.
3. A sign restriction is associated with each variable. for
any variable Xi, the sign restriction specifies either
that Xi must be nonnegative (Xi > 0) or that Xi may be
unrestricted in sign.
13
Linear Programming Assumptions
(i) proportionality
(ii) additivity
linearity
(iii) divisibility
(iv) certainty
14
Explanation of LP Assumptions
(i) activity j’s contribution to objective function is cjxj
and usage in constraint i is aijxj
both are proportional to the level of activity j
(volume discounts, set-up charges, and nonlinear
efficiencies are potential sources of violation)
(ii) “cross terms” such as x x may not
1 5
appear in the objective or constraints.
15
Explanation of LP Assumptions
(iii) Fractional values for decision variables are permitted
(iv) Data elements aij , cj , bi , uj are known with certainty
• Nonlinear or integer programming models should be
used when some subset of assumptions (i), (ii) and
(iii) are not satisfied.
• Stochastic models should be used when a problem
has significant uncertainties in the data that must be
explicitly taken into account [a relaxation of
assumption (iv)].
16
Applications Of LP
1. Product mix problem
2. Diet problem
3. Blending problem
4. Media selection problem
5. Assignment problem
6. Transportation problem
7. Portfolio selection problem
8. Work-scheduling problem
9. Production scheduling problem
10. Inventory Problem
11. Multi period financial problem
12. Capital budgeting problem
17
1. Product Mix Problem
Example
Formulate a linear programming model for this
problem, to determine how many containers of
each product to produce tomorrow in order to
maximize the profits. The company makes four
types of juice using orange, grapefruit, and
pineapple. The following table shows the price and
cost per quart of juice (one container of juice) as
well as the number of kilograms of fruits required
to produce one quart of juice.
Product
Price/quart
Cost/quart
Fruit needed
Orange juice
3
1
1 Kg.
Grapefruit juice
2
0.5
2 Kg.
Pineapple juice
2.5
1.5
1.25 Kg.
4
2
18
0.25 Kg each
All –in - one
Product Mix Problem
Example (cont.)
On hand there are 400 Kg of orange, 300 Kg.
of grapefruit, and 200 Kg. of pineapples.
The manager wants grapefruit juice to be used
for no more than 30 percent of the number of
containers produced. He wants the ratio of the
number of containers of orange juice to the
number of containers of pineapples juice to be
at least 7 to 5. pineapples juice should not
exceed one-third of the total product.
19
Product Mix Problem
Solution
Decision variables
X1 = # of containers of orange juice
X2 = # of containers of grapefruit juice
X3 = # of containers of pineapple juice
X4 = # of containers of All-in-one juice
Objective function
Max Z = 2 X1 + 1.5 X2 + 1 X3 + 2 X3
Constraints
X 1  0.25 X 4  400
Orange constraints
2 X 2  0.25 X 4  300
Grapefruit constraint
1.25 X 3  0.25 X 4  200
Pineapple constraints
X 2  0.3( X 1  X 2  X 3  X 4)
Max. of grapefruit
X1
7

X3 5
1
X 2  ( X 1  X 2  X 3  X 4)
3
X 1, X 2, X 3, X 4  0
Ratio of orange to pineapple
Max. of pineapple
20
Non-negativity constraints
2. Diet problem
Example
My diet requires that all the food I eat come from one of
the four “basic food groups” (chocolate cake, ice
cream, soda, and cheesecake). At present, the
following four foods are available for consumption:
brownies, chocolate ice cream, cola, and pineapple
cheesecake. Each brownie costs 50 cents, each scoop
of chocolate ice cream costs 20 cents, each bottle of
cola costs 30 cents, and each piece of pineapple
cheesecake costs 80 cents. Each day, I must ingest at
least 500 calories, 6 oz of chocolate, 10 oz of sugar,
and 8 oz of fat. The nutritional content per unit of
each food is shown in the following table. Formulate a
linear programming model that can be used to satisfy
my daily nutritional requirements at minimum costs.
21
Diet problem
Calories
Chocolate
Sugar
Fat
Brownie
400
3 ounce
Chocolate ice cream
(1 scoop)
200
2
2
4
Cola (1 bottle)
150
0
4
1
Pineapple cheesecake
(1piece)
500
0
4
5
2 ounce 2 ounce
22
Diet problem
Solution
• Decision variables: as always, we begin by
determining the decisions that must be made
by the decision maker: how much of each food
type should be eaten daily. Thus, we define the
decision variables:
X1 = number of brownies eaten daily
X2 = number of scoops of chocolate ice cream
eaten daily
X3 = number of bottles of cola drunk daily
X4 = number of pieces of pineapple cheesecake
eaten daily
23
Diet problem
• Objective function: my objective
function is to minimize the cost of my
diet. The total cost of my diet may be
determined from the following relation:
Total cost of diet = (cost of brownies) +
(cost of ice cream) + (cost of cola) + (cost
of cheesecake)
Thus, the objective function is:
Min Z = 50 X1 + 20 X2 + 30 X3 + 80 X4
24
Diet problem
• Constraints: the decision variables must satisfy the
following four constraints:
Constraint 1: daily calorie intake must be at least 500
calories.
Constraint 2: daily chocolate intake must be at least 6
oz.
Constraint 3: daily sugar intake must be at least 10 oz.
Constraint 4: daily fat intake must be at least 8 oz.
To express constraint 1 in terms of the decision
variables, note that (daily calorie intake) = (calorie in
brownies) + (calories in chocolate ice cream) + (calories
in cola) + (calories in pineapple cheesecake)
Therefore,
the daily calorie intake = 400 X1 + 200 X2 + 150 X3 + 500
X4 must be greater than 500 ounces
By the same way the other three constraints can be
25
formulated.
Diet problem
The four constraints are:
400 X1 + 200 X2 + 150 X3 + 500 X4  500
3 X1 + 2 X2  6
2 X1 + 2 X2 + 4 X3 + 4 X4  10
2 X1 + 4 X2 + X3 + 5 X4  8
Nonnegativity constraints: it is clear that
all decision variables are restricted in
sign, i.e., Xi  0, for all i = 1, 2, 3, and 4
26
Diet problem
• Combining the objective function, constraints,
and nonnegativity constraints, the LP model is
as follows:
Min Z = 50 X1 + 20 X2 + 30 X3 + 80 X4
st.
400 X1 + 200 X2 + 150 X3 + 500 X4  500
3 X1 + 2 X2  6
2 X1 + 2 X2 + 4 X3 + 4 X4  10
2 X1 + 4 X2 + X3 + 5 X4  8
Xi  0, for all i = 1, 2, 3, and 4
The optimal solution to this LP is X1 = X4 = 0,
X2 = 3, X3 = 1 and Z = 90 cents
27
3. Blending problem
Example
The Low Knock Oil company produces two grades of cut rate
gasoline for industrial distribution. The grades, regular and
economy, are produced by refining a blend of two types of crude
oil, type X100 and type X220. each crude oil differs not only in
cost per barrel, but in composition as well. The accompanying
table indicates the percentage of crucial ingredients found in each
of the crude oils and the cost per barrel for each. Weekly demand
for regular grade of Low Knock gasoline is at least 25000 barrels,
while demand for the economy is at least 32000 barrels per week.
At least 45% of each barrel of regular must be ingredient A. At
most 50% of each barrel of economy should contain ingredient B.
the Low Knock management must decide how many barrels of
each type of crude oil to buy each week for blending to satisfy
demand at minimum cost .
Crude oil
type
X100
X220
Ingredient Ingredient Cost/barrel
A%
B%
($)
35
55
30
60
25
34.8
28
Blending problem
Solution
Let
X1 = # of barrels
X2 = # of barrels
economy
X3 = # of barrels
X4 = # of barrels
economy
of crude X100 blended to produce the refined regular
of crude X100 blended to produce the refined
of crude X220 blended to produce the refined regular
of crude X220 blended to produce the refined
Min Z = 30 X1 + 30 X2 + 34.8 X3 + 34.8 X4
St.
X1 + X3  25000
X2 + X4  32000
-0.10 X1 + 0.15 X3  0
0.05 X2 – 0.25 X4  0
Xi  0, i = 1, 2, 3, 4
The optimal solution is: X1 = 15000, X2 = 26666.6, X3 = 10000,
X4 = 5333.3, and Z = 1,783,600
29
4. Media selection problem
Example
A company has budgeted up to $8000 per week
for local advertisement. The money is to be
allocated among four promotional media: TV
spots, newspaper ads, and two types of radio
advertisements. The company goal is to reach
the largest possible high-potential audience
through the various media. The following
table presents the number of potential
customers reached by making use of
advertisement in each of the four media. It
also provides the cost per advertisement
placed and the maximum number of ads that
can be purchased per week.
30
Media Selection
Medium
Audience
Reached per
ad
Cost per
ad
Maximum
ads per
week
TV spot (1 minute)
5000
800
12
Daily newspaper
(full-page ad)
8500
925
5
Radio spot
(30 second, prime time)
2400
290
25
Radio spot
(1 minute, afternoon)
2800
380
20
The company arrangements require that at least five radio spots
be placed each week. To ensure a board-scoped promotional
campaign, management also insists that no more than $1800 be
spent on radio advertising every week.
31
Media selection
Solution
Let
X1 = number
X2 = number
X3 = number
X4 = number
of
of
of
of
1-miute TV spots taken Each week
full-page daily newspaper ads taken each week.
30-second prime-time radio spots taken each week.
1-minute afternoon radio spots taken each week.
Max Z = 5000 X1 + 8500 X2 + 2400 X3 + 2800 X4
st
X1  12
(maximum TV spots/week)
X2  5
(maximum newspaper ads/week)
X3  25
(maximum 30-second radio spots/week)
X4  20
(maximum 1-minute radio spots/week)
800 X1 + 925 X2 + 290 X3 + 380 X4  8000
(weekly budget)
X3 + X4  5
(minimum radio spots contracted)
290 X3 + 380 X4  1800
(maximum dollars spent on radio)
X1, X2, X3, X4  0
The optimal solution is: X1 = 1.97, X2 = 5, X3 =6.2, and X4 = 0, Z = 67240
32
5. Assignment problem
Example
A law firm maintains a large staff of young attorneys
who hold the title of junior partner. The firm
concerned with the effective utilization of this
personnel resources, seeks some objective means of
making lawyer-to-client assignments. On march 1,
four new clients seeking legal assistance came to the
firm. While the current staff is overloads and
identifies four junior partners who, although busy,
could possibly be assigned to the cases. Each young
lawyer can handle at most one new client.
Furthermore each lawyer differs in skills and
specialty interests.
Seeking to maximize the overall effectiveness of the new
client assignment, the firm draws up the following
table, in which he rates the estimated effectiveness
(of a scale of 1 to 9) of each lawyer on each new case.
33
Assignment problem
Client case
Lawyer
Divorce
Corporate
merger
embezzlement
exhibitionism
Adam
6
2
8
5
Brook
9
3
5
8
Carter
4
8
3
4
Darwin
6
7
6
4
34
Assignment problem
Solution
Decision variables:
1 if attorney i is assigned to case j
Let Xij =
0 otherwise
Where : i = 1, 2, 3, 4 stands for Adam, Brook,
Carter, and Darwin respectively
j = 1, 2, 3, 4 stands for divorce,
merger, embezzlement, and exhibitionism
respectively.
The LP formulation will be as follows:
35
Assignment problem
Max Z = 6 X11 + 2 X12 + 8 X13 + 5 X14 + 9 X21 + 3 X22 +
5 X23 + 8 X24 + 4 X31 + 8 X32 + 3 X33 + 4 X34 +
6 X41 +7 X42 + 6 X43 + 4 X44
St.
X11 + X21 + X31 + X41 = 1
(divorce case)
X12 + X22 + X32 + X42 = 1
(merger)
X13 + X23 + X33 + X43 = 1
(embezzlement)
X14 + X24 + X34 + X44 = 1
(exhibitionism)
X11 + X12 + X13 + X14 = 1
(Adam)
X21 + X22 + X23 + X24 = 1
(Brook)
X31 + X32 + X33 + X34 = 1
(Carter)
X41+ X42 + X43 + X44 = 1
(Darwin)
The optimal solution is: X13 = X24 = X32 = X41 = 1. All other
variables are equal to zero.
36
6. Transportation problem
Example
The Top Speed Bicycle Co. manufactures and markets a
line of 10-speed bicycles nationwide. The firm has
final assembly plants in two cities in which labor
costs are low, New Orleans and Omaha. Its three
major warehouses are located near the larger market
areas of New York, Chicago, and Los Angeles.
The sales requirements for next year at the New York
warehouse are 10000 bicycles, at the Chicago
warehouse 8000 bicycles, and at the Los Angeles
warehouse 15000 bicycles. The factory capacity at
each location is limited. New Orleans can assemble
and ship 20000 bicycles; the Omaha plant can
produce 15000 bicycles per year. The cost of shipping
one bicycle from each factory to each warehouse
differs, and these unit shipping costs are:
37
Transportation problem
New Orleans
New
York
$2
Chicago
3
Los
Angeles
5
Omaha
3
1
4
The company wishes to develop a shipping
schedule that will minimize its total annual
transportation cost
38
Transportation problem
Solution
To formulate this problem using LP, we again employ the
concept of double subscribed variables. We let the first
subscript represent the origin (factory) and the second
subscript the destination (warehouse). Thus, in
general, Xij refers to the number of bicycles shipped
from origin i to destination j. Therefore, we have six
decision variables as follows:
X11
X12
X13
X21
X22
X23
=
=
=
=
=
=
#
#
#
#
#
#
of bicycles shipped from New Orleans to New York
of bicycles shipped from New Orleans to Chicago
of bicycles shipped from New Orleans to Los Angeles
of bicycles shipped from Omaha to New York
of bicycles shipped from Omaha to Chicago
of bicycles shipped from Omaha to Los Angeles 39
Transportation problem
Min Z = 2 X11 + 3 X12 + 5 X13 + 3 X21 + X22 + 4
X23
St
X11 + X21 = 10000 (New York demand)
X12 + X22 = 8000 (Chicago demand)
X13 + X23 = 15000 (Los Angeles demand)
X11 + X12 + X13  20000 (New Orleans Supply
X21 + X22 + X23  15000 (Omaha Supply)
Xij  0 for i = 1, 2 and j = 1, 2, 3
The optimal solution is: X11 = 10000, X12 = 0, X13 = 8000, X21 =
0, X22 = 8000, X23 = 7000, and Z = $96000
40
7. Portfolio selection
Example
The International City Trust (ICT) invests in
short-term trade credits, corporate bonds, gold
stocks, and construction loans. To encourage
a diversified portfolio, the board of directors
has placed limits on the amount that can be
committed to any one type of investment. The
ICT has $5 million available for immediate
investment and wishes to do two things: (1)
maximize the interest earned on the
investments made over the next six months,
and (2) satisfy the diversification requirements
as set by the board of directors. The specifics
of the investment possibilities are:
41
Portfolio selection
Investment
Interest
earned
%
Trade credit
Corporate bonds
7
11
Maximum
investment
($ Million)
1
2.5
Gold stocks
Construction loans
19
15
1.5
1.8
In addition, the board specifies that at least
55% of the funds invested must be in gold
stocks and construction loans, and that no less
than 15% be invested in trade credit.
42
Portfolio selection
Solution
To formulate ICT’s investment problem as
a linear programming model, we
assume the following decision variables:
X1 = dollars invested in trade credit
X2 = dollars invested in corporate bonds
X3 = dollars invested in gold stocks
X4 = dollars invested in construction
loans
43
Portfolio selection
Max Z = 0.07 X1 + 0.11 X2 + 0.19 X3 + 0.15 X4
St.
X1
 1
 2.5
X3
 1.5
X4
 1.8
X3 + X4
 0.55(X1 + X2 + X3 + X4)
X1
 0.15(X1 + X2 + X3 + X4)
X1 + X2 + X3 + X4  5
Xi  0 , i = 1, 2, 3, 4
X2
The optimal is: X1 = 75,000, X2 = 950,000, X3 =
1,500,000, and X4 = 1,800,000, and total interest
Z = 712,000
44
8. Work Scheduling Problem
Example
Microsoft has a 24-hour-a-day, 7-days-a-week toll
free hotline that is being set up to answer questions
regarding a new product. The following table
summarizes the number of full-time equivalent
employees (FTEs) that must be on duty in each time
block.
Shift
1
2
3
4
5
6
Time
0-4
4-8
8-12
12-16
16-20
20-0
FTEs
15
10
40
70
40
35
45
Work Scheduling problem
• Microsoft may hire both full-time and part-time
employees. The former work 8-hour shifts and the
latter work 4-hour shifts; their respective hourly
wages are $15.20 and $12.95. Employees may start
work only at the beginning of one of 6 shifts.
Part-time employees can only answer 5 calls in the
time a full-time employee can answer 6 calls. (i.e., a
part-time employee is only 5/6 of a full-time
employee.)
Formulate an LP to determine how to staff
the hotline at minimum cost.
46
Decision Variables
xt = # of full-time employees that begin work in shift t
yt = # of part-time employees that work shift t
(4  12.95)
(8  15.20)
Min
s.t.
121.6 (x1 +
•••
+ x6) + 51.8 (y1 +
x6
+
x1 +
x1
+
x2 +
x2
+
x3 +
x3
+
x4 +
x4
+
x5 +
x5
+
x6 +
5
6
5
6
5
6
5
6
5
6
5
6
•••
+ y 6)
y1

15
y2

10
y3

40
y4

70
y5

40
y6

35
All shifts
must be
covered
xt, yt  0
PT employee is 5/6 FT employee
47
Terminology for solution of LP
• A feasible solution is a solution for which all
the constraints are satisfied.
• A corner point feasible solution (CPF) is a
feasible solution that lies at a corner point.
• An infeasible solution is a solution for which
at least one constraint is violated.
• The feasible region is the collection of all
feasible solution.
• An Optimal solution is a feasible solution that
has the most favorable value of the objective
function. (it is always one of the CPF solution
• The most favorable value is the largest
(smallest) value if the objective function is to
be maximized (minimized).
48
Graphical solution
A Graphical Solution Procedure (LPs with 2 decision variables
can be solved/viewed this way.)
1. Plot each constraint as an equation and then decide which
side of the line is feasible (if it’s an inequality).
2. Find the feasible region.
3. find the coordinates of the corner (extreme) points of the feasible
region.
4. Substitute the corner point coordinates in the objective function
5. Choose the optimal solution
49
Example 1: A Minimization Problem
• LP Formulation
Min z = 5x1 + 2x2
s.t.
2x1 + 5x2 > 1
4x1 - x2 > 12
x1 + x2 > 4
x1, x2 > 0
50
Example 1: Graphical Solution
• Graph the Constraints
•
Constraint 1: When x1 = 0, then x2 = 2; when x2 =
0, then x1 = 5. Connect (5,0) and (0,2). The ">" side
is above this line.
•
Constraint 2: When x2 = 0, then x1 = 3. But
setting x1 to
0 will yield x2 = -12, which is not on
the graph.
Thus, to get a second point on this
line, set x1 to any
number larger than 3 and solve
for x2: when x1 = 5, then x2 = 8. Connect (3,0) and
(5,8). The ">" side is to the right.
•
Constraint 3: When x1 = 0, then x2 = 4; when x2 =
0, then x1 = 4. Connect (4,0) and (0,4). The ">" side
is above this line.
51
Example 1: Graphical Solution
• Constraints Graphed
x2
Feasible Region
4x1 - x2 > 12
5
4
x1 + x2 > 4
3
2x1 + 5x2 > 10
2
(16/5,4/5)
(10/3, 2/3)
1
1
2
3
4
5
(5,0)
6
x1
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Example 1: Graphical Solution
•
Solve for the Extreme Point at the Intersection of the second and third
Constraints
4x1 - x2 = 12
x1+ x2 = 4
Adding these two equations gives:
5x1 = 16 or x1 = 16/5.
Substituting this into x1 + x2 = 4 gives: x2 = 4/5
• Solve for the extreme point at the intersection of the first and third constraints
2x1 + 5x2 =10
x1 + x2= 4
Multiply the second equation by -2 and add to the first equation, gives
3x2 = 2 or x2 = 2/3
Substituting this in the second equation gives x1 = 10/3
Point
(16/5, 4/5)
(10/3, 2/3)
(5, 0)
Z
88/5
18
25
53
Example 2: A Maximization Problem
Max
s.t.
z = 5x1 + 7x2
x1
< 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1, x2 > 0
54
Example 2: A Maximization Problem
• Constraint #1 Graphed
x2
8
7
6
x1 < 6
5
4
3
2
(6, 0)
1
1
2
3
4
5
6
7
8
9
10
x1
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Example 2: A Maximization Problem
• Constraint #2 Graphed
x2
(0, 6 1/3)
8
7
2x1 + 3x2 < 19
6
5
4
3
2
(9 1/2, 0)
1
1
2
3
4
5
6
7
8
9
10
x1
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Example 2: A Maximization Problem
• Constraint #3 Graphed
x2
(0, 8)
8
7
x1 + x2 < 8
6
5
4
3
2
(8, 0)
1
1
2
3
4
5
6
7
8
9
10
x1
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Example 2: A Maximization Problem
• Combined-Constraint Graph
x2
x1 + x2 < 8
8
7
x1 < 6
6
5
4
3
2x1 + 3x2 < 19
2
1
1
2
3
4
5
6
7
8
9
10
x1
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Example 2: A Maximization Problem
• Feasible Solution Region
x2
8
7
6
5
4
3
Feasible
Region
2
1
1
2
3
4
5
6
7
8
9
10
x1
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Example 2: A Maximization Problem
• The Five Extreme Points
8
7
(0, 19/3)
5
6
5
(5, 3)
4
4
3
1
(0, 0)
3
Feasible
Region
2
1
(6, 2)
2
1
2
3
4
5
6
(6, 0)
7
8
9
10
x1
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Example 2: A Maximization Problem
• Having identified the feasible region for the
problem, we now search for the optimal
solution, which will be the point in the
feasible region with the largest (in case of
maximization or the smallest (in case of
minimization) of the objective function.
• To find this optimal solution, we need to
evaluate the objective function at each one of
the corner points of the feasible region.
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Example 2: A Maximization Problem
• Optimal Solution
Point
x2
Z
(0,0)
0
8
(6,0)
30
7
(6,2)
44
(5,3)
46
(0,19/3)
44.33
6
Optimal Solution
5
4
3
2
1
1
2
3
4
5
6
7
8
9
10
x1
62
Extreme Points and the Optimal Solution
• The corners or vertices of the feasible region
are referred to as the extreme points.
• An optimal solution to an LP problem can be
found at an extreme point of the feasible
region.
• When looking for the optimal solution, you do
not have to evaluate all feasible solution
points.
• You have to consider only the extreme points
of the feasible region.
63
Feasible Region
• The feasible region for a two-variable linear
programming problem can be nonexistent, a single
point, a line, a polygon, or an unbounded area.
• Any linear program falls in one of three categories:
– is infeasible
– has a unique optimal solution or alternate optimal
solutions
– has an objective function that can be increased
without bound
• A feasible region may be unbounded and yet there
may be optimal solutions. This is common in
minimization problems and is possible in
maximization problems.
64
Special Cases
• Alternative Optimal Solutions
In the graphical method, if the objective function
line is parallel to a boundary constraint in the
direction of optimization, there are alternate
optimal solutions, with all points on this line
segment being optimal.
• Infeasibility
A linear program which is overconstrained so that
no point satisfies all the constraints is said to be
infeasible.
• Unbounded
For a max (min) problem, an unbounded LP occurs
in it is possible to find points in the feasible region
with arbitrarily large (small) Z
65
Example with Multiple Optimal
Solutions
x2
z1
z2
z3
Maximize z = 3x1 – x2
4
subject to 15x1 – 5x2  30
3
10x1 + 30x2  120
2
1
x1  0, x2  0
0
0
1
2
3
4
x1
66
Example: Infeasible Problem
• Solve graphically for the optimal solution:
Max
s.t.
z = 2x1 + 6x2
4x1 + 3x2 < 12
2x1 + x2 > 8
x1 , x2 > 0
67
Example: Infeasible Problem
• There are no points that satisfy both
constraints, hence this problem has no
feasible region, and no optimal solution.
x2
2x1 + x2 > 8
8
4x1 + 3x2 < 12
4
3 4
x1
68
Example: Unbounded Problem
• Solve graphically for the optimal solution:
Max
s.t.
z = 3x1 + 4x2
x1 + x2 > 5
3x1 + x2 > 8
x1 , x2 > 0
69
Example: Unbounded Problem
• The feasible region is unbounded and the objective
function line can be moved parallel to itself without
bound so that z can be increased infinitely.
x2
3x1 + x2 > 8
8
5
x1 + x2 > 5
2.67
5
x1
70
Solve the following LP graphically
max
s.t.
45 x1
10 x1
28 x1
6 x1
15 x1
+
+
+
+
+
60 x2
20 x2
12 x2
15 x2
10
0 x2
x1  40
x2  100
Objective Function




1800
1440
2040
2400
A
B
C
D
Structural
constraints
E
F
x1 ≥ 0, x2 ≥ 0
Where x1 is the quantity produced
from product Q, and x2 is the
quantity of product P
nonnegativity
Are the LP assumptions
valid for this problem?
71
The graphical solution
P
240
Assignment:
Complete the
problem to find the
optimal solution
Max Q
E
200
D
160
120
Max P
F
80
A
40
0
B
0
40
(
1
)
80 120
C
160
200 240 280 320
360
Q
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Possible Outcomes of an LP
1. Infeasible –
feasible region is empty; e.g., if the
constraints include
x1+ x2  6 and x1+ x2  7
2. Unbounded -
Max 15x1+ 7x2
s.t. x1 + x2  1
x1, x2  0
(no finite optimal
solution)
3. Multiple optimal solutions - max 3x1 + 3x2
s.t. x1+ x2  1
x1, x2  0
4. Unique Optimal Solution
Note: multiple optimal solutions occur in many practical (real-world) LPs.
73