Linear programming graphical method

19 March 2020 Dr. Abdulfatah A. Salem 2
• A linear equation are usually shown on a coordinate plane and its
graph forms a straight line.
• Standard form for linear equations is ax + by= c.
Example: 2x+3y=6
1.) Substitute in a zero for x. 2(0)+3y=6 y=2
2.) One point of the line is (0,2). Plot the point.
3.) Substitute y with zero. 2x+3(0)=6 x=3
4.) The second point is (3,0). Plot the point.
5. ) Connect the two points drawing a line through them.
What is a Linear Equation?
Linear Programming
Graphical solution

Linear Programming
Graphical solution
Mathematical
models
An equation of the form
ax + by = c
defines a straight line in
the x--y plane.
An inequation of the form
ax + by ≠ c
defines an area bounded by a
straight line In
the x--y plane .

There are three possible solutions to a system of linear equations in two variables :
1) The two lines intersect at a single point. The coordinates give the solution of the system. In this case, the
solution is “consistent” and the equations are “independent”.
2) The two lines are parallel. In this case the system is “inconsistent” and the solution set is 0 or null.
3) The two lines are the same line. In this case, the equations are “dependent” and the solution set is an infinite
set of ordered pairs.
Solving Linear Systems by Graphing
(Geometric Approach)
Linear Programming
Graphical solution

Linear Programming
Graphical solution
5x + 7y = 100
X=0 y=100/7 Point 1 (0 , 100/7)
Y=0 x= 20 Point 2 (20 , 0)
5x + 7y ≥ 100
X=0 y=100/7 Point 1 (0 , 100/7)
Y=0 x= 20 Point 2 (20 , 0)
5x + 7y ≤ 100
X=0 y=100/7 Point 1 (0 , 100/7)
Y=0 x= 20 Point 2 (20 , 0)

To solve an LP, the graphical method includes two major steps:
• The determination of the solution space that defines the feasible solution. Note
that the set of values of the variable x y which satisfy all the constraints and also the
non-negative conditions is called the feasible solution of the LP.
• The determination of the optimal solution from the feasible region. the optimal
solution is the value of the variable x y which satisfy the feasible area and the
objective function.
To determine the feasible solution of an LP, we have the following steps.
Step 1: Since the two decision variable x and y are non-negative, consider only the first quadrant of xy-
coordinate plane
Step 2: Each constraint is of the form ax + by ≤ c or ax + by ≥ c .
For each constraint draw the line ax + by = c,
The line divides the first quadrant in to two regions say R1 and R2, suppose the point x = 0 and y = 0.
If this point satisfies the inequation, then shade the region R1 else shade the region R2.
Step 3: Corresponding to each constraint, we obtain a shaded region. The intersection of all these shaded
regions is the feasible region or feasible solution of the LP.
Linear Programming
Graphical solution

19 March 2020 Dr. AbdulfatahA. Salem 7
Graphing Systems of Linear Inequalities
To determine the optimal solution There are two techniques
Linear Programming
Graphical solution
Corner-point method is a method for solving graphical linear programming
problems. The mathematical theory behind linear programming states that
an optimal solution to any problem will lie at a corner point, or an
extreme point, of the feasible region. Hence, it is necessary to find only
the values of the variables at each corner; the optimal solution will lie at
one or more of them. This is the corner-point method.
ISO-profit line method is an approach to identifying the optimum point in a
graphic linear programming problem. The line representing the objective
function and touches a particular point of the feasible region will pinpoint
the optimal solution. Once the feasible region has been graphed, one can
find the optimal solution to the problem. The optimal solution is the point
lying in the feasible region that produces the highest profit.
2. ISO- Profit (or Cost) method
1. Corner Point Method

19 March 2020 Dr. AbdulfatahA. Salem 8
• Corner Point Method
Step 1: Find the feasible region of the LLP.
Step 2: Find the co-ordinates of each vertex of the feasible region.
These co-ordinates can be obtained from the graph or by solving the
equation of the lines.
Step 3: At each vertex (corner point) compute the value of the objective
function.
Step 4: Identify the corner point at which the value of the objective function
is maximum (or minimum depending on the LP)
The co-ordinates of this vertex is the optimal solution and the value of Z
is the optimal value
Linear Programming
Graphical solution

19 March 2020 Dr. Abdulfatah Salem 9
Problem Statement:
Toys Manufacturing company produces 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 labor and
overhead cost by $14. A train sells for $21 and uses $9 worth of raw
materials. Each train built increases 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, the company 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. The company wants to maximize weekly profit.
Formulate and solve graphically a linear programming model of
the situation that can be used to maximize the weekly profit.
Linear Programming
Graphical solution

1) Model formulation
– Max Z = 3 X1 + 2 X2 (objective function)
– S.T.
 2 X1 + X2  100 (finishing constraint)
 X1 + X2  80 (carpentry constraint)
 X1  40 (soldier demand constraint)
 X1  0 and X2  0 (no negativity constraint)
2) Model solution
II. Drawing the feasible region
• 2 X1 + X2  100 2 X1 + X2 = 100 (0,100) (50,0)
• X1 + X2  80 X1 + X2 = 80 (0,80) (80,0)
• X1  40 X1 = 40 (40,0) (40,1)
• X1  0 X1 = 0 (0,0) (0,1)
• X2  0 X2 = 0 (0,0) (1,0)
Solution
Linear Programming
Graphical solution

2 X1 + X2 = 100 (0,100) (50,0)
X1 + X2 = 80 (0,80) (80,0)
X1 = 40 (40,0) (40,1)
Linear Programming
Graphical solution

•Point 1 : (0 , 80)
•Point 2 :
The intersection point between :
2 X1 + X2 = 100
X1 + X2 = 80
By solving the two equations Simultaneously
we find that the point is ( 20,60)
•Point 3 :
The intersection point between :
2 X1 + X2 = 100
X1 = 40
By solving the two equations Simultaneously
we find that the point is ( 40,20)
•Point 4 : (40 , 0)
Linear Programming
Graphical solution

Referring to the objective function
Max Z = 3 X1 + 2 X2
Determining the value of Z at each point of the corners of the polygon
as indicated in the table below :
The # of soldiers to be produced weekly is 20
The # of trains to be produced weekly is 60
The Max. profit to be obtained weekly is $180
Point Co-ords X1 X2 Z
Point 1 ( 0 , 80 ) 0 80 160
Point 2 ( 20 , 60 ) 20 60 180
Point 3 ( 40 , 20 ) 40 20 160
Point 4 ( 40 , 0 ) 40 0 120
Linear Programming
Graphical solution

Linear Programming
Graphical solution
Woodek Co. manufactures two products:
Tables
chairs.
Resources are limited to
1000 cubic meters woods.
40 hours of production time per week.
Marketing requirement
Total production cannot exceed 650 dozens.
Number of dozens of tables cannot exceed number of dozens of chairs by more than 350.
Technological input
Tables requires 2 cubic meters of wood and 3 minutes of labor per dozen.
 chairs requires 1 cubic meters of wood and 4 minutes of labor per dozen.
The current production plan calls for:
Producing as much as possible of the more profitable product:
table sells ($80 profit per dozen).
Chair sells ($50 profit per dozen).

Linear Programming
Graphical solution
Solution
Management is seeking a production schedule that will increase the
company’s profit.
A linear programming model can provide an insight and an intelligent
solution to this problem.
Decisions variables:
X1 = Weekly production level of tables (in dozens)
X2 = Weekly production level of chairs (in dozens).
Objective Function:
Max Z = 80X1 + 50X2
Constraints:
2X1 + X2  1000 (Wood)
3X1 + 4X2  2400 (Production Time)
X1 + X2  650 (Total production)
X1 - X2  350 (Mix)
X1 , X2 ≥ 0 (Non negativity)

Linear Programming
Graphical solution
2X1 + X2  1000
3X1 + 4X2  2400
X1 + X2  650
X1 - X2  350
X1 , X2 ≥ 0
2X1 + X2 = 1000
3X1 + 4X2 = 2400
X1 + X2 = 650
X1 - X2 = 350
X1 , X2 = 0
(0,1000) (500,0)
(0,600) (800,0)
(0,650) (650,0)
(0,-350) (0,350)

Linear Programming
Graphical solution
(0,1000)
(500,0)
(0,600)
(800,0)
(0,650)
(0,650) (0,350)

Linear Programming
Graphical solution

Linear Programming
Graphical solution
Point (5)
( 350 , 0 )
5
4
3
2
1
Point (1)
( 0 , 600 )
Point (2)
The intersection point between:
3X1 + 4X2 = 2400
X1 + X2 = 650
Which is ( 200 , 450 )
Point (3)
2X1 + X2 = 1000
X1 + X2 = 650
Which is ( 350 , 300 )
Point (4)
2X1 + X2 = 1000
X1 - X2 = 350
Which is ( 450 , 100 )

Linear Programming
Graphical solution
Points ( X1 , X2 ) Z
Point (1) ( 0 , 600 ) 30000
Point (2) ( 200 , 450 ) 38500
Point (3) ( 350 , 300 ) 43000
Point (4) ( 450 , 100 ) 41000
Point (5) ( 350 , 0 ) 28000
Max Z = 80X1 + 50X2
The optimal solution is :
The quantity of tables to be manufactured is 350
The quantity of chairs to be manufactured is 300
The maximum profit obtained is $ 43000

Linear Programming
Graphical solution
Solve the following linear programming model graphically.
Max. Z = 25x1 + 20x2
s.t.
12x1 + 10x2 ≤ 150
8x1 + 4x2 ≤ 80
2x1 + 4x2 ≤ 50
x1 + 2x2 ≥ 8
x1 ≥ 0 , x2 ≥ 0
Solution
12x1 + 10x2 ≤ 150
8x1 + 4x2 ≤ 80
2x1 + 4x2 ≤ 50
x1 + 2x2 ≥ 8
x1 ≥ 0
x2 ≥ 0
12x1 + 10x2 = 150
8x1 + 4x2 = 80
2x1 + 4x2 = 50
x1 + 2x2 = 8
x1 = 0
x2 = 0
(0 , 15) (12.5 , 0)
(0 , 20) (10 , 0)
(0 , 12.5) (25 , 0)
(0 , 4) (8 , 0)
(0 , 0) (0 , 1)
(0 , 0) (1 , 0)

Linear Programming
Graphical solution
12x1 + 10x2 ≤ 150
8x1 + 4x2 ≤ 80
2x1 + 4x2 ≤ 50
x1 + 2x2 ≥ 8
x1 ≥ 0
x2 ≥ 0
12x1 + 10x2 = 150
8x1 + 4x2 = 80
2x1 + 4x2 = 50
x1 + 2x2 = 8
x1 = 0
x2 = 0
(0 , 15) (12.5 , 0)
(0 , 20) (10 , 0)
(0 , 12.5) (25 , 0)
(0 , 4) (8 , 0)
(0 , 0) (0 , 1)
(0 , 0) (1 , 0)

Linear Programming
Graphical solution
Point (1) (0 , 12.5)
Point (2)
The intersection point between
• 12x1 + 10x2 = 150
• 2x1 + 4x2 = 50
( 43/12 , 75/7 )
Point (3)
The intersection point between
• 12x1 + 10x2 = 150
• 8x1 + 4x2 = 80
( 6.25 , 7.5 )
Point (4) (10 , 0)
Point (5) (8 , 0)
Point (6) (0 , 4)

Linear Programming
Graphical solution
Points (X1 , X2) Z
Point (1) (0 , 12.5) 250
Point (2) ( 43/12 , 75/7 ) 303.86
Point (3) ( 6.25 , 7.5 ) 306.25
Point (4) (10 , 0) 250
Point (5) (8 , 0) 200
Point (6) (0 , 4) 80
It is clear from the table that the greatest value of Z is at the 3rd corner
“Point (3)” with values
X1 = 6.25
X2 = 7.5
Z = 306.25
Max. Z = 25x1 + 20x2

Max Z = 5X1 + 7X2
S.T. X1 < 6
2X1 + 3X2 < 19
X1 + X2 < 8
X1, X2 > 0
Linear Programming
Graphical solution
Graph the Constraints
Constraint 1:
X1 < 6
x2
x1
X1 < 6
(6, 0)
Solution

Constraint 2:
2x1 + 3x2 < 19
when x1 = 0, then x2 = 19/3 (0 , 19/3)
when x2 = 0, then x1 = 19/2 (19/2 , 0) Connect (0 , 19/3) and (19/2 , 0) .
Linear Programming
Graphical solution

Constraint 3:
x1 + x2 < 8
when x1 = 0, then x2 = 8 (0 , 8)
when x2 = 0, then x1 = 8 (8 , 0)
Linear Programming
Graphical solution

Linear Programming
Graphical solution
Determining the feasible solution
Feasible
Region

1. After drawing the feasible region, find the co-ordinates (x1,x2) for each corner
point
2. Solve for the corner Point (4)
at the Intersection of the second and third line
2x1 + 3x2 < 19
x1 + x2 < 8
solving these two equations gives:
x1 = 5 and x2 = 3 (5 , 3)
3. Solve for the corner point (3)
at the intersection of the first and third line
x1 < 6
x1 + x2 < 8
solving these two equations gives:
x1 = 6 and x2 = 2 (6 , 2)
Linear Programming
Graphical solution
Determining the optimal solution
The corner points Method
5
1 2
3
4
(0 , 6.33)
(6 , 0)(0, 0)
(5 , 3)
(6 , 2)

Objective function is
Max z = 5x1 + 7x2
• Find the value of Z at each corner
• Select the corner with max value of Z to
be the optimal solution
The optimal solution is x1 = 5 , x2 = 3
Z = 46
Pts. X1,X2 Z
1 0,0 0
2 6,0 30
3 6,2 44
4 5,3 46
5 0,19/3 44.33
Linear Programming
Graphical solution
Determining the optimal solution
The corner points Method

Product Mix Example
A company wishes to produce two types of products: type-A will result in a profit of $1.00,
and type-B in a profit of $1.2, to manufacture a type-A product requires 2 minutes on
machine I and 1 minute on machine II. A type-B product requires 1 minute on machine I
and 3 minutes on machine II. There are 3 hours available on machine I and 5 hours
available on machine II.
How many products of each type should the company make in order to maximize its
profit?
Solution
Type-A Type-B Time Available
Profit/Unit $1.00 $1.20
Machine I 2 min 1 min 180 min
Machine II 1 min 3 min 300 min
Linear Programming
Graphical solution

Decision Variables
X1 be the quantity produced from type-A
X2 be the quantity produced from type-B
Objective Function
Max Z = X1 + 1.2X2
Constraints
• The total amount of time that machine I used is (2 X1 + X2) and must not exceed 180
thus, we have the inequality 2X1 + X2 < 180
• The total amount of time that machine II used is (X1 + 3X2) and must not exceed 180
thus, we have the inequality 1X1 + 3X2 < 300
Finally, neither X1 nor X2 can be negative, so X1 , X2 0
Linear Programming
Graphical solution

200
100
100 200 300
 We first graph the feasible set for the constraint 2X1 + X2 < 180
considering only positive values for X1 , X2
X1
X2
(90, 0)
(0, 180)
2X1 + X2 < 180
Linear Programming
Graphical solution

200
100
Graphing the feasible set for the constraint X1 + 3X2 < 300
100 200 300
(0, 100)
(300, 0)
X1 + 3X2 < 300
X1
X2
Linear Programming
Graphical solution

200
100
Graph the intersection of the solutions to the
inequalities, yielding the feasible set .
100 200 300
S
X1
X2
Linear Programming
Graphical solution

200
100
100 200 300
S
C(48, 84)
D(0, 100)
B(90, 0)A(0, 0)
Next, find the vertices of the feasible set .
The vertices are A(0, 0), B(90, 0), C(48, 84), and D(0, 100).
X2
X1
Linear Programming
Graphical solution
Vertex Z = X1 + 1.2 X2
A(0, 0) 0
B(90, 0) 90
C(48, 84) 148.8
D(0, 100) 120
Vertex Z = X1 + 1.2 X2
A(0, 0) 0
B(90, 0) 90
C(48, 84) 148.8
D(0, 100) 120
Finally, identify the vertex with the highest value for Z:
We can see that Z is maximized at the vertex C(48, 84) with value 148.8.

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Linear Programming
Graphical solution
ISO- Profit (or Cost) method
ISO-profit line method is an approach to identifying the optimum point in a graphic linear programming
problem. The line representing the objective function and touches a particular point of the feasible region
will pinpoint the optimal solution. Once the feasible region has been graphed, one can find the optimal
solution to the problem. The optimal solution is the point lying in the feasible region that produces the
highest profit.
Step 1: Draw the half planes of all the constraints
Step 2: Shade the intersection of all the half planes which is the feasible region.
Step 3: Since the objective function is Z = ax + by, draw a dotted line for the equation ax + by = k,
where k is varies between 0 and the maximum value( max x and max y).
Step 4: To maximize Z draw a line parallel to ax + by = k and farthest from the origin. This line
should contain at least one point of the feasible region. Find the coordinates of this point by
solving the equations of the lines on which it lies.
To minimize Z draw a line parallel to ax + by = k and nearest to the origin. This line should contain
at least one point of the feasible region. Find the co-ordinates of this point by solving the equation
of the line on which it lies.
Step 5: If (x1, y1) is the point found in step 4, then x = x1, y = y1, is the optimal solution of the LPP
and Z = ax1 + by1 is the optimal value.
Dr. Abdulfatah Salem

3/19/2020 9:02 PM 38
Linear Programming
Graphical solution
2X1 + X2  1000
3X1 + 4X2  2400
X1 + X2  650
X1 - X2  350
X1 , X2 ≥ 0
2X1 + X2 = 1000
3X1 + 4X2 = 2400
X1 + X2 = 650
X1 - X2 = 350
X1 , X2 = 0
(0,1000) (500,0)
(0,600) (800,0)
(0,650) (650,0)
(0,-350) (0,350)
Example

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Linear Programming
Graphical solution
Max Z = 80X1 + 50X2
Let k = 50*80 = 4000
80X1 + 50X2 = 4000
350
300
Z = 80X1 + 50X2
= 80(350) + 50(300)
= 43000

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Example
Linear Programming
Graphical solution
Max z = 50x + 18y
s.t.
2x + y ≤ 100
x + y ≤ 80
x ≥ 0 y ≥ 0
Solution

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Linear Programming
Graphical solution
(0,80)
1
4
3
2
(20,60)
(50,0)
(0,0)
Max z = 50x + 18y
50x + 18y = k = 900
The optimal solution will be at
x = 50 y = 0
Z = 50*50 + 18*0
= 2500

Max z = 5x1 + 7x2
s.t. x1 < 6
2x1 + 3x2 < 19
x1 + x2 < 8
x1, x2 > 0
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Linear Programming
Graphical solution
Solution

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Linear Programming
Graphical solution
Max z = 5x1 + 7x2
5x1 + 7x2 = k = 35
(5 , 3)The optimal solution will be at
x1 = 5 x2 = 3
Z = 5*5 + 7*3 = 46
3
2
5
6
6.33

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Sensitivity Analysis (What if analysis)
Linear Programming
Graphical solution
Investigation of the changes in the optimum solution resulting from
changes in the basic parameters of linear programming model
Changes in
Objective
function
coefficients
Right hand
side
constants
Addition of
new
variables
Deletion of
variables
Addition of
new
constraints
Deletion of
constraints
linear programming
modal parameters
objective function constraints

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Linear Programming
Graphical solution
X2
X1
a
X2 = a + bX1
Slope b

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Linear Programming
Graphical solution
S.T. X1 < 6
2X1 + 3X2 < 19
X1 + X2 < 8
X1, X2 > 0
MAX Z = 5X1 + 7X2
Solution
5X1 + 7X2 = Z
X1 = 5 X2 = 3 Z = 46
5X1 + 7X2 = 46
C1X1 + C2X2 = 46

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Linear Programming
Graphical solution
C1X1 + C2X2 = 46
2X1 + 3X2 = 19
X1 + X2 = 8
X2 = (46/C2) – (C1/C2) X1
X2 = (19/3) – (2/3)X1
X2 = 8 – X1
b0 = – (C1/C2)
b1 = – (2/3)
b2 = – 1
b1 ≥ b0 ≥ b2
– (2/3) ≥ – (C1/C2) ≥ – 1
(2/3) ≤ (C1/C2) ≤ 1
Now, If you want to alter the value of C2 from 7 to 9
So, the value of C1 can be changed to :
(2/3) ≤ (C1/9) ≤ 1
6 ≤ C1 ≤ 9
while, If you want to alter the value of C1 from 5 to 9
So, the value of C2 can be changed to :
(2/3) ≤ (9/C2) ≤ 1
3/2 ≥ C2 /9 ≥ 1
27/2 ≥ C2 ≥ 9
b0
b1
b2

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Linear Programming
Graphical solution
If C1 is the unit price for Product X1 and C2 is the unit price for Product X2 , we want to
maintain the unit price for product X2 at value 12 , what is the allowed change in the unit
price of product X1 .
(2/3) ≤ (C1 / C2) ≤ 1
(2/3) ≤ (C1 / 12) ≤ 1
(24/3) ≤ (C1 ) ≤ 12
8 ≤ C1 ≤ 12
So the unit price for product C1 can vary from 8 to 12 without affecting the optimal point

Linear Programming
Graphical solution
Objective Function:
Max Z = 80X1 + 50X2
Constraints:
2X1 + X2  1000
3X1 + 4X2  2400
X1 + X2  650
X1 - X2  350
X1 , X2 ≥ 0
Points ( X1 , X2 ) Z
Point (0) ( 0 , 0 ) 0
Point (1) ( 0 , 600 ) 30000
Point (2) ( 200 , 450 ) 38500
Point (3) ( 350 , 300 ) 43000
Point (4) ( 450 , 100 ) 41000
Point (5) ( 350 , 0 ) 28000

Linear Programming
Graphical solution
80X1 + 50X2 = 43000
C1X1 + C2X2 = 43000 X2 = 43000/C2 - (C1/C2)X1
X1 + X2 = 650 X2 = 650 - X1
2X1 + X2 =1000 X2 = 1000 - 2X1
1 ≤ (C1/C2) ≤ 2
b0 = – (C1/C2)
b1 = – 1
b2 = – 2
If you know that C1 , C2 are prices of two products, for marketing reasons
the manager wants to make little reduction with C2 to be 40 instead of 50,
what is the allowable values for C1 .
1 ≤ (C1/C2) ≤ 2
1 ≤ (C1/40) ≤ 2
40 ≤ (C1) ≤ 80

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Linear Programming
Graphical solution
Objective Function line:
Max Z = 80X1 + 50X2
80X1 + 50X2 = Z
X2 = Z - (80/50)X1
The slope of objective: = - (C1/C2)
= - (80/50)
= - 1.6
Constraints lines:
2X1 + X2 = 1000 with slope = -2
3X1 + 4X2 = 2400 with slope = -3/4
X1 + X2 = 650 with slope = -1
X1 - X2 = 350 with slope = 1
X1 = 0 with slope = ∞
X2 = 0 with slope = 0

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Linear Programming
Graphical solution
Constraints Slope L1 L2 L3 L4 L5 L6 Co-ords Z
L1 X1 = 0 -∞ 0 0,0 0
L2 2X1 + X2 = 1000 -2 1 0,600 30000
L3 X1 + X2 = 650 -1 2 200,450 38000
L4 3X1 + 4X2 = 2400 -3/4 3 350,300 43000
L5 X1 - X2 = 350 1 4 450,100 41000
L6 X2 = 0 0 5 350,0 28000

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Special cases in Graphical method
Linear Programming
Graphical solution
There are many difficulties arise at times when solving LP problems using the
graphical approach.
1. Multiple Optimal Solutions
2. No feasible solution
3. Unboundedness
4. Redundancy

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Linear Programming
Graphical solution
Multiple Optimal Solutions
Max Z = 4X1 + 3X2
s.t. 4X1+ 3X2 ≤ 24
X1 ≤ 4.5
X2 ≤ 6
X1 ≥ 0 , X2 ≥ 0
at (0, 6)
Z = 4(0) + 3(6) = 18
at (1.5, 6)
Z = 4(1.5) + 3(6) = 24
at (4.5, 2)
Z = 4(4.5) + 3(2) = 24
at (4.5, 0)
Z = 4(4.5) + 3(0) = 18
Max Z = 4X1 + 3X2
Ex.

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Linear Programming
Graphical solution
Multiple Optimal Solutions
Max Z = X1 + 2X2
s.t. X1 ≤ 80
X2 ≤ 60
5X1 + 6X2 ≤ 600
X1 + 2X2 ≤ 160
X1 ≥ 0 , X2 ≥ 0
Ex.

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Linear Programming
Graphical solution
No Optimal Solutions
Max Z = 3X1 + 2X2
S.t.
X1+ X2 ≤ 1
X1+ 2X2 ≥ 3
X1 ≥ 0 , X2 ≥ 0
Ex.
No Feasible Solutions

3/19/2020 9:02 PM 57
Linear Programming
Graphical solution
No Optimal Solutions
Max Z = 200X1 + 300X2
S.t.
X1+ X2 ≤ 400
2X1+ 1.5X2 ≥ 900
2X1+ 3X2 ≥ 1200
X1 ≥ 0 , X2 ≥ 0
Ex.
No Feasible Solutions

3/19/2020 9:02 PM 58
Linear Programming
Graphical solution
Unbounded Solutions
Max Z = 3X1 + 5X2
s.t. 2X1+ X2 ≥ 7
X1 + X2 ≥ 6
X1+ 3X2 ≥ 9
X1 ≥ 0 , X2 ≥ 0
Ex.

3/19/2020 9:02 PM 59
Linear Programming
Graphical solution
Max Z = 4X1 + 3X2
s.t. 4X1+ 3X2 ≤ 24
X1 ≤ 4.5
X2 ≤ 6
X1 ≤ 10
X1 ≥ 0 , X2 ≥ 0
Ex.
Redundancy
X1≤10

3/19/2020 9:02 PM 60
Linear Programming
Graphical solution
Ex.
Redundancy
Max Z = 4x1 + 3x2
S.t

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