Linear Programming (graphical method)

Faculty of Economics and Business Administration
Lebanese University
Chapter 1: Linear Programming
(Graphical Method)
Dr. Kamel ATTAR
attar.kamel@gmail.com
Lecture #2 F Thursday 4/MAR/2021 F

2Ú42
Linear Programming
The Graphical Solution of Two-Variable L.P Problems
Examples
Special Cases
1 Linear Programming
Definition
Feasible Region and Optimal Solution
2 The Graphical Solution of Two-Variable L.P Problems
The characteristics of Graphical method
Finding the Feasible Solution
Finding the Optimal Solution
3 Examples
The Graphical Solution of Maximization Problems
The Graphical Solution of Minimization Problems
4 Special Cases
Alternative or Multiple Optimal Solutions
Infeasible solutions or No solutions
Unbounded problem
Dr. Kamel ATTAR | Chapter 1: Linear Programming-Graphical Method: | Graphical Method

3Ú42
Linear Programming
Examples
Special Cases
Definition
H Linear programming H
Definition
A linear programming problem (LP) is an optimization problem for which we do the
following:
À We attempt to maximize (or minimize) a linear function of the decision variables.
The function that is to be maximized or minimized is called the objective function.
A function f(x1, x2, · · · , xn) is a linear function ⇐⇒ ∃c1, c2, · · · , cn , s.t.
f(x1, x2, · · · , xn) = c1x1 + c2x2 + · · · + cnxn

4Ú42
Linear Programming
Examples
Special Cases
Definition
Definition
following:
f(x1, x2, · · · , xn) = c1x1 + c2x2 + · · · + cnxn
Á The values of the decision variables must satisfy a set of constraints. Each
constraint must be a linear equation or linear inequality.
c1x1 + c2x2 + · · · + cnxn ≤ b is a linear inequality

5Ú42
Linear Programming
Examples
Special Cases
Definition
Definition
following:
f(x1, x2, · · · , xn) = c1x1 + c2x2 + · · · + cnxn
Á The values of the decision variables must satisfy a set of constraints. Each
constraint must be a linear equation or linear inequality.
c1x1 + c2x2 + · · · + cnxn ≤ b is a linear inequality
Â A sign restriction is associated with each variable. For any variable xi , the sign
restriction specifies that xi must be either nonnegative xi ≥ 0 or unrestricted in
sign.

6Ú42
Linear Programming
Examples
Special Cases
Definition
Definition (Feasible region)
The feasible region for an LP is the set of all points that satisfies all the LP’s constraints
and sign restrictions.
Example

7Ú42
Linear Programming
Examples
Special Cases
Definition
Example
The point (x1 = 40, x2 = 20) is in the feasible region of the
following LP’s constraints
On the other hand, the point (x1 = 15, x2 = 70) is not in the
feasible region,







2x1 + x2 ≤ 100
x1 + x2 ≤ 80
x1 ≤ 40
x1, x2 ≥ 0
Definition (Optimal Solution)

8Ú42
Linear Programming
Examples
Special Cases
Definition
Example
The point (x1 = 40, x2 = 20) is in the feasible region of the
following LP’s constraints
On the other hand, the point (x1 = 15, x2 = 70) is not in the
feasible region,







2x1 + x2 ≤ 100
x1 + x2 ≤ 80
x1 ≤ 40
x1, x2 ≥ 0
Definition (Optimal Solution)
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.

9Ú42
Linear Programming
Examples
Special Cases
H The Graphical Solution of Two-Variable L.P Problems H
À Generally the method is used to solve the problem, when it involves two
decision variables.

10Ú42
Linear Programming
Examples
Special Cases
decision variables.
Á For three or more decision variables, the graph deals with planes and
requires high imagination to identify the solution area.

11Ú42
Linear Programming
Examples
Special Cases
decision variables.
Â Always, the solution to the problem lies in first quadrant.

12Ú42
Linear Programming
Examples
Special Cases
decision variables.
Â Always, the solution to the problem lies in first quadrant.
Ã This method provides a basis for understanding the other methods of
solution.
We always label x1 and x2 and the coordinate axes the x1 and x2 axes.

13Ú42
Linear Programming
Examples
Special Cases
Example
Max : z = x1 + 4x2
Under the constraints:







x1 + x2 ≤ 8
x1 + 3x2 ≤ 18
2x1 + x2 ≤ 14
x1 , x2 ≥ 0
Solution

14Ú42
Linear Programming
Examples
Special Cases
Example
Max : z = x1 + 4x2







x1 + x2 ≤ 8
x1 + 3x2 ≤ 18
2x1 + x2 ≤ 14
x1 , x2 ≥ 0
Solution
The feasible region will be the set of points in the first quadrant that satisfies the
first three inequalities.
x1 + x2 = 8
x1 0 8
x2 8 0
x1 + 3x2 = 18
x1 0 18
x2 6 0
2x1 + x2 = 14
x1 0 7
x2 14 0

15Ú42
Linear Programming
Examples
Special Cases

x1 + x2 = 8
x1 + 3x2 = 18
B(x1 = 3, x2 = 5)

2x1 + x2 = 14
x1 + x2 = 8
C(x1 = 6, x2 = 2)

2x1 + x2 = 14
x2 = 0
D(x1 = 7, x2 = 0)

x1 + 3x2 = 18
x1 = 0
A(x1 = 0, x2 = 6)

16Ú42
Linear Programming
Examples
Special Cases
We need to find the extremity points of the feasible solution. Here, the extremity
points are O(0, 0), A(0, 6) and D(7, 0) for the last two points B and C, we see
that B is the intersection between x1 + 3x2 = 18 and x1 + x2 = 8, and C is the
intersection between 2x1 + x2 = 14 and x1 + x2 = 8, by comparison we find
B = (3, 5) and C = (6, 2).
Now the optimal solution must be one of these extremity points, so we
substitute each point in the objective function. In a max problem the largest
value corresponds to the optimal value (in a min problem, the smalest value).
We obtain, for:
O(0, 0) ; z = 0 + 4(0) = 0 5
A(0, 6) ; z = 0 + 4(6) = 24 3
B(3, 5) ; z = 3 + 4(5) = 23 5
C(6, 2) ; z = 6 + 4(2) = 14 5
D(7, 0) ; z = 7 + 4(0) = 7 5
Thus point A has the greatest value so it’s the optimal solution with max value z = 24.

17Ú42
Linear Programming
Examples
Special Cases
H Examples H
Example (À)
A company produces two types of products, A and B. The products A and B passes
through three types of work before they are in their final state.
Machines nb. of hrs needed nb. of hrs needed Cost in $ per hr.
by product A by product B
Machine 1 2 1 10
Machine 2 1 4.5 12
Machine 3 4 3 14
The company can afford 200 hrs for machine 1 and 540 hrs for machine 2 and 480 hrs
for machine 3.
What are the quantities produced of types A and B that maximize the profit of the
company if the price of type A is 138$ and the price of type B is 136$.

18Ú42
Linear Programming
Examples
Special Cases
Solution
Cost A : 2 × 10 + 12 × 1 + 4 × 14 = 88$
Cost B : 1 × 10 + 12 × 4.5 + 3 × 14 = 106$
x1 : nb. of units produced of type A
x2 : nb. of units produced of type B
Max z : (138 − 88)x1 + (136 − 106)x2 = 50x1 + 30x2

19Ú42
Linear Programming
Examples
Special Cases
Solution
Cost A : 2 × 10 + 12 × 1 + 4 × 14 = 88$
Cost B : 1 × 10 + 12 × 4.5 + 3 × 14 = 106$
Max z : (138 − 88)x1 + (136 − 106)x2 = 50x1 + 30x2
Therefore
max z = 50x1 + 30x2







2x1 + x2 ≤ 200
x1 + 4.5x2 ≤ 540
4x1 + 3x2 ≤ 480
x1, x2 ≥ 0

20Ú42
Linear Programming
Examples
Special Cases
Solution
Cost A : 2 × 10 + 12 × 1 + 4 × 14 = 88$
Cost B : 1 × 10 + 12 × 4.5 + 3 × 14 = 106$
Max z : (138 − 88)x1 + (136 − 106)x2 = 50x1 + 30x2
Therefore
max z = 50x1 + 30x2







2x1 + x2 ≤ 200
x1 + 4.5x2 ≤ 540
4x1 + 3x2 ≤ 480
x1, x2 ≥ 0
Now let’s draw the linear equations:
2x1 + x2 = 200
x1 0 100
x2 200 0
x1 + 4.5x2 = 540
x1 0 90
x2 120 100
4x1 + 3x2 = 480
x1 0 120
x2 160 0

21Ú42
Linear Programming
Examples
Special Cases

x1 + 4.5x2 = 540
4x1 + 3x2 = 480
B(x1 = 36, x2 = 112)

2x1 + x2 = 200
4x1 + 3x2 = 480
A(x1 = 60, x2 = 80)

x1 + 4.5x2 = 540
x1 = 0
C(x1 = 0, x2 = 120)

4x1 + 3x2 = 480
x2 = 0
D(x1 = 100, x2 = 0)

22Ú42
Linear Programming
Examples
Special Cases
The coordinates of the corners are:
C = (0, 120), O = (0, 0), D = (100, 0), B = (36, 112) and A = (60, 80) .
Substituting these values in objective function. We obtain, for:
O(0, 0) ; z = 50(0) + 30(0) = 0 5
A(60, 80) ; z = 50(60) + 30(80) = 5400 3
B(36, 112) ; z = 50(36) + 30(112) = 5160 5
C(0, 120 ; z = 50(0) + 30(120) = 3600 5
D(100, 0) ; z = 50(100) + 30(0) = 5000 5
The Max profit occures at the point A, so we obtain a Max profit of 5400$ when
we produce 60 items of product A and 80 items of product B.

23Ú42
Linear Programming
Examples
Special Cases
Example (Á)
A company Produces 2 types of chocolate (light chocolate dark chocolate).
• For light chocolate; we need 5g of sugar, 4g of cacao and 0.5g of butter.
• For dark chocolate; we need 10g of sugar and 3g of cacao.
The company must use at least 90g of sugar, 48g of cacao and 1.5g of butter.
If the cost of one unit of light chocolate is 2$ and the cost of one unit of dark chocolate is 3$. Find
the combination of products that gives the minimal cost.
Solution

24Ú42
Linear Programming
Examples
Special Cases
Example (Á)
A company Produces 2 types of chocolate (light chocolate dark chocolate).
• For light chocolate; we need 5g of sugar, 4g of cacao and 0.5g of butter.
• For dark chocolate; we need 10g of sugar and 3g of cacao.
The company must use at least 90g of sugar, 48g of cacao and 1.5g of butter.
If the cost of one unit of light chocolate is 2$ and the cost of one unit of dark chocolate is 3$. Find
the combination of products that gives the minimal cost.
Solution
The details given in the problem is given in the table below:
Ingredients Products Quantities in g
Type 1 (light) Type 2 (dark)
Sugar 5 10 90
Cacao 4 3 48
Butter 0.5 0 1.5
Profit per unit in $ 2 3

25Ú42
Linear Programming
Examples
Special Cases
The L.P. model is:
Min : z = 2x1 + 3x2



5x1 + 10x2 ≥ 90
4x1 + 3x2 ≥ 48
0.5x1 ≥ 1.5
x1, x2 ≥ 0

26Ú42
Linear Programming
Examples
Special Cases
Here we find the coordinates of corners and substitute the values in the
objective function. In minimization problem, we select the coordinates giving
minimum value. The coordinates of the corners are:
A = (3, 12), B = (8.4, 4.5) and C = (18, 0) .
A(3, 12) ; z = 2(3) + 3(12) = 42 5
B(8.4, 4.5) ; z = 2(8.4) + 3(4.5) = 30.3 3
C(18, 0) ; z = 2(18) + 3(0) = 36 5
Hence the optimal solution for the problem is company has to manufacture 18
units of product of type 1 and 9 units of product of type 2.

27Ú42
Linear Programming
Examples
Special Cases
Example (Â)
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 has decided to purchase 1-minute commercial spots on two types of
programs: comedy shows and football games.
• Each comedy commercial is seen by 7 million high-income women and 2
million high-income men.
• Each football commercial is seen by 2 million high-income women and 12
million high-income men.
A 1-minute comedy ad costs 50, 000$, and a 1-minute football ad costs
100, 000$. Dorian would like the commercials to be seen by at least 28 million
high-income women and 24 million high-income men. Use linear programming
to determine how Dorian Auto can meet its advertising requirements at
minimum cost.

28Ú42
Linear Programming
Examples
Special Cases
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 purchased
x2 number of 1-minute football ads purchased
Then Dorian wants to minimize total advertising cost (in thousands of dollars).
Total advertising cost=cost of comedy ads+cost of football ads=50x1 + 100x2.
Thus, Dorian’s objective function is
Min z = 50x1 + 100x2 .
Dorian faces the following constraints:
• Constraint 1 Commercials must reach at least 28 million high-income
women.
• Constraint 2 Commercials must reach at least 24 million high-income men.

29Ú42
Linear Programming
Examples
Special Cases
To express Constraints 1 and 2 in terms of x1 and x2, let HIW stand for
high-income women viewers and HIM stand for high-income men viewers (in
millions).
Ingredients Products of 1min Number of customers
comedy ads football ads
HIW 7 2 28
HIM 2 12 24
Cost per 1minute in $ 50, 000 100, 000
So the Dorian LP is given by:
Min : z = 50x1 + 100x2
Under the constraints: 






7x1 + 2x2 ≥ 28
2x1 + 12x2 ≥ 24
x1 ≥ 0
x2 ≥ 0

30Ú42
Linear Programming
Examples
Special Cases

31Ú42
Linear Programming
Examples
Special Cases
The coordinates of the corners are:
A = (0, 14), B = (18/5, 7/5) and C = (12, 0) .
A(0, 14) ; z = 50(0) + 100(14) = 1400 5
B(7/5, 18/5) ; z = 50(3.6) + 100(1.4) = 320 3
C(12, 0) ; z = 50(12) + 100(0) = 600 5

32Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
H Special Cases H
In this section, we encounter three types of LPs that do not have unique optimal solutions.
1. Some LPs have an infinite number of optimal solutions (alternative or multiple optimal
solutions).
2. Some LPs have no feasible solutions (infeasible LPs).
3. Some LPs are unbounded: There are points in the feasible region with arbitrarily large (in a
max problem) z-values.
Example
An auto company manufactures cars and trucks. Each vehicle must be processed in the paint shop
and body assembly shop. If the paint shop were only painting trucks, then 40 per day could be
painted. If the paint shop were only painting cars, then 60 per day could be painted. If the body
shop were only producing cars, then it could process 50 per day. If the body shop were only
producing trucks, then it could process 50 per day. Each truck contributes 300$ to profit, and each
car contributes 200$ to profit. Use linear programming to determine a daily production schedule
that will maximize the company’s profits.

33Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
Solution
The company must decide how many cars and trucks should be produced daily.
This leads us to define the following decision variables:
x1 is the number of trucks produced daily
x2 is number of cars produced daily
The company’s daily profit (in hundreds of dollars) is 3x1 + 2x2, so the
company’s objective function may be written as
Maxz = 3x1 + 2x2
The company’s two constraints are the following:
• Constraint 1 The fraction of the day during which the paint shop is busy is
less than or equal to 1.
• Constraint 2 The fraction of the day during which the body shop is busy is
less than or equal to 1.

34Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
Thus, Constraint 1 may be expressed by
1
40
x1 +
1
60
x2 ≤ 1 (Paint shop constraint)
Constraint 2 may be expressed by
1
50
x1 +
1
50
x2 ≤ 1 (Body shop constraint)
Because x1 ≥ 0 and x2 ≥ 0 must hold, the relevant LP is
Max : z = 3x1 + 2x2







1
40
x1 + 1
60
x2 ≤ 1
1
50
x1 + 1
50
x2 ≤ 1
x1 ≥ 0
x2 ≥ 0

35Ú42
Linear Programming
Examples
Special Cases
Unbounded problem

36Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
The coordinates of the corners are:A = (0, 50), B = (20, 30) and
C = (40, 0). Substituting these values in objective function. We obtain, for:
A(0, 50) ; z = 3(0) + 2(50) = 100 5
B(20, 30) ; z = 3(20) + 2(30) = 120 3
C(40, 0) ; z = 3(40) + 2(0) = 120 3
From our current example, it seems reasonable (and can be shown to be true)
that if two points (B and C here) are optimal, then any point on the line
segment joining these two points will also be optimal.
If an alternative optimum occurs, then the decision maker can use a secondary
criterion to choose between optimal solutions. The auto company’s managers
might prefer point C because it would simplify their business (and still allow
them to maximize profits) by allowing them to produce only one type of product
(trucks).

37Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
Example
Suppose that auto dealers require that the auto company in the previous
example produce at least 30 trucks and 20 cars. Find the optimal solution to
the new LP.
Solution
After adding the constraints x1 ≥ 30 and x2 ≥ 20 to the LP of the previous
example, we obtain the following LP:
Max : z = 3x1 + 2x2






1
40
x1 + 1
60
x2 ≤ 1
1
50
x1 + 1
50
x2 ≤ 1
x1 ≥ 30
x2 ≥ 20

38Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
From the figure it is clear that no
point satisfies all the constraints.
This means that the problem has
an empty feasible region and is
an infeasible LP.

39Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
Unbounded problem
Example
Graphically solve the following LP:
Max : z = 2x1 − x2


x1 − x2 ≤ 1
2x1 + x2 ≥ 6
x1, x2 ≥ 0

40Ú42
Linear Programming
Examples
Special Cases
Unbounded problem
Solution

41Ú42
Exercise
À Graphically solve the following LP:
Max : z = 3x1 + x2






2x1 + x2 ≤ 6
x1 + 3x2 ≤ 9
x1 + x2 ≤ 4
x1, x2 ≥ 0

Linear Programming (graphical method)

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