LP Graphical Solution

Linear Programming – Graphical Method

1. Maximise: Z = 7X1 + 5X2

s.t. X1 + 2X2 6

4X1 + 3X2 12

X1,X2 0

Equation When X1 = 0 When X2 = 0

X1 + 2 X2 = 6 0 + 2 X2 = 6

X2 = 3

Point is (0,3)

X1 + 0 = 6

X1 = 6

Point is (6,0)

4X1 + 3 X2 = 12 0 + 3 X2 = 12

X2 = 4

Point is (0,4)

4X1 + 0 = 12

X1 = 3

Point is (3,0)

1 2 3 4 5 6

1

2

3

4

X1 + 2 X

2 ≤ 6

4X1 + 3 X

2 ≤ 12

A

BC

D

X1

X2 ≤ Shaded area towards the origin

To find C X1 + 2 X2 = 6 ----- (1)

4X1 + 3 X2 = 12 ----- (2)

(1)X 4 4X1 + 8 X2 = 24

(-) -5X2 = - 12

X2 = 12/5 = 2.4

Substitute X2 value in Eq (1)

X1 + 2 (2.4) = 6

Therefore X1 = 1.2

Point Co-Ordinate Zmax = 7 X1 + 5 X2

A 0, 0 7 x 0 + 5 x 0 = 0

B 0, 3 7 x 0 + 5 x 3 = 15

C 1.2, 2.4 7 x 1.2 + 5 x 2.4 = 20.4

D 3, 0 7 x 3 + 5 x 0 = 21

Conclusion: X1 = 3 and X2 = 0 will give Zmax = 21


2) Maximise: Z=3X1 + 2X2

s.t. 2X1 + X2 2

3X1 + 4X2 12

X1, X2 0


2X1 + X2 = 2 0 + X2 = 2

X2 = 2

Point is (0,2)

2 X1 + 0 = 2

X1 = 1

Point is (1,0)

3X1 + 4 X2 = 12 0 + 4 X2 = 12

X2 = 3

Point is (0,3)

3X1 + 0 = 12

X1 = 4

Point is (4,0)

1 2

1

2

3X1 + 4 X

2 ≥ 12

2X1 + X

2 ≤ 2

X1

X2

≤ Shaded area towards the origin

Conclusion: Since there is no common area, there is feasible solution for the given problem

≥ Shaded area away from the origin

3

3 4

Identify the common area for the following examples

a) 2 X + Y 3 Y – 3 X 1 X, Y 0

Equation When X = 0 When Y = 0

2X + Y = 3 0 + Y = 3

Y = 3

Point is (0, 3)

2 X + 0 = 3

X = 1.5

Point is (1.5, 0)

Y - 3 X = 1 Y - 3 (0) = 1

Y = 1

Point is (0, 1)

0 – 3 X = 1

X = -1/3

Point is (-1/3, 0)

1 2

1

2

2X + Y

≥ 3

Y -

3X ≤

1

X1

X2

≤ Shaded area towards the origin≥ Shaded area away from the origin3

3 4

Identify the common area for the following examples

b) 2 X + 3Y 6 3 X– 2 Y 4X,Y 0

Equation When X = 0 When Y = 0

2X + 3Y = 6 0 + 3Y = 6

Y = 2

Point is (0, 2)

2 X + 0 = 6

X = 3

Point is (3, 0)

3 X – 2 Y = 4 0 - 2 Y = 4

Y = - 2

Point is (0, -2)

3 X – 0 = 4

X = 4/3

Point is (4/3, 0)

1 2

-2

1

2X + 3Y ≤ 3

3X -2

Y ≥

4

X1

X2

≤ Shaded area towards the origin≥ Shaded area away from the origin2

3 4

-1

3



s.t.X1 + X2 = 200

X1 80

X2 60

X1,X2 0


X1 + X2 = 200 0 + X2 = 200

X2 = 200

Point is (0,200)

X1 + 0 = 200

X1 = 200

Point is (200,0)

X1 = 80 Point is (80,0)

50 100 150

50

100

X1 + X

2 = 200

X1 ≤ 80

A

B

X1

X2



A 0, 200 3 x 0 + 8 x 200 = 1600

B 80, 120 3 x 80 + 8 x 120 = 1200

X2 = 60 Point is (0, 60)

200

150

200

= Shaded area only on the line≥ Shaded area away from the origin

X2 ≥ 60

(0, 200)

(80, 120)




s.t.-2X1 + X2 1

X1 2

X1 + X2 3

X1,X2 0

1 2 3

1

2

X1 + X

2 ≤ 3X

1 ≤ 2

AX1

X2



4

3

4- 2

X1 +

X2 ≤

1

- 1


- 2X1 + X2 = 1 0 + X2 = 1

X2 = 1

Point is (0,1)

- 2 X1 + 0 = 1

X1 = - 0.5

Point is (-0.5,0)

X1 = 2 Point is (2,0)

X1 + X2 = 3 Point is (0, 3) Point is (3, 0)

B

C

D

E

To find C -2 X1 + X2 = 1 ----- (1)

X1 + X2 = 3 ----- (2)

(1)-(2) -3 X1 = - 2

X1 = 2/3


2/3 + X2 = 3

Therefore X2 = 7/3


A 0, 0 3 x 0 + 2 x 0 = 0

B 0, 1 3 x 0 + 2 x 1 = 2

C 2/3, 7/3 3 x 2/3 + 2 x 7/3 = 20/3

D 2, 1 3 x 2 + 2 x 1 = 8

E 2, 0 3 x 2 + 2 x 0 = 6

Summary :

8. Let us assume that you have inherited Rs.1,00,000 from your father-in-law that can be invested in a combination of two stock portfolios, with the maximum investment allowed in either portfolio set at Rs.75,000. The first portfolio set has an average rate of return of 10% whereas the second has 20%. In terms of risk factors associated with these, the first has a risk rating of 4 (on a scale from 0 to 10) and the second has 9. Since you wish to maximize your return, you will not accept an average rate of return below 12% or risk factor above 6. Hence, you then face an important question. How much should you invest in each portfolio. Formulate this as a linear programming problem and solve it by the graphical method.

Input Portfolio I Portfolio II

Average rate of return

10% 20% Availability

Investment 1 1 100000

Max investment 1 - 75000

- 1 75000

Risk rating 4 9 6 (Max.)

Other Information 1. Will not accept an average rate of return below 12%

Step 0: Tabular Presentation

Step 1: Variable Definition

Let X1 be the amount invested in Portfolio I

Let X2 be the amount invested in Portfolio II

Step 2: Objective Function

Maximise return Z = 0.10 X1+ 0.20 X2

Step 3: Constraints

Amount inherited : X1 + X2 ≤ 100000

Maximum investment : X1 ≤ 75000 ; X2 ≤ 75000

Risk Factors : 4X1 + 9 X2 ≤ 6 (X1 + X2)

(OR) - 2 X1 + 3 X2 ≤ 0

Average rate of return not to be below 12 % :

0.10 X1 + 0.20 X2 ≥ 0.12 (X1 + X2)

- 0.02 X1 + 0.08 X2 ≥ 0

This linear programming problem is summarised as follows: Maximise return Z = 0.10 X1+ 0.20 X2

s.t.X1 + X2 ≤ 100000

X1 ≤ 75000 ; X2 ≤ 75000 - 2 X1 + 3 X2 ≤ 0

-0.02 X1 + 0.08 X2 ≥ 0

X1, X2 ≥ 0

25 50 75

25

50

X1 + X

2 ≤ 100000

X1 ≤ 75000

AX1

X2


100

75

100


X1 + X2 = 100000 Point is (0,100000) Point is (100000,0)

X2 = 75000 Point is (0, 75000)

-2X1 + 3X2 = 0 When X1 = 75000;

-2(75000)+3X2 = 0

3 X2 = 150000 (or) X2 = 50000

Point is (75000, 50000)

B

C

Maximise return Z = 0.10 X1+ 0.20 X2

s.t.X1 + X2 ≤ 100000

X1 ≤ 75000 ; X2 ≤ 75000

- 2 X1 + 3 X2 ≤ 0

-0.02 X1 + 0.08 X2 ≥ 0

X1, X2 ≥ 0

X1 = 75000 Point is (75000,0)

Point is (75000, 50000) and (0, 0)

-0.02X1 + 0.08X2 = 0 When X1 = 75000;

-0.02(75000)+0.08X2 = 0

0.08 X2 = 1500 (or) X2 = 18750

Point is (75000, 18750)

Point is (75000, 18750) and (0, 0)

≥ Shaded area away frrom the origin

’000

’000

D

X2 ≤ 75000

-2 X 1 + 3 X 2

≤ 0

-0.02 X1 + 0.08 X2

≥ 0

To find B X1 + X2 = 100000 ----- (1)

-2 X1 + 3 X2 = 0 ---- (2)

(1) X 2 2 X1 + 2 X2 = 200000

(add) 5 X2 = 200000

X2 = 40000


X1 = 60000

To find C; X1 = 75000

X1 + X2 = 100000

X2 = 25000

(75000, 25000)

To find D; X1 = 75000

0.08 X2 -0.02 X1 = 0

0.08 X2 = 1500

X2 = 18750

(75000, 18750)

Point Co-Ordinate Zmax = 0.1 X1 + 0.2 X2

A 0, 0 0.1 x 0 + 0.2 x 0 = 0

B 60000, 40000 0.1 x 60000 + 0.2 x 40000 = 14000

C 75000, 25000 0.1 x 75000 + 0.2 x 25000 = 12500

D 75000, 18750 0.1 x 75000 + 0.2 x 18750 = 11250


Minimise: Z = 4X1 + 2X2

s.t.X1 + 2 X2 ≥ 2

3 X1 + 2 X2 ≥ 3

4 X1 + 3 X2 ≥ 6

X1, X2 ≥ 0


X1 + 2 X2 = 2 0 + 2 X2 = 2

X2 = 1

Point is (0, 1)

X1 + 0 = 2

X1 = 2

Point is (2, 0)

3X1 + 2 X2 = 3 0 + 2 X2 = 3

X2 = 1.5

Point is (0, 1.5)

3X1 + 0 = 3

X1 = 1

Point is (1,0)

4X1 + 3 X2 = 6 0 + 3 X2 = 6

X2 = 2

Point is (0, 2)

4X1 + 0 = 6

X1 = 1.5

Point is (1.5, 0)

1

1

2

X2

2 3X1


3X1 + 2 X

2 ≥ 3

4X1 + 3 X

2 ≥ 6

B

A

C

X1 + 2 X

2 ≥ 2

To find B X1 + 2X2 = 2 ----- (1)

4 X1 + 3 X2= 6 ---- (2)

(1) X 4 4 X1 + 8 X2 = 8

(Sub) -5 X2 = - 2

X2 = 2/5 = 0.4


X1 + 2 (0.4) = 2

Therefore X1 = 1.2


A 2, 0 4 x 2 + 2 x 0 = 8

B 1.2, 0.4 4 x 1.2 + 2 x 0.4 = 5.6

C 0, 2 4 x 0 + 2 x 2 = 4

Conclusion: X1 = 0 and X2 = 2 will give Zmin = 4

9. A local travel agent is planning a chartered trip to a major sea resort. The eight day/seven night package includes the fare for round trip, surface transportation, board and lodging and selected tour options. The chartered trip is restricted to 200 persons and past experience indicates that there will not be any problem for getting 200 persons. The problem for the travel agent is to determine the number of Deluxe, Standard and Economy tour packages to offer for this charter. These three plans each differ according to seating and service for the flight, quality of accommodation , meal plans and tour options. The following table summarises the estimated prices for the 3 packages and the corresponding expenses for the travel agent. The travel agent has hired an aircraft for the flat fee of Rs.2,00,000 for the entire trip.

PRICES AND COST FOR TOUR PACKAGES / PERSONS

Tour Plan Price Rs. Hotel Cost Meals & other expenses

Deluxe 10,000 3,000 4,750

Standard 7,000 2,200 2,500

Economy 6,500 1,900 2,200

In planning the trip the following considerations must be taken into account

a) atleast 10% of packages must be of the deluxe typeb) atleast 35% but not more than 70% must be of the standard typec) atleast 30% must be of the economy typed) the maximum number of deluxe packages available in any aircraft is restricted to 60e) the hotel desires that atleast 120 of the tourists should be on the deluxe and the standard packages together.

The travelling agent wishes to determine the number of packages to offer in each type so as to maximise the total profit. a)Formulate the above as a linear programming problemb) Restate the above linear programming problem in terms of two decision variables taking advantage of the fact that 200 packages will be sold.c) Find the optimal solution using graphical method for the restated linear programming problem and interpret your results.

200111No. of Package

120 (Min.)-11Hotel Constraint

60--1

140 (70% of 200)-1-Max. Package

60 (30% of 200)1--

70 (35% of 200)-1-

20 (10% of 200)--1Min. Package

240023002250Profit

220025004750Meals & Other expenses

190022003000Hotel Cost

6500700010000Price

EconomyStandardDeluxe

Summary

Max Z = 2250 X1 + 2300 X2 + 2400 X3 – 200000

S.t.

(i) X1 ≥ 20 ; (ii) X2 ≥ 70 and X2 ≤ 140

(iii) X3 ≥ 60 ; (iv) X1 ≤ 60

(v) X1 + X2 ≥ 120

(vi) X1 + X2 + X3 = 200

X1, X2, X3 ≥ 0 In order to solve this problem by graphical method, reduce the problem to 2 variables. Since X1 + X2 + X3 = 200 or X3 = 200 – (X1 +X2).

Substitute the value of X3 in the relations

mentioned above.

Max Z = -150 X1 - 100 X2 + 2800000

S.t.(i) X1 ≥ 20 ; (ii) X2 ≥ 70 and X2 ≤ 140

(iii) X1+ X2 ≤ 140 ; (iv) X1 ≤ 60

(v) X1 + X2 ≥ 120

X1, X2 ≥ 0


X1 = 20 Point is (20, 0)

X2 = 70 Point is (70, 0)


Max Z = -150 X1 + 100 X2 + 2800000

S.t.(i) X1 ≥ 20 ; (ii) X2 ≥ 70 and X2 ≤ 140

(iii) X1+ X2 ≤ 140 ; (iv) X1 ≤ 60

(v) X1 + X2 ≥ 120

X1, X2 ≥ 0 X1 = 60 Point is (60, 0)

X2 = 140 Point is (140, 0)


20 X1

X2

40 60 80 100 120 140

20

40

60

80

100

120

140



X1 ≥

20

X2 ≥ 70

X1

≤ 6

0

X2 ≤ 140

X1 + X

2 ≥ 120

X1 + X

2 ≤ 140

A

B

C

DE

Point Co-Ordinate Zmax= - 150 X1 - 100 X2+280000

A 20, 100 -150 x 20 - 100 x 100+ 280000 = 267000

B 20, 120 -150 x 20 - 100 x 120 + 280000 = 265000

C 60, 80 -150 x 60 - 100 x 80 + 280000 = 263000

D 60, 70 -150 x 60 - 100 x 70 + 280000 = 264000

E 50, 70 -150 x 50 - 100 x 70 + 280000 = 265500

Conclusion: X1 = 20 , X2 = 100 and X3 = 80 will give Zmax = 267000

X1 = 20 ; X2 = 100 ; Therefore X3 = 200-(X1+X2) = 200 – (20+100) = 80

LP Graphical Solution

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