SVKM’s NIMIS University
Linear Programming – Graphical Solution
A graphical solution procedure is one method of solving two-variable LPP (Linear Programming
Problem). The primary purpose of this method is in helping to provide an understanding of
1. What is involved in solving an LPP?
2. What information is available in the solution?
The step-by-step procedure is explained with the help of following example:
Ex: Two types of Television sets are produced with a profit of 6 units from each
television of Type I and 4 units from each television of Type II. In addition, 2 and 3
units of raw materials are needed to produce one television of Type I and Type II,
respectively & 4 and 2 units of time are required to produce one television of Type I
and Type II, respectively. If 100 units of raw materials and 120 units of time are
available, how many units of each type of television should be produced to maximize
profit and still meet all constraints of the problem?
Step I: Write down the given information in a table format.
Given:
Type I
Type II
Available
2
3
100
Raw Material
4
2
120
Time
6
4
Profit/Unit
Here, the profit is to be maximized within the limited resources. This is the
maximization problem.
Step II: Initially the problem should be formulized i.e. objective function & constraints
should be written.
In this example, the main objective is to produce more number of units of
Type I & II within limited resources.
Let X1, X2 be the number of units of Type I and Type II respectively.
The objective function (the function to be optimized/maximized) is
Z = 6X1 + 4X2
Since each unit of Type I and Type II yields a profit of 6 and 4 units, this implies
that
Max Z = 6X1 + 4X2
Subject to constraints
2X1 + 3X2 ≤ 100 ………… Raw material
4X1+ 2X2 ≤ 120 …………. Time
X1, X2 ≥ 0
...………. Non-negativity
Since only two variables are involved in this example, the problem can be solved
graphically.
Step III: To solve it graphically i.e. to draw a graph, for all constraints draw the straight
line for where the constraints are satisfied exactly. Graph the constraints as if it
were equality.
Therefore, the 1st constraint i.e. 2X1 + 3X2 ≤ 100 becomes
2X1 + 3X2 = 100
st
To graph the 1 constraint, determine the set of points that satisfy this constraint.
Check whether the origin (0,0) satisfies the constraint.
In this case, 2*0 + 3*0 = 0 ≤ 100
This implies that all points below the line 2X1 + 3X2 = 100 satisfy this
constraint.
Now, to determine (X1, X2) co-ordinates, set X1=0 and find out X2 and viceversa.
Therefore, putting X1 = 0 in 2 X1 + 3X2= 100
We get, 3X2 = 100
X2 = 33.33
And putting X2 = 0 in 2X1 + 3X2 = 100
We get, 2X1 = 100
X1 = 50
Therefore, set of points is (50, 33.33).
Draw a line with (X1, X2) co-ordinates as (50, 33.33).
Step IV: Now in the similar way determine the set of points by substituting X1 = 0 and
X2 = 0 simultaneously in the second constraint.
The 2nd constraint is 4X1 + 2X2 = 120
Putting X1 =0 in this constraint, we get
2X2 =120
X2=60
Putting X2 = 0, we get
4X1 = 120
X1 = 30
Therefore, set of points is (30, 60).
Draw a straight line with (X1, X2) co-ordinates as (30, 60).
Step V: Determine the region where X1≥0 & X2≥0.
X2
(0,60)
4X1 + 2X2=120
D (0,33,33)
Region of “Feasible Solutions”
C (20,20)
2X1 + 3X2= 100
B (30,0)
(30,0)
A (0,0)
Step VI: Take the region which is common to all the lines i.e. 2X1 + 3X2 ≤ 100;
4X1 + 2X2 ≤ 120 and X1, X2 ≥ 0. That region is called as “Feasible Region”.
X1
Feasible region is a set of points that make all linear inequalities in the system true
simultaneously. That is, each point in this region satisfies all of the constraints and is a candidate for
providing the maximum profit. But the region contains infinity of feasible solutions. Therefore, there
is a need to find out the optimal solution. Find out the intersection points.
Step VII: Therefore, to find out the optimal solution, substitute the set of points of
intersection in the objective function i.e. Z = 6X1 + 4X2
The points of intersection are:
A (0, 0)
B (30, 0)
C (20, 20)
D (0, 33.33)
Substitute these values in Z = 6X1 + 4X2
A (0, 0)
Z = 6*0 + 4*0 = 0
Z=0
B (30, 0)
Z = 6*30 + 4*0 = 180
Z = 180
C (20, 20)
Z = 6* 20 + 4*20 = 200
Z = 200
D (0, 33.33)
Z = 6*0 + 4*33.33 = 133.32
Z = 133.32
Z = 200 is the maximum value & is the optimal solution.
Therefore, 20 units of Type I and 20 units of Type II should be
produced to yield a maximum profit of 200 units.
Minimization case
Illustration:
Minimize Z= 10X1+4.5X2
Subject to constraints
2X1+3X2≥3500
6X1+2X2≥7000
X1≥0 X2≥0
We first find the points to determine the line For constraint 1 2X1+3X2≥3500
When X1=0, X2=3500/3. The coordinates of the point are (0,3500/3)
When X2=0, X1=1750. The coordinates of the point are (1750,0)
We plot these points on the graph paper to show this constraint. Similarly for the second constraint
When X1=0, X2=7000/2. The coordinates of the point are (0,3500)
When X2=0, X1=7000/6. The coordinates of the point are (7000/6,0)
We plot these points on the graph paper lines to show this constraint.
3500
A
3000
2500
6X1+2X2≥7000
2000
2X1+3X2≥3500
1500
feasible region
1000
500
B
C
500
1000
1500
2000
The shaded region is the feasible region. Infinitely many points satisfy the constraints. The objective is to
minimize the value of the objective function. The feasible region is bounded below. The points A, B and C
are the points where the minimum value of the objective function may lie.
The feasible points are
Point Coordinates
A
(0,3500)
B
(1000,500)
C
(0,3500/3)
The value of the objective function at these points is
Point Coordinates Value of Z
A
(0,3500)
15750
B
(1000,500)
12250* Minimum
C
(1750,0)
17500
The solution lies at point B. X1=1000 and X2=500 value of Z=12250
Problems
1.
Use the graphical method to solve the following LP problem.
Maximize Z = 15x1 + 10x2
subject to the constraints
4x1 + 6x2 ≤ 360
3x1 + 0x2 ≤ 180
0x1 + 5x2 ≤ 200
and
x1, x2  0.
2.
Use the graphical method to solve the following LP problem.
Maximize Z = 2x1 + x2
subject to the constraints
x1 + 2x2 ≤ 10
x1 + x2 ≤ 6
x1 - x2 ≤ 2
x1 - 2x2 ≤ 1
and
x1, x2  0.
3.
Use the graphical method to solve the following LP problem.
Maximize Z = 10x1 + 15x2
subject to the constraints
2x1 + x2 ≤ 26
2x1 + 4x2 ≤ 56
- x 1 + x2 ≤ 5
and
x1, x2  0.
4. The ABC Company has been a producer of picture tubes for television sets and certain printed circuits
for radios. The company has just expanded into full scale production and marketing of AM and AMFM radios. It has built a new plant that can operate 48 hours per week. Production of an AM radio in
the new plant will require 2 hours and production of an AM-FM radio will require 3 hours. Each AM
radio will contribute Rs 40 to profits while an AM-FM radio will contribute Rs 80 to profits. The
marketing department, after extensive research has determined that a maximum of 15 AM radios and
10 AM-FM radios can be sold each week.
(a)
Formulate a linear programming model to determine the optimum production mix of AM
and FM radios that will maximize profits.
(b)
Solve the above problem using the graphic method.
5. Anita Electric Company produces two products P1 and P2. Produced and sold on a weekly basis. The
weekly production cannot exceed 25 for product P1 and 35 for product P2 because of limited available
facilities. The company employs total of 60 workers. Product P1 requires 2 man-weeks of labour,
while P2 requires one man-week of labour. Profit margin on P1 is Rs 60 and on P2 is Rs 40. Formulate
this problem as an LP problem and solve for maximum profit.
6. A local travel agent is planning a charter trip to a major sea resort. The eight day/seven-night package
includes the fare for round-trip travel, surface transportation, boarding and loading and selected tour
options. The charter 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 differ
according to seating and service for the fight, quality of accommodations, meal plans and tour options.
The following table summarizes the estimated prices for the three 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 Costs for Tour Packages per Person
Tour Plan
Delux
Standard
Economy
Price
(Rs)
10,000
7,000
6,500
Hotel Costs
(Rs)
3,000
2,200
1,900
Meals & Other Expenses
(Rs)
4,750
2,500
2,200
In planning the trip, the following considerations must be taken into account:
(i)
At least 10 per cent of the packages must be of the delux type.
(ii)
At least 35 per cent but no more than 70 per cent must be of the standard type.
(iii) At least 30 per cent must be of the economy type.
(iv)
The maximum number of delux packages available in any aircraft is restricted to 60.
(v)
The hotel desires that at least 120 of the tourists should be on the deluxe and standard
packages together.
The travel agent wishes to determine the number of packages to offer in each type so as to maximize
the total profit.
(a)
Formulate this problem as a linear programming problem.
(b)
(c)
Restate the above linear programming problem in terms of two decision variables, taking
advantage of the fact that 200 packages will be sold.
Find the optimum solution using graphical method for the restated linear programming
problem and interpret your results.
Ans : 20 – D , 100 – S, 80 – E
Profit = 280000 – 13000 = 267000.
7. Use the graphical method to solve the following LP Problem.
Minimize Z = 3x1 + 2x2
Subject to the constraints
5x1 + x2  10
x1 + x2  6
x1 + 4x2  12
and
x1, x2  0.
8.
Use the graphical method to solve the following LP problem.
Maximize Z = x1 + 2x2
subject to the constraints
- x1 + 3x2 ≤ 10
x1 + x2 ≤ 6
x1 - x2 ≤ 2
and
x1, x2  0.
9.
Cashewco has two grades of cashew nuts: Grade I – 750 kg and Grade II – 1,200 kg. These are to be
mixed in two types of packages of one kilogram each- economy and special. The economy pack
consists of grade I and grade II cashews in the proportion of 1:3, while the special pack combines the
two in equal proportion. The profit margin on the economy and special packs is, respectively, Rs. 5
and Rs. 8 a pack.
(a) Formulate this as a linear programming problem.
(b) Ascertain graphically the number of packages of economy and special types to be made that will
maximize the profits.
Would your answer be different if the profit margin on a special pack be Rs. 10?
10. Alpha Radio Manufacturing company must determine production quantities for this month for two
different models, A and B. Data per unit are given in the following table:
Model
Revenue
Sub-assembly
Final assembly Quality
(Rs)
time (hr)
time (hr)
inspection (hr)
A
250
1.0
0.8
0.5
B
300
1.2
2.0
0
The maximum time available for these products is 1,200 hours for sub-assembly, 1,600 hours for final
assembly, and 500 hours for quality inspection. Orders outstanding require that at least 200 units of A and
100 units of B be produced. Determine the quantities of A and B that maximize the total revenue.
11. Attempt graphically the following problem:
Maximise 3x1 + 2x2
Subject to 2x1 + x2  12; x1 + x2  10; - x1 + 3x2  6; and x1, x2  0
12. Obtain graphically the solution to the following LPP:
Maximise
Z = x1 + 3x2
Subject to
x1 + 2x2  9
x1 + 4x2  11
x1 – x 2  2
x1 , x 2
0
13. A firm is engaged in producing two products : P1 and P2. The relevant data are given here:
Per Unit
Product P1
Product P2
(i) Selling price
Rs 200
Rs 240
(ii) Direct materials
Rs. 45
Rs. 50
(iii) Direct wages
Deptt A
8 hrs @ Rs. 2 / hr
10 hrs @ Rs 2 /hr
Deptt B
10 hrs @ Rs 2.25 / hr
6 hrs @ Rs 2.25 /hr
Deptt C
4 hrs @ Rs 2.5 / hr
12 hrs @ Rs 2.5 / hr
(iv) Variable overheads
Rs. 6.50
Rs 11.50
Fixed overhead = Rs 2,85,000 per annum
No of employees in the three departments:
Deptt A = 20
Deptt B = 15
Deptt C = 18
No. of hours / employee / week = 40 in each department
No. of weeks per annum = 50
(a) Formulate the given problem as a linear programming problem and solve graphically to determine
(i) the product mix as will maximize the contribution margin of the firm
(ii) the amount of contribution margin and profit obtainable per year
(a) From the graph, do you observe any constraint that is redundant? Which one, if yes?
Yes the constraint 8X1 + 10 X2 <=40000 is redundant.
14. A local business firm is planning to advertise a special sale on radio and television during a particular
week. A maximum budget of Rs. 16,000 is approved for this purpose. It is found that radio
commercials cost Rs. 800 per 30 second spot with a minimum contract of five spots. Television
commercials, on the other hand, costs Rs. 4,000 per spot. Because of heavy demand, only four
television spots are still available in the week. Also, it is believed that a TV spot is six times as
effective as a radio spot in reaching consumers. How should the firm allocate its advertising budget to
attract the largest number of consumers? How will the optimal solution be affected if the availability
of TV spot is not constrained?
15. M/s. P.M.S. Industries makes two kinds of leather purses for ladies. Purse type ‘A’ is of
high quality and purse type ‘B’ is of lower quality. Contributions per unit were Rs. 4 and
Rs. 3 for purse type ‘A’ and type ‘B’ respectively. Each purse of type ‘A’ requires two
hours of machine time per unit & that of type ‘B ’ requires one hour per unit. The
company has 1000 hours per week of maximum available machine time. Supply of
leather is sufficient for 800 purses of both types combined per week. Each type requires
the same amount of leather. Purse type ‘A’ requires a fa ncy zip and 400 such zips are
available per week, there are 700 zips for purse type ‘B’ available per week. Assuming
no market or finance constraints recommend an optimum product mix.
Solution:
Decision variables:
Let X 1 be the number of units of purse type A
Let X 2 be the number of units of purse type B
The LP model
Maximise Z = 4x 1 + 3x 2
subject to the constraints
2X 1 +
X 2 ≤ 1000 (Machine time)
X 2 ≤ 800 (Leather)
X1 +
X1
≤ 400
X2
≤ 700 (Ordinary zip for Purse B)
(Special zip for Purse A)
X i ≥ 0 (Non negativity restriction)

X2 Units
1000
X1  400
800
D
X2  700
E
C
600
Feasible Solution Area OABCDE
X1 + X2  800
400
2X1 + X2  1000
200
B
A
O
0.0
100
200
300
400
500
600
700
800
900
1000
X1 Units 
Scale
X-Axis 1 cm = 100 units
Y-Axis 1 cm = 100 units
Draw the graph as indicated in Fig 4.1
The optimum solution will lie on one of the corner points of the shaded area OABCDE
Calculation of Z (Z = 4X 1 + 3X 2 ) at corner points
O (0,0);
ZO = 4 x 0 + 3 x 0 = 0
A (400, 0);
Z A = 4 x 400
B (400, 200);
Z B = 4 x 400 + 3 x 200 = Rs. 2200
C (200, 600);
Z C = 4 x 200 + 3 x 600 = Rs. 2600
= Rs. 1600
D (100, 700); Z D = 4 x 100 + 3 x 700 = Rs. 2500
E (0, 700);
Z E = 3 x 700
= Rs. 2100
Corner point C signifies the optimum solution because the corresponding value of Z is
maximum.
The optimum solution is 200 units of Purse Type ‘A’ & 600 units of Purse Type B
The corresponding Total Contribution = Rs. 2600
16. A catering manager is in the process of replacing the furniture in a canteen. He wishes
to determine how many tables of type ‘S’ (seating 6) and how many of type ‘T’ (seati ng
10) to buy.
He has to work under the following constraints:
(1) The canteen must be able to accommodate at least 60000 people.
(2) The available floor space of the canteen is at most 63000 sq. meters.
He estimates that each type ‘S’ table needs 7 meters sq. of floor space while each type ‘T’
needs 9.
Advise the manager on how many tables of each type to buy if each type ‘S’ costs Rs. 100
and each type ‘T’ costs Rs. 190.
Solution:

X2 Units
10000
8000
Feasible Solution Area ABC
B
6000
C
7X1 + 9X2  63000
4000
A
6X1 + 10X2  60000
2000
O
(0, 0)
1000
2000
3000
4000
5000
6000
7000
8000
9000
10000
X1 Units 
Decision variables:
Let X 1 be the number of units of table type S
Let X 2 be the number of units of table type T
The LP model
Minimise Z = 100x 1 + 190x 2
subject to the constraints
6X 1 +
10X 2 ≥60000 (People)
7X 1 +
9X 2 ≤ 63000 (Floor space)
Xi
≥0
Draw the Graph as indicated in Figure.
The optimum solution will lie on one of the corner points of the shaded area ABC
Calculation of Z (Z = 100X 1 + 190X 2 ) at corner points
A (5625, 2625);
Z A = 100 x 5625 + 190 x 2625
B (0, 7000);
Z B = 190 x 7000
= Rs. 1330,000
C (0, 6000);
Z C = 190 x 6000
= Rs. 1140,000
= Rs. 1061,250
Corner point A signifies the optimum solution because the corresponding value of Z is
minimum.
The optimum solution is 5625 units of Table type S & 2625 units of Table type T
The corresponding Total Cost = Rs. 1061250
SOME SPECIAL CASES
Unbounded Solutions
When the value of a decision variable is permitted to increase infinitely without violating
any of the feasibility conditions, then the solution is said to be unbounded. Thus an
unbounded LPP occurs if it is possible to find arbitrarily large valu es of Z within the
feasible region. It is not possible to find a single optimal solution to an unbounded problem
though there are infinite feasible solutions for the same.
17. Use graphical method to solve the following LP problem
Maximise Z = 3x 1 + 3x 2
subject to the constraints
x1 - x2 ≤ 1
x1 + x2 ≥ 3
and
x1, x2 ≥ 0
Solution:
The problem is depicted graphically in Fig. 4.3. The solution space is shaded and is bound
by A and B from below.
X2
3
A (0,3)
Unboun
ded
Feasible
Region
2
X1 - X2 = 1
B (2,1)
1
X1 + X2 = 3
0
1
2
3
4
X1
It is noted here that the shaded convex region (solution space) is unbounded. The two
corners of the region are A = (0, 3) and B = (2, 1). The values of the objective function at
these corners are:
Z(A) = 6 and Z(B) = 8.
But there exist number of points in the shaded region for which the value of the objective
function is more than 8. For example the point (10,12) lies in the region and the function
value at this point is 70 which is more than 8. Thus both the variables x 1 and x 2 can be
made arbitrarily large and the value of Z also increased. Hence, the problem has an
unbounded solution.
Remark: An unbounded solution does not mean that there is no solut ion to the given LP
problem, but implies that there exist an infinite number of solutions.
Infeasible Solution
Sometimes it is not possible to find a single solution that satisfies all the feasibility
constraints. Such a problem does not have a feasible solution.
Please note that in the case of infeasiblity it is not possible to have a single feasible
solution whereas in the case of unboundedness, it is possible to have infinite feasible
solutions, but not a single optimal solution.
18. (Problem with inconsistence system of constraints) Use graphical method to solve the
following LP problem:
Maximise Z = 6x 1 - 4x 2
subject to the constraints
2x 1 + 4x 2 ≤ 4
4x 1 + 8x 2 ≥ 16
x1, x2 ≥ 0
and
Solution:
The problem is shown graphically in Fig. 4.4. The two inequalities that form the constraint
set are inconsistent. Thus, there is no set of points that satisfies all the constraints. Hence,
there is no feasible solution to this problem.
X2
3
4X1 + 8X2 = 16
2
2X1 + 4X2 = 4
1
0
1
2
3
4
X1
Multiple Optimal Solutions
Can exist for a linear programming problem if
i.
The given objective function is parallel to a constraint that forms the boundary of the
feasible solutions region
ii.
The constraint should form a boundary on the feasible region in the direction of the
optimal movement of the objective function. In other words, the constraint must be a
binding constraint.
In case the slope of the objective function (represented by the iso -profit lines) is the same
as that of a constraint, then multiple optimal solutions might exist. For multiple optimal
solutions to exist, the following two conditions need to be satisfied:
(a)
The objective function should be parallel to a constraint that for ms an edge or
boundary on the feasible region; and
(b)
The constraint should form a boundary on the feasible region in the direction of
optimal movement of the objective function. In other words, the constraint must be a
binding constraint.
19. Solve graphically the following LPP:
Maximise Z = 8x 1 + 16x 2
Subject to
x 1 + x 2 ≤ 200
x 2 ≤ 125
3x 1 + 6x 2 ≤ 900
x1, x2 ≥ 0
Solution:
The constraints are shown plotted on the graph in Figure 4.5. Also, iso -profit lines have
been graphed. We observe that iso-profit lines are parallel to the equation for third
constraint 3x 1 + 6x 2 = 900. As we move the iso-profit line farther from the origin, it
coincides with the portion BC of the constraint line that forms the boundary of the fe asible
region. It implies that there are an infinite number of optimal solutions represented by all
points lying on the line segment BC, including the extreme points represented by B (50,
125) and C (100, 100). Since the extreme points are also included in the solutions, we may
disregard all other solutions and consider only these ones to establish that the solution to a
linear programming problem shall always lie at an extreme point of the feasible region.
The extreme points of the feasible region are give n and evaluated here.
Point
x1
x2
0
0
0
Z = 8x 1 + 16x 2
0
A
B
0
50
125
125
2000
2400
C
D
100
200
100
0
2400 }
1600
The point B and C clearly represent the optima.
In this example, the constraint to which the objective function was parallel, was the one
which formed a side of the boundary of the space of the feasible region. As mentioned in
condition (a), if such a constraint (to which the objective function is parallel) does not form
an edge or boundary of the feasible region, then multiple s olutions would not exist.
X2
200
Optimal Solutions
150
X2 = 125
A
B
100
50
C
FEASIBLE
Iso-profit lines
REGION
D
0
100
200
E
300
X1
20. Solve graphically:
Minimise Z = 6x 1 + 14x 2
Subject to
5x 1 + 4x 2 ≥ 60
3x 1 + 7x 2 ≤ 84
x 1 + 2x 2 ≥ 18
x1, x2 ≥ 0
Solution:
The restrictions in respect of the given problem are depicted graphically in Figure 4.6. The
feasible area has been shown shaded. It may be observed here that although the iso-cost line
is parallel to the second constraint line represented by 3x 1 + 7x 2 = 84, and this constraint
does provide a side of the area of feasible solutions, yet the problem has a unique optimal
solution, given by the point D. Here co ndition (b) mentioned earlier, is not satisfied. This is
because, being a minimisation problem, the optimal movement of the objective function
would be towards the origin and the constraint forms a boundary on the opposite side. Since
the constraint is not a binding one, the problem does not have multiple optima.
X2
15
12
B
9
6
A
3
0
FEASIBLE
REGION
D
6
12
18
C
24
30
X1
We can show the uniqueness of the solution by evaluating various extreme points as done
here.
Point
x1
x2
Z = 6x 1 + 14x 2
A
B
C
8
84/23
28
5
240/23
0
118
168
148
D
18
0
108
The optimum solution is 108 units for x 1 = 18 and x 2 = 0.