SOLVED EXAMPLES
EXAMPLE 1: (Production allocation Problem)
A company produces three different products. These products are processed on three different
machines. The time required to manufacture one unit of each of the three products and the daily
capacity of the three machines are given in the table below.
MACHINE
M1
M2
M3
TIME PER UNIT (Minutes)
Product 1
2
4
2
Product 2
3
5
Product 3
2
3
-
MACHINE
CAPACITY(minutes/day)
440
470
430
It is required to determine the daily number of units to be manufactured for each product.
The profit per unit for product 1,2 and 3 is Rs. 4, Rs. 3 and Rs. 6 respectively. It is assumed that
all the amounts produced are consumed in the market. Formulate the linear programming model
that will maximize the daily profit.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to decide the number of units of each product to be produced.
Let x1 be the number of units of product 1 to be produced
x2 be the number of units of product 2 to be produced and
x3 be the number of units of product 3 to be produced.
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to maximize the profit.
Now Total profit= profit on product 1+profit on product 2+profit on product 3
Given that profit on product 1=4x1, profit on product 2=3x2 and profit on product 3=6x3
Therfore MAXIMIZE Z= 4x1+3x2+6x3
STEP 3: (CONSTRAINTS)
Time consumed on M1 440
 Time consumed by P1 on M1 + Time consumed by P2 on M1+Time consumed by P3 on
M1 440
 2x1+3x2+2x3 440
Time consumed on M2 470
 Time consumed by P1 on M2 + Time consumed by P2 on M2+Time consumed by P3 on
M2 470
 4x1+3x3 470
Time consumed on M3 430
 Time consumed by P1 on M3 + Time consumed by P2 on M3+Time consumed by P3 on
M3 430
 2x1+5x2 430
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
x1 0, x2 0 and x3 0
Therefore the final model is
MAXIMIZE Z= 4x1+3x2+6x3
Subjected to
2x1+3x2+2x3 440
4x1+3x3 470
2x1+5x2 430
Where
x1 0, x2 0 and x3 0
EXAMPLE 2: (DIET PROBLEM)
A pet store has determined that each hamster should receive at least 70 units of protein, 100 units
of carbohydrates and 20 units of fats daily. If the store carries the six types of feed as shown in
the table what blend of feeds satisfies the requirements at minimum cost to the store?
FEED
A
B
C
D
E
F
PROTEINS
Units/oz
20
30
40
40
45
30
CARBOHYDRATES
Units/oz
50
30
20
25
50
20
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to determine the quantity of each feed.
Let
be the quantity of feed A to be used.
be “
“
“
B “
“
be “
“
“
C “
“
be “
“
“
D “
“
be “
“
“
E “
“
be “
“
“
F “
“
FAT
Units/oz
4
9
11
10
9
10
COST
Per oz
2
3
5
6
8
8
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to minimize the total cost.
MINIMIZE Z= 2
+3
+5
+6
+8
+8
STEP 3: (CONSTRAINTS)
Given that hamster should receive at least 70 units of proteins, 100 units of carbohydrates and 20
units of fat.
 20 +30 +40
& 50 +30 +20
& 4 + 9 +11
+40
+25
+10
+45 +30
+50 +20
+ 9 +10
70
100
0
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
0 i=A,B,C,……,F
Therefore the final model is
MINIMIZE Z= 2 +3
Subjected to
20 +30 +40 +40
50 +30 +20 +25
4 + 9 +11 +10
Where
0 i=A,B,C,……,F
+5 +6 +8 + 8
+45 +30
70
+50 +20
100
+ 9 +10
0
EXAMPLE 3: (BLENDING PROBLEM)
A firm produces an alloy having the following specifications:
i) Specific gravity 0.98
ii) Chromium 8%
iii) Melting point 450 C
Raw materials A, B and C having the properties shown in the table can be used to make the
alloy.
PROPERTY
Specific Gravity
Chromium
Melting point
PROPERTIES OF RAW MATERIAL
A
B
C
0.92
0.97
1.04
7%
13%
16%
440 C
490 C
480
Costs of the various raw materials per ton are : Rs. 90 for A, Rs. 280 for B and Rs. 40 for C.
Formulate the L.P model to find the proportions in which A, B and C be used to obtain an alloy
of desired properties while the cost of raw materials is minimum.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to decide the percentages of each raw materials A, B and C to be used in
the alloy.
Let , and
be the percentage contents of raw materials A, B and C respectively.
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to minimize the total cost.
MINIMIZE Z= 90 +280
+40
STEP 3: (CONSTRAINTS)
0.92 +0.97 +1.04
7 +13 +16
440 +490 +480
0.98
450
Also since we have to find the percentage contents therefore there will be an additional
constraint i.e.
+ +
100
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
x1 0, x2 0 and x3 0
Therefore the final model is
Subjected to
Where
MINIMIZE Z= 90 +280 +40
0.92 +0.97 +1.04
0.98
7 +13 +16
440 +490 +480
450
+ +
100
x1 0, x2 0 and x3 0
EXAMPLE 4: (CONSTRUCTION MANAGEMNET PROBLEM)
A contractor has been assigned the excavation work for a canal and of a head works on a project.
In order to ensure a balanced progress on the entire work the management has imposed certain
conditions on his working. The excavation of the canal is more profitable than that of head
works, but he has to follow the conditions of contract. In addition, he has his own limitations on
manpower and equipment. It is desired to find the optimum amount of excavation on the two
works which he should undertake so that his profits are maximized and he satisfies the
constraints. These constraints are given below:
i) The difference in the quantity of earthwork done on the two works does not exceed 2
units in a day.
ii) The difference between the quantity of a canal excavation and of twice the head works
excavation does not exceed 1 unit in a day.
iii) Each unit of canal excavation done in a day requires one unit of manpower and one unit
of machine; and each unit of head works excavation requires 2 units of manpower and
one unit of machine. Maximum available manpower is 10 units and machines 6 units.
The contractor stipulates a profit of 2 units for each unit of canal excavation and half of this for
each unit of head works excavation.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to find the optimum amount of excavation work of canal and head
works to be done on the project.
Let
be the units of excavation work of canal/day
And
be the units of excavation of head works/day
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to maximize the profit.
Therefore MAXIMIZE Z=2x1+x2
STEP 3: (CONSTRAINTS)
Now given that the difference between the two works should not exceed two units in a day
i.e.
2
Also the difference between canal excavation and twice the head works should not exceed 1 unit
in a day
i.e.
2
1
Canal excavation requires 1 unit of manpower and head works require 2 units of manpower and
maximum available manpower is 10 units
i.e. +2
10
Also both canal excavation and head works excavation requires 1 unit of machine and maximum
available machines is 6 units
i.e. +
6
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
x1 0 and x2 0
Therefore the final model is
Subjected to
Where
MAXIMIZE Z=2x1+x2
2
2
1
+2
10
+
6
x1, x2 0
EXAMPLE 5: (MEDIA SELECTION PROBLEM)
The owner of Metro sports wishes to determine how many advertisements to place in three
selected monthly magazines A, B, C. His objective is to advertise in such a way that total
exposure to principal buyers of expensive sports goods is maximized. Percentages of readers for
each magazines are known. Exposure in any particular magazine is the number of advertisements
placed multiplied by the number of principal buyers. The following data may be used:
EXPOSURE CATEGORY
Readers
Principal buyers
Cost/advertisement(Rs)
A
1 lakh
10%
5000
MAGAZINES
B
0.6 lakh
15%
4500
C
0.4 lakh
7%
4250
The budgeted amount is Rs 1 lakh for the advertisements. The owner has already decided that
magazine A should have no more than six advertisements and that B and C each have at least
two advertisements. Formulate a L.P model for the problem.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to determine the number of advertisements to be placed in the
magazines.
Let
be the number of advertisements placed in magazine A
be the number of advertisements placed in magazine B
be the number of advertisements placed in magazine C
STEP 2: (OBJECTIVE FUNCTION)
Here objective is to advertise in such a way that total exposure to principal buyers of expensive
sports goods is maximized.
The number of advertisements seen by readers of magazine A is 1,00,000, that of magazine B is
60,000 and that of magazine C is 40,000 . But since we have to maximize the principal
buyers and the principal buyers of magazines A, B and C are 0.10, 0.15 and 0.07 respectively
Therefore MAXIMIZE Z=0.10 1,00,000 +0.15 60,000 +0.07 40,000
STEP 3: (CONSTRAINTS)
Given that the total budget for the magazines is Rs. 1,00,000
 5000 +4500 +4250
1,00,000
Magazine A should not have more than 6 advertisements

6
Magazines B and C must have at least 2 advertisements

2,
2
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
x1 0 , x2 0 and
0
Therefore the final model is
MAXIMIZE Z=0.10 1,00,000 +0.15 60,000
Subjected to
5000 +4500 +4250
1,00,000
0
6
2,
2
+0.07 40,000
EXAMPLE 6: (INSPECTION PROBLEM)
A company has two grades of inspectors, I and II to undertake quality control inspection. At least
1,500 pieces must be inspected in an 8-hour day. Grade I inspector can check 20 pieces in an
hour with an accuracy of 96%.Grade II inspector checks 14 pieces an hour with an accuracy of
92%.
Wages of grade I inspector are Rs. 5 per hour while those of grade II inspector are Rs. 4 per
hour. Any error made by an inspector costs Rs. 3 to the company. If there are, in all, 10 grade I
inspectors and 15 grade II inspectors in the company, find the optimal assignment of inspectors
that minimizes the daily inspection cost.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to find the number of grade I and grade II inspectors that are assigned
the job.
Let
and
denote the number of grade I and grade II inspectors that may be assigned the job
of quality control inspection.
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to minimize the daily cost of inspection.
Now there are two types of costs in this problem: wages paid to the inspectors and the cost of
their inspection errors.
Now the wages grade 1 inspector are Rs. 5/hour and the company will pay an additional cost of
Rs. 3 incase of error. Also the grade I inspector can check the 20 pieces in an hour with 96%
accuracy therefore there will be an additional cost of 0.04 20.
 The cost of grade I inspector is:
Rs. (5+3 0.04 20) = Rs. 7.40
Similarly the cost of grade II inspector/hour is
Rs. (4+3 0.08 14) = Rs.7.36
Now since the pieces must be inspected in an 8-hour day therefore the objective function is:
MINIMIZE Z= 8(7.40 +7.36 ) =59.20 +58.88
STEP 3: (CONSTRAINTS)
Given that there are 10 grade I inspectors in the company
10
Also there are 15 grade II inspectors in the company
15
Since at least 1500 pieces must be inspected in an 8-hour day
20
+14
1500
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
x1 0 and x2 0
Therefore the final model is
Subjected to
20
Where
MINIMIZE Z= 59.20 +58.88
10
15
+14
1500
x1 0 and x2 0
EXAMPLE 7: (FLEET ASSIGNMENT PROBLEM)
Jyoti Tourist Corporation operates three different types of coaches along four routes. The number
of passengers, the number of daily trips that each coach can make on a given route, the number
of coaches available of each type and the daily number of customers expected for each route
during a season are given below
COACH
CAPACITY NO. OF COACHES
(Type)
(passengers)
AVAILABLE
A
60
4
B
40
6
C
20
8
DAILY NO OF CUSTOMERS
NUMBER OF DAILY TRIPS ON ROUTE
I
II
III
IV
2
1
2
1
3
3
4
2
4
5
6
4
125
130
250
100
The following table gives the operating costs per trip on each route as well as the penalty cost if
a customer is not given service:
TABLE II
COACH TYPE
A
B
C
PENALTY
CUSTOMER
COST
PER
OPERATING COST PER TRIP ON GIVEN ROUTE (Rs.)
I
II
III
IV
500
600
600
700
400
500
400
500
200
300
250
350
30
40
35
60
Formulate the L.P model to determine the assignment of tourist coaches to dofferent routes so as
to minimize overall cost.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to decide the assignment of tourist coaches to different routes.
Let the no. of trips made by coach type A along routes I,II,III be
no. of trips made by coach type B along routes I,II,III be
no. of trips made by coach type C along routes I,II,III be
,
,
,
,
,
,
respectively.
respectively.
respectively.
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to minimize the total cost.
Now there are two types of costs in this problem: operating costs per trip on each route and the
penalty cost in case a customer is not given any service.
The total operating cost per trip on each route is (from table II)
500 +600
+600
+700 +400 +500
+400
+500
200 +300
+250
+350
The penalty cost per customer is
30(125 60
40
20
) +40(130 60
35(250 60
40
20
+60(100 60
Therefore the objective function is
40
40
+
20 ) +
20 )
MINIMIZEZ=500 +600
+600
+700 +400 +500
+400
+500 +
200 +300
+250
+350 +30(125 60
40
20
)+40(130 60
40
20 ) +35(250 60
40
20
+60(100 60
40
20 )
Or MINIMIZEZ=23700 1300
800
400
1800
1100
500
1500
1000
450
2900
1900
850
STEP 3: (CONSTRAINTS)
Given that the maximum number of customers for route I is 125
60
40
20
125
Maximum number of customers for route II is 130
60
40
20
130
Maximum number of customers for route III is 250
60
40
20
250
Maximum number of customers for route IV is 100
60
40
20
100
Also each type of coach can make limited number of trips daily along any stated route and the
total number of trips made by a given type of coach along a stated route is limited by the product
of the maximum number of trips possible and the number of coaches available in that type.
4
4
4
4
2
1
2
1
6
6
6
6
3
3
4
2
8
8
8
8
4
5
6
4
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
0
i=A,B,C j=1,2,3,4
EXAMPLE 8: (PRODUCT MIX PROBLEM)
A firm manufactures three products A, B and C. time taken to manufacture product A is twice
that for B and thrice that for C and if the entire labour is engaged in making product A, 1600
units of this product can be produced. These products are to be manufactured in the ratio of
3:4:5. There is demand for at least 300, 250 and 200 units of products A, B and C and the profit
earned per unit is Rs. 90, Rs. 40 and Rs. 30 respectively.
The requirement and total availability of the product is given in the following table:
RAW
MATERIAL
P
Q
REQUIREMENT PER UNIT
OF PRODUCT (kg)
A
B
C
6
5
2
4
7
3
TOTAL
AVAILIBILTY (kg)
5,000
6,000
Formulate the L.P model.
SOLUTION:
STEP 1: (KEY DECISION)
Here the key decision is to find the units of products X and Y produced.
Let and
be the number of units of products X and Y produced.
Also let
be the number of units of by-product Z produced.
Where =number of units of Z sold + number of units of Z destroyed
= +
(say)
STEP 2: (OBJECTIVE FUNCTION)
Our objective is to maximize the profit.
MAXIMIZE Z=10 +20 +6
4
.
STEP 3: (CONSTRAINTS)
Given that each unit of product X requires 3 hours on operation I and 4 hours on operation II and
total available time is 20 hours.
3 +4
20
Also each unit of product Y requires 4 hours on operation I and 5 hours on operation II and total
available time is 26 hours
4 +5
26
Maximum 5 units of products Z should be sold
5
Each unit of product Y results in 2 units of product Z
2Y=Z
Or 2 = +
STEP 4: (NON-NEGATIVITY RESTRICTIONS)
0
i=1,2,3,4.
Therefore the final model is
Subjected to
Where
MAXIMIZE Z=10 +20 +6
+4
20
4 +5
26
5
2 = +
0
i=1,2,3,4.
4