Operation Research (15ME81) – CBCS Scheme

Operation Research
(15ME81) – CBCS Scheme

Linear Programming Problem

Module-1

By

Dr A K Murthy

Formulation of Linear Programming Problems Page 1 of 16

 Operation Research (15ME81) – CBCS Scheme

[1] A Person should provide 100, 120 and 120 units of chemicals C1, C2 and C3 respectively to a Gardner. These
Chemicals can be manufactured by a Liquid and a Dry product. The liquid product contains 500, 200 and 100
units of chemical C1, C2 and C3 per Jar and costs Rs 500.00 / Jar. The Dry product contains 100, 200, and 400
units of chemical C1, C2 and C3 per packet and each packet costs Rs 400.00 / packet. How many of these should
produce to meet the requirement with least cost. Formulate as LPP. Jan-Feb2003

Solution [1]: X1 & X2 denotes number of units of Liquid and Dry products respectively.
Objective function, Minimize Z = 500 X1 + 400 X2
Subjected to the conditions, 500X1 +100X2 ≥ 100, ( Chemical A requirements)
200X1 +200X2 ≥ 120, ( Chemical B requirements)
100X1 +400X2 ≥ 120, ( Chemical C requirements)
and X1 & X2 ≥ 0

[2] A Firm makes 2 types of furniture chair and tables. The contribution for each product, calculated by the
accounting department is Rs.200/- per chair and Rs.300/- per table. Both products are processed on 3 machines
M1, M2, M3. The time required in hours by each product and total time available in hours per week on each
machine are given as follows,
Machine Chairs Tables Available time
M1 3 3 36
M2 5 2 50
M3 2 6 60
Determine the product mix graphically. Dec-2010

Solution [2]: XC & XT denotes the number of chairs and tables,

Objective function, Maximize Z = 200 XC + 300 XT
Subjected to the conditions, 3XC + 3XT ≤ 36,
5XC + 2XT ≤ 50,
2XC + 6XT ≤ 60,
and XC & XT ≥ 0

[3] A manufacturer of a line of patent medicines is preparing a production plan on medicines A and B. There are
sufficient ingredients available to make 20,000 bottles of A and 40,000 bottles of B, but there are only 45,000
bottles into which either of the medicines can be put. Furthermore, it takes 3 hours to prepare enough material to
fill 1,000 bottles of A, it takes one hour to prepare enough material to fill 1,000 bottles of B and there are 66 hours
available for this operation. The profit is Rs.80 per bottle of A and Rs.70 per bottle of B.
i) Formulate this problem as an L.P.P to maximize the profit.
ii) Solve it graphically. Jul-2006

Solution [3]: X1 & X2 denotes number of bottles of A and B type respectively.

Objective function, Maximize Z = 80 X1 + 70 X2
Subjected to the conditions , X1 ≤ 20000, (Bottle A limitations)
X2 ≤ 40000, (Bottle B limitations)
X1 + X2 ≤ 45000, (Bottle Limitations)
3X1/1000 + 1X2/1000 ≤ 66, (Number hours available)
and X1 & X2 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[4] A plant manufactures washers and dryers. The major manufacturing departments are stamping department and
final assembly department. Stamping department fabricates a large number of metal parts for both washers and
dryers. Monthly dept. capacities are as follows:
Stamping dept. : 10000 washers or 10000 dryers
Motor and transmission dept. : 16000 washers or 7000 dryers
Dryer assembly dept. : only 5000 dryers, Washer assembly dept.: only 9000 washers.
Stamping dept. can produce parts for 10000 washers or 10000 dryers per month as well as for some suitable
combinations. It is assumed that there is no change over cost from washers to dryers. A similar situation exists in
motor and transmission dept. but assembly lines are separate. The contribution to monthly profit is Rs.900/- per
washer and Rs.1200/- per dryer. Determine the number of washer and dryers to be produced. Dec-2010

Solution [4]: X1 & X2 denotes the number of units of washers and dryers.
Objective function, Maximize, Z = 900 X1+1200 X1.
Subjected to the conditions, X1 /10000 + X2 /10000 ≤ 1, (Time in unit month) ,
X1/16000 + X2/7000 ≤ 1 ,
X1 ≤ 9000, (Capacity limitations)
X2 ≤ 5000,
and X1 & X2 ≥ 0

[5] Old hens can be brought at Rs 40 each and young ones at Rs 60 each. The old hens lay 15 eggs per week and
young ones lay 25 eggs per week. Each egg is being worth of Rs 4.00. A hen (young or old) costs Rs 18 per week
to feed. If the person has only Rs 2400/- to spend for hens, formulate the problem to decide how many of each
kind of hen should he buy?, Assume that he cannot house more than 50 hens altogether and only 20 young hens
available in the market. Dec-2013
Solution [5] : X1 & X2 be the number of old and young hens to be brought.
Revenue => 4(15 X1+25 X2) = 60 X1+100 X2
Feeding Cost => 8(X1+ X2) = 8 X1+8 X2
Profit = Revenue – Total Cost = (60 X1+100 X2) – (8 X1+8 X2) = 42 X1+82 X2

Objective function, Maximize, Z = 42 X1+82 X2

Subjected to the conditions 40X1 + 60X2 ≤ 2400, (One time investment) ,
X1 + X2 ≤ 50 , (Housing capacity limitations)
X1 ≤ 20, (Availability limitations)
and X1 & X2 ≥ 0

[6] A student has two final exams to prepare for. Each hour of study he devotes to course A is expected to return
him Rs.600/- in terms of long range benefits and each hour of study he devotes to course B is expected to return
him Rs.300/- in terms of long range benefits. The shops are all closed; the student has only 15 chocolates,
remaining. He feels he needs one chocolate every 20 minute while studying for course B and 1 every 12 minutes
while studying for course A. time is running short and only 4 hours remain to prepare for exam. It is necessary
that the student must devote at least 2 hours for studying. The student would like to maximize his returns for the
effort expended. Solve this problem as an L.P.P. and determine the optimum policy for the student by solving the
problem. Dec08-Jan09 & Jan/Feb2004
Solution [6]: XA & XB denotes the number of hours of spent to study course A and B.
Objective function, Maximize Z = 600 XA + 300 XB
Subjected to the conditions , 5XA +3XB ≤ 15, (5 chocolates per hour for A and 3 per hour for B)
XA + XB ≤ 4, (4 hours remain)
XA + XB ≥ 2, (Minimum study of 2 hours)
and XA & XB ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[7] Formulate a linear programming model for the problem given below. The apex television company need to
decide on the number of 27-inch and 20-inch sets to be produced at one of its factories. Market research indicates
that at most 40 of the 27-inch sets and 10 of 20-inch sets can be sold per month. The maximum number of work
hours available is 500 per month. A 27-inch set requires 20 work hours and 20-inch set requires 10 work hours.
Each 27-inch set sold produces a profit of $12000 and each 20-inch produces a profit of $8000. A wholesaler
agreed to purchase all the television sets produced if the numbers do not exceed the maxima indicated by market
research. Dec09/Jan10

Solution [7]: X1 & X2 denotes the number of number of 27-inch and 20-inch sets produced respectively.
Objective function, Maximize Z = 12000X1 + 8000X2
Subjected to the conditions , X1 ≤ 40, (27-inch type can be sold)
X2 ≤ 10, (20-inch type can be sold)
20X1 + 10X2 ≤ 500, ( Number of work hours available)
and X1 & X2 ≥ 0

[8] A firm plans to purchase at least 200 quintals of scrap containing high quality (x) and low quality (y) metals. It
decides that scrap purchased must contain at least 100 quintals of x-metal and not more than 35 quintals of y-
metal. The firm can purchase metals from two supplier’s A and B in unlimited quantities. The purchase of x and y
in terms of weight of scrap supplied by A and B are given below:
Metals Supplier A Supplier B
High quality (X) 25% 75%
Low quality (Y) 10% 20%
The price of scrap supplied by A and B is Rs.200/quintal and Rs.400/quintal. Formulate the problem as LP model
and solve graphically to determine the quantities purchased from A and B suppliers. (2002 SCH) Dec09/Jan10

Solution [8]: XA & XB denotes quantity (in Quintals) of scrap material purchased from Supplier A and B,

Objective function, Minimize Z = 200XA + 400XB

Subjected to the conditions , XA + XB ≥ 200, (Minimum purchase of scrap)
0.25XA + 0.75XB ≥ 100, (Must contain x-metal type)
0.10XA + 0.20XB ≤ 35, (Not more than y-metal type)
and XA & XB ≥ 0

[9] A former need to plant two kinds of trees P and Q in a land of 4000 sq. m. area. Each P tree requires at least 25
sq. m and Q tree requires at least 40 sq. m of land. The annual water requirements P tree is 30 units and of Q tree
is 15 units per tree, while at most 3000 units of water is available. It is also estimated that the ratio of the number
Q trees to the number of P trees should not be less than 6/19 and should not be more than 17/8. The return per tree
from P is expected to be one and half times as much as from Q tree. Formulate the problem as a LP model.
May-Jun2010

Solution [9]: XP & XQ denotes number P and Q type of plants to be planted,

Return per tree from P is expected to be one and half times as much as from Q tree, So
Objective function, Maximize Z = 1.5 XP + XQ

Subjected to the conditions , 25XP + 40XQ ≤ 4000, (Available land)

30XP + 15XQ ≤ 3000, (Available water)
XQ / XP ≥ 6/19, (Q to P ratio Not less than)
XQ / XP ≤ 17/8, (Q to P ratio Not more than)
and XP & XQ ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[10] The manager of an oil refinery must decide on the optimal mix of 2 possible blending process of which the
input for production run as follows.
Inputs (units) Output (units)
Process
Crude A Crude B Gasolene-X Gasolene-Y
1 5 3 5 8
2 4 5 4 4
The maximum amount of crude available is 200 units of crude A and 150 units of crude B. the market requirement
shows that at least 100 units of gasoline X and units of gasoline Y must be produced. The profit form production
run of process 1 and 2 are Rs.300 and Rs.400. formulate a suitable mathematical model and solve the same by
using graphical method. Dec06-Jan07

Solution [10]: X1 & X2 denotes number of production runs from Process-1 and Process-2.

Objective function, Maximize Z = 300 X1 + 400 X2

Subjected to the conditions , 5X1 + 4X2 ≤ 200, (Crude A limitations)
3X1 + 5X2 ≤ 150, (Crude B limitations)
5X1 + 4X2 ≥ 100, (Gas -X Requirements)
8X1 + 4X2 ≥ 100, (Gas -Y Requirements)
and X1 & X2 ≥ 0

[11] ABC food’s company is developing a calorie high protein diet supplement called HI-Pro. The specification
along with calorie, protein and vitamin content of three basic foods are given in the following table.
Nutritional elements Basic foods Hi-Pro. Specifications
No. 1 No. 2 No. 3
Calories 350 250 200 At most 300
Protein 250 300 150 At least 200
Vitamin - A 100 150 75 At least 100
Vitamin – B 75 125 150 At least 100
Cost of serving (Rs) 1.5 2.0 1.2
Revenue from serving(Rs) 5.5 7.0 4.2
Formulate the LPP model.

Solution [11]: Solution: X1 & X2 & X3 denotes the number of units of basic food number 1, 2 and 3.

Profit = (Revenue - cost)

Objective function: Maximize Z = 4X1+ 5X2+ 3X3.

Subjected to the conditions 350 X1+250 X2+200 X3 ≤ 300,
250 X1+300 X2+150 X3 ≥ 200,
100 X1+150 X2+75 X3 ≥ 100,
75 X1+125 X2+150 X3 ≥ 100
and X1, X2 & X3 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[12] The following table gives the data for a problem. Formulate the problem as a LP model.
Jun-July2009 & Jun2012

Requirement/Unit
Raw Materials Availability
I II III
A 2 3 5 4000
B 4 2 7 6000
Min Demand 200 200 150
Profit/ Unit 30 20 50

Solution [12]: X1, X2 & X3 denotes the number of required units of I, II and III.

Objective function, Maximize Z = 30X1 + 20X2+ 50X3

Subjected to the conditions , 2X1 + 3X2 + 5X3 ≤ 4000, ( Raw material A available)
4X1 + 2X2 + 7X3 ≤ 6000, ( Raw material B available)
X1 ≥ 200, ( Minimum requirements of I)
X2 ≥ 200, ( Minimum requirements of II)
X3 ≥ 150, ( Minimum requirements of III)
and X1, X2 & X3 ≥ 0

[13] The diet for a broiler to be formulated. The required daily batch of the feed is 100kgs. The diet must contain
(i) At least 0.8% but not more than 1.2% calcium. (ii) At least 22% protein. (iii) At least 5% crude fiber. The main
ingredients used include limestone, Corn and Soyabean. The nutritive content of these ingredients is summarized
below Formulate the model to minimize the cost. Dec-2010

Ingredients Calcium Protein Fiber Cost/kg in Rs

Limestone 0.3 0.0 0.00 1.65
Corn 0.001 0.09 0.02 4.65
Soyabean 0.002 0.5 0.08 12.5

Solution [13]: XL, XC & XS denotes the quantity (in Kgs) of Lime stone, Corm and Soya bean respectively,

Objective function, Minimize, Z = 1.65 XL+ 4.65 XC + 12.5 XS

Subjected to the conditions , (XL + XC + XS) ≥ 100

(0.3 XL + 0.001XC+ 0.002XS ) ≥ 0.008 (at least Calcium)
(0.3 XL + 0.001XC+ 0.002XS ) ≤ 0.012 (at most Calcium)
(0.0 XL + 0.09XC+ 0.5XS ) ≥ 0.22 (at least Protein)
(0.0 XL + 0.02XC+ 0.08XS ) ≥ 0.05 (at least Fiber)
and XL , XC & XS ≥ 0 (Non negativity constraints)

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 Operation Research (15ME81) – CBCS Scheme

[14] A boat manufacturer builds two types: type A and type B boats. The boats built during the months January-
June go on sale in the months July-December at a profit of Rs.2000/- per type A and Rs.1500/-per type B. Those
built during the months July-December go on sale in the months January-June at a profit of Rs.4000/-per type A
and Rs.3300/-per type B. Each type A required 5 hours in the carpentry shop and 3 hours in the finishing shop.
Each type B required 6 hours in the carpentry shop and 1 hour in the finishing shop. During each half year period,
a maximum of 12000 hours and 15000 hours are available in the carpentry and finishing shops respectively.
Sufficient material is available to built no more than 3000 type A and 3000 type B boats a year. How many of
each type of boat should be built during each half year in-order to maximize the profit. Formulate as the LPP.
June-July09
Solution [14]:
XAD & XBD denotes number of boats of A and B types sold during July-December and
XAJ & XBJ denotes number of boats of A and B types sold during January-June.

Objective function, Maximize Z = 2000 XAD + 1500 XBD + 4000 XAJ + 3300 XBJ
Subjected to the conditions , 5 XAD + 6 XBD ≤ 12000, (Available carpentry period January-June)
5 XAJ + 6 XBJ ≤ 12000, (Available carpentry period July-December)
3 XAD + 1XBD ≤ 15000, (Available finishing period January-June)
3 XAJ + 1 XBJ ≤ 15000, (Available finishing period July-December)
XAD + XAJ ≤ 3000, (Material available)
XBD + XBJ ≤ 3000, (Material available)
and XAD , XAJ, XBD & XBJ ≥ 0

[15] METRO Sports wishes to determine how many advertisements to place in the selected three-monthly
Magazines A, B and C. Objective is to advertise in such a way that total exposure to principle buyers of expensive
sports goods is maximized. Percentages of readers for each magazine are known. Exposure in any magazine is the
number of advertisements placed multiplied by the number of principle buyers. The following date may be used.

Exposure category Magazines A Magazines B Magazines C

Readers (in Lakh’s) 1 0.6 0.4
Principal buyers 10% 15% 7%
Cost per advertisement (Rs) 5000 4500 4250

The budgeted amount is at most Rs 1 Lakh for the advertisements. The owner has already decided that magazine
‘A’ should have no more than 6 advertisements and that ‘B’ and ‘C’ each have at least two advertisements.
Formulate LPP model for the problem.

Solution [15]: X1, X2 and X3 be the number of advertisement insertions in magazine A, B and C
respectively.

Objective function: (Find the total Exposure),

Maximize Z = (10% of 100000 X1+ 15% of 60000 X2+7% of 40000 X3)

Subjected to the conditions,

5000 X1+4500 X2+4250 X3 ≤ 100000, (Budget con)
X1 ≤ 6,
X2 ≥ 2 ,
X3 ≥ 2 (Advertisement con)
and X1 X2 & X3 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[16] A city hospital has the following daily requirements for nurses.
Period Clock time (24 hrs a day) Minimum number of NURSES required
1 6 AM – 10 AM 2
2 10 AM – 2 PM 7
3 2 PM – 6 PM 15
4 6 PM – 10 PM 8
5 10 - PM – 2 AM 20
6 2 AM – 6 AM 6
Nurses report to the hospital at the beginning of each period and work for 8 consecutive hours. The hospital wants
to determine the minimum number of nurses to be employed so that there will be sufficient number nurses
available for each period. Formulate this as a linear programming problem by setting up appropriate constraints
and objective function.
Solution [16]:
Let X1, X2, X3, X4, X5 and X6 be the number of nurses on duty at 6 AM, 10 AM, 2 PM, 6 PM, 10
PM, 2 AM and 6 AM respectively.
Objective function: Minimize Z = x1+x2+x3+x4+x5+x6
Subjected to the constraints
X1 + X 2 ≥ 7 ,
X2 + X3 ≥ 15,
X3 + X4 ≥ 8,
X4 + X5 ≥ 20,
X5 + X 6 ≥ 6
X6 + X 1 ≥ 2
and X1, X2, X3, X4, X5 and X6 ≥ 0

[17] A manufacturer of biscuits is considering three types of gift packs containing three types of biscuits; Orange
(O), Chocolate (C) and Wafers (W) Cream. Market research conducted to access the preferences of consumer
shows the following assortments to be in good demand.
Assortments Contents Selling price per kg
A Not less than 40% of O, Not more than 20% of C and any 220
B quantity
Not of W50% of O, Not more than 30% of C
less than 200
C No restrictions 120
Biscuits Plant capacity and Cost of manufacturing is given below.
Biscuit Varity : Orange Chocolate Wafers
Plant capacity (kg/day) : 200 200 150
Manufacturing Cost (Rs/kg) : 80 90 70
Formulate as LP Model to maximize the profit assuming there are no market restrictions.
Solution [17]:
Assortment 'A'- Consider X1a, X2a and X3a denote quantity in kg. of O, C & W biscuits respectively.
Assortment 'B'- Consider X1b and X2b denote quantity in kg. of O& C biscuits respectively.
Assortment 'C'- Consider X1c, X2c and X3c denote quantity in kg. of O, C & W biscuits respectively.
Profit = Revenue - Total cost
Profit = [220(X1a + X2a + X3a) + 200(X1b + X2b) + 120(X1c + X2c + X3c)] -
[80(X1a + X1b + X1c) + 90(X2a + X2b + X2c) + 70(X3a + X3c)]
Objective function:
Maximize Z = 140 X1a + 120 X1b + 40 X1c + 130 X2a + 110 X2b + 30 X2c + 150 X3a + 50 X3c
Subjected to the constraints,
Gift pack -A- Assortment X1a ≥ 0.4 (X1a + X2a + X3a), X2a ≤ 0.2 (X1a + X2a + X3a),
Gift pack -B- Assortment X1b ≥ 0.5 (X1b + X2b), X2b ≤ 0.3 (X1b + X2b),
Orange constraints - (X1a + X1b + X1c) ≤ 200
Chocolate constraints - (X2a + X2b + X2c) ≤ 200
Wafers constraints - (X3a + X3c) ≤ 150 Also X1a, X2a ,X3a, X1b ,X2b, X1c, X2c and X3c ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[18] An oil refinery wishes to blend 3 petroleum constitutes to make 2 grades of petrol A and B. The availability
and costs of the 3 constituents are given below:

Constituents Max. Available Barrels/Day Costs Rs/Barrel

1 3500 3000
2 2000 6000
3 3000 4000

To maintain the required quality of each grade of petrol, the following specifications are given along with the
selling price each grade.
Grade Specification Selling price(Rs/Barrel)
Not more than 30% of 1
A 5000
and Not more than 50% of 3
Not more than 50% of 1
B 4500
and Not more than 10% of 2
Setup a linear programming model for determining the amount of blend in each grade of petrol. Only formulate.
May 2007 (10 marks)

Solution [18]:
X1A & X3A denotes the quantity of Constituents 1 and 3 respectively in A.
X1B & X2B denotes the quantity of Constituents 1 and 2 respectively in B.

Profit = Revenue – Cost

Revenue = 5000 (X1A + X3A) + 4500 (X1B + X2B) (Revenue from A and B)
Cost = 3000 (X1A + X1B) + 6000 (X2B) + 4000 (X3A) (Cost constituents 1,2 and 3 in A and B)
Profit = 2000 X1A + 1500 X1B + 1000 X3A - 1500 X2B (Revenue - Cost)

Objective function, Maximize, Z = 2000 X1A + 1500 X1B + 1000 X3A - 1500 X2B

Subjected to the conditions , (X1A + X1B ) ≤ 3500,

X2B ≤ 2000,
X3A ≤ 3000,
X1A ≤ 0.3 (X1A + X3A),
X3A ≤ 0.5 (X1A + X3A)
X1B ≤ 0.5 (X1B + X2B)
X2B ≤ 0.1 (X1B + X2B)
and X1A , X3A , X1B & X2B ≥ 0.

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 Operation Research (15ME81) – CBCS Scheme

[19] A farmer has a 125 acre farm. He produces Radish, Lettuce and Potato. Whatever he raises is fully sold. He
gets Rs. 5 per kg for radish, Rs. 4 per kg for Lettuce and Rs. 5 per kg for potato. The average yield per acre is
1500 kg for radish, 1800 kg for Lettuce and 1200 kg for potato. Cost of manure per acre is Rs. 187.50, Rs. 225
and Rs. 187.50 for radish, Lettuce and potato respectively. Labor required per acre is 6 man-days each for radish
and potato and 5 man days for Lettuce. A total of 500 man-days of labor is available at the rate of Rs. 40 per man-
day. Formulate this as an LPP model to maximize the profit.

Solution [19]:
Consider XR, XL and XP denote quantity in ACRE land for Radish, Lettuce and Potato respectively.

Profit = Revenue - Total cost

Profit earned /acre = Profit/Kg * Yield /acre – manure cost – Labor
Profit from Radish = (5*1500 -187.5 – 6*40)
Profit from Lettuce = (4*1800 – 225 - 5*40)
Profit from Potato = (5*1200 – 187.5 – 6*40)

Total Profit = [(5*1500 -187.5 – 6*40) + (4*1800 – 225 - 5*40) + (5*1200 – 187.5 – 6*40) ]

Objective function:
Maximize Z = 7072.5 XR + 6775 XL +5572 XP
Subjected to the constraints,
Land constraints - (XR + XL + XP) ≤ 125
Labor constraints - (6XR + 5XL +6 XP) ≤ 500
Also, XR ,XL & XP ≥ 0

[20] The vitamins V and W are found in two different foods, F1 and F2. The respective prices per unit of each food
are Rs. 4 and Rs. 3. One unit of F1 contains 2 units of vitamin V and 3 units of vitamin W. One unit of F2 contains
4 units of vitamin V and 2 units of vitamin W. The daily requirements of V and W are at least 60 units and 75
units respectively. Formulate an LPP to meet the daily requirement of the vitamins at minimum cost.

Solution [20]:
Consider X1 and X2 denote quantity of Food F1 and F2 respectively.

Objective is to Minimize Z = 4X1 + 3X2

Subjected to the Constraints:

Requirement of V: 2X1 + 4 X2  60
Requirement of W: 3X1 + 2 X2  75
Non-negativity: X1 and X2 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[21] A mutual fund has Rs. 2 million available for investment in Government bonds, blue chip stocks, speculative
stocks and short-term bank deposits. The annual expected return and the risk factor are as shown.

Investment Return% Risk factor (0- 100)

Bonds 14 12
Blue Chip 19 24
Speculative 23 48
Short-term 12 6
The fund is required to keep at least Rs. 200,000 in short-term deposits and not to exceed an average risk factor of
42. Speculative stocks must not exceed 20% of the money invested. Formulate the LPP maximizing expected
annual return.

Solution [21]:
Consider X1, X2, X3 and X4 denote the amounts invested in Govt. Bonds, Blue Chip, Speculative and Short-
term respectively.

Maximize the Return on investments,

Maximize, Z = 0.14 X1 + 0.19X2 + 0.23X3 + 0.12X4

Subjected to the constraints,

Average risk factor = [(12 x1+24 x2 + 48 x3 + 6 x4)]/[(x1+ x2+ x3+ x4)]  42.
 This gives 30x1+18x2-6x3+36x4  0.
Also x3  0.2 (x1+ x2+ x3+ x4) - Maximum limit on speculative stock not exceeding 20% of Money invested
 This gives: 0.2 x1 + 0.2 x2 – 0.8 x3 + 0.2 x4  0

LPP formulation is as follows:

Maximize, Z = 0.14 X1 + 0.19X2 + 0.23X3 + 0.12X4

Subjected to the constraints,
30x1+18x2-6x3+36x4  0 (Avg Risk factor)
0.2 x1 + 0.2 x2 – 0.8 x3 + 0.2 x4  0 (limit on speculative stock)
x1 + x2+ x3 + x4  2,000,000 (Available to Invest)
x4  200,000 (Short Term Deposit)
also , x1, x2, x3, x4  0 (Non-negativity)

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 Operation Research (15ME81) – CBCS Scheme

[22] Medical experts and dieticians opine that it is necessary for an adult to consume at least 75g proteins, 85g of
Fats and 300g of Carbohydrates daily. The following table lists 6 types of food items and their respective
nutritional values and the corresponding costs per Kg. Formulate the LP so that the total cost of food satisfying
min. requirements of balanced diet is lowest

Food Type (Gms) per 100g

Food Type Cost/Kg(Rs)
Proteins Fats Carbs
1 8 1.5 35 1
2 18 15 3
3 16 4 7 4
4 4 20 2.5 2
5 5 8 40 1.5
6 2.5 25 3
Min. daily requirements 75 85 300

Solution [22]:
Consider X1,X2,X3,X4,X5, and X6 denote each FOOD types respectively to be used per day.
Objective Function is to minimize the cost of foods while meeting the minimum requirements of the nutrition.

Minimize Z = 1X1 + 3X2 + 4X3 + 2X4 + 1.5X5 + 3,X6

Subjected to the constraints,
Daily requirements of Proteins (75 gms)
→ 8X1 + 18X2 + 16X3 + 4X4 + 5X5 + 2.5,X6 ≥ 75
Daily requirements of Fats (85gms )
→ 1.5X1+ 15X2 + 4X3 + 20X4 + 8X5 + 0,X6 ≥ 85
Daily requirements of Carbohydrates (300gms )
→ 35X1 + 0X2 + 4X3 + 2.5X4 + 40X5 + 25,X6 ≥ 300
Also, X1,X2,X3,X4,X5, and X6 ≥ 0

[23] A company manufactures two products X and Y, which require, the following resources. Which undergoes
operation three machines M1, M2, and M3. The available capacities are 50, 25, and 15 hours respectively in the
planning period. Product X requires 1 hour of machine M2 and 1 hour of machine M3. Product Y requires 2 hours
of machine M1, 2 hours of machine M2 and 1 hour of machine M3. The profit contribution of products X and Y
are Rs.50/- and Rs.40/-respectively

Solution [23]:
Consider X1 and X2 denote quantity of products X and Y respectively.
Objective is to Minimize Z = 50X1 + 40X2
Subjected to the Constraints:
Operation on M1 : 0X1 + 2 X2  50
Operation on M2 : 1X1 + 2 X2  25
Operation on M3 : 1X1 + 1 X2  15
Non-negativity: X1 and X2 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[24] A retail store stocks two types of shirts A and B. These are packed in attractive cardboard boxes. During a
week the store can sell a maximum of 400 shirts of type A and a maximum of 300 shirts of type B. The storage
capacity, however, is limited to a maximum of 600 of both types combined. Type A shirt fetches a profit of Rs.
200/- per unit and type B a profit of Rs. 500/- per unit. How many of each type the store should stock per week to
maximize the total profit? Formulate a mathematical model of the problem

Solution [24]:
Consider X1 and X2 denote quantity of Shirts A and B respectively.
Objective is to Maximize, Z = 200X1 + 500X2
Subjected to the Constraints:
Maximum sales of A, X1  400
Maximum sales of B, X2  300
Storage Capacity : X1 + X2  600
Non-negativity: X1 and X2 ≥ 0

[25] A computer company manufactures laptops and desktops that fetch them a profit of Rs. 7000 and Rs. 5000
respectively. Each unit of Laptop takes 4 hrs to assemble and 2 hrs to test where as each unit of Desktop takes 3
hrs to assemble and 1 hr for testing. In a given month, the total assembly time available is 2100 hrs and total
testing time available is 900 Hrs. Market can absorb only 2000 laptops and 3500 desktops. Formulate the LPP
such that the profit is maximum

Solution [25]:
Consider X1 and X2 denote quantity of laptops and desktops respectively.
Objective is to Maximize, Z = 7000X1 + 5000X2

Subjected to the Constraints:

Maximum Assembly time, 4X1 + 3X2  2100
Maximum Testing time, 2X1 + 1X2  900
Market constraints : X1  2000
Market constraints : X2  3500

Non-negativity: X1 and X2 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

[26] A garment manufacturer has a production line making two styles of shirts. Style I needs 200 g of cotton
thread, 300 g of Dacron thread and 300 g of linen thread. Corresponding requirements of style II are 200g, 200g
and 100g. The net contributions are Rs. 519.50 for style I and Rs. 515.90 for style II. The available inventory of
cotton thread, Dacron thread and linen thread are, respectively, 240 kg, 260 kg and 220 kg. The manufacturer
wants to determine the number of each style to be produced with the given inventory. Formulate the LPP model

Solution [26]:
Consider X1 and X2 denote quantity of Style I and Style II respectively.
Objective is to Maximize, Z = 519.50X1 + 515.90X2

Subjected to the Constraints:

Maximum Cotton, 0.2X1 + 0.2X2  240
Maximum Dacron, 0.3X1 + 0.2X2  260
Maximum Linen, 0.3X1 + 0.1X2  220
Non-negativity: X1 and X2 ≥ 0

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 Operation Research (15ME81) – CBCS Scheme

Q1) Explain briefly explain the areas of management decision making, where OR techniques can be applied.
Q2) List the various phases of OR problems.
Q3) Define, i.) Feasible solution ii) Feasible region iii) optimal solution iv) Infeasible solution
v) CPF solution vi) Degeneracy.
Q4) Define Operation Research.
Q5) Explain limitations of OR models.
Q6) Enumerate and briefly explain applications and limitations of OR to engineering problems.
Q7) Discuss the areas of management where operation research techniques are applied.
Q8) State the assumptions made in LPP and explain in brief any one of them.
Q9) Explain briefly the scope of Operation Research.
Q10) Describe the phases of OR.
Q11) Discuss the areas of managements where OR techniques are applied.
Q12) Give the classification of models used in OR. Explain the mathematical modeling process.
Q13) Explain the components involved in the formulation of LPP, with a simple example.
Q14) Explain with few points about Variables, Objective function, constraints and non-negativity

1. Solve by the following LPP by simplex method

Maximize z =3x1 + 2x2
Subject to, x1 +x2 <_8
x1 -x2<_2
x1.x2>_0
2. Solve the following LPP by simplex method:
Maximize z =6x1 + 8x2
Subject to, 2x1 +8x2 <_16
2x1 +4x2<_8
x1.x2>_0
3. Use graphical method to solve LPP
Minimize z =3x1 + 5x2
Subject to, -3x1 +4x2<_12----------- (1)
2x1 +3x2>_12----------- (2)
2x1 -x2>_0----------- (3)
x1>_4----------- (4)
x2>_2----------- (5)
x1.x2>_0
Write the dual of the problem. 10 Marks

4. Solve the following LPP using simplex method

Minimize z =2x1 + 3x2 +x3
Subject to, x1 +4x2 +2x3 <_8
3x1 +2x2>_6
x1.x2>_0

5. Solve the following LPP using dual simplex method

Minimize z =2x1 + x2
Subject to, 4x1 +2x2 >_6
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 Operation Research (15ME81) – CBCS Scheme

x1 +2x2>_3
x1.x2>_0
6. Solve the following LPP using Big M method
Minimize z =2x1 + 5x2
Subject to, x1 +x2 =100
x1 <_40
x2>_30
x1.x2>_0
7. Using graphical method, Solve the LPP
Maximize z =5x1 + 4x2
Subject to, 6x1 + 4x2 <_24
x1 +2x2<_6
-x1 + x2<_1
x1.x2>_0
8. Using simplex method, Solve the following LPP
Maximize z =4x1 + 3x2 +6x3
Subject to, 2x1 + 3x2 +2x3 <_440
4x1 +3x2<_470
2x1 +5x2<_430
x1.x2.x3>_0

9. Using graphical method, Solve the LPP

Maximize z =3x1 + 5x2
Subject to, x1 <_4
2x2<_12
3x1 + 2x2<_18
x1.>_0
x2>_0
10. Solve by simplex method
Maximize z =3x1 + 9x2
Subject to, x1 +4x2 <_8
x1 +2x2<_4
x1.x2>_0
11. Solve the following LPP:
Maximize z =3x1 + 9x2
Subject to, x1 +4x2 <_8
x1 +2x2<_4
x1.x2>_0

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