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Formulation Of Lp Problem

Example 2.1.A manufacturing company is producing two products A and B. Each of the products A and B requires the use of two machines P and Q. A requires 4 hours of processing on machine P and 3 hours of processing on machine Q. Product B requires 3 hours of processing on machine P and 6 hours of processing on machine Q. The unit profits for products A and Bare Rs 20 and Rs 30 respectively. The available time in a given quarter on machine P is 1000 hours and on machine Q is 1200 hours. The market survey has predicted 250 units of products A and 300 units of product B can be consumed in a quarter. The company is interested in deciding the product mix to maximize the profits. Formulate this problem as LP
model.

Sol. Formulating the problem in mathematical Equations

Let XA= the quantity of product of type A manufactured in a quarter

XB= the quantity of products of type B manufactured in a quarter

Z = the profit earned in a quarter.

(Objective function, which is to be maximized)

Therefore Z = 20 XA+ 30 XB

Z is to maximized under the following conditions:

4XA+ 3XB ≤ 1000 (Time constraint of machine P)

3XA+ 6XB ≤ 1200 (Time constraint of machine Q)

XA ≤  250 (Selling constraint of product A)

XB ≤ 300 (Selling constraint of product B)

XAand XB  ≥0 (Condition of non-negativity)

Example 2.2.MIS Steadfast Lid produces both interior and exterior house paints for wholesale distribution. Two types of raw materials A and B are used to manufacture the paints. The maximum availability of A is 10 tons a day and that of B is 15 tons a day daily requirement of raw material per ton of interior and exteriorpaints are as follow:

Requirement of Raw material Per ton of paint

Raw material A

3

2

10

Raw material B

2

3

15

 

The market survey indicates that daily demand of interior paints cannot exceed that of exterior paints by 2 tons. The survey also shows that maximum demand of interior paints is only 3 tons daily.

The wholesale price per ton is Rs 75,000 for exterior paints and Rs 50,000 for interior paints.

Problem.How much interior and exterior paints MIS Steadfast should produce to maximize its profits?

Sol. Let XE – tons of external paints to be produced daily.

X1- tons of internal paints to be produced daily.

Z – profit earned (objective function which is to be maximized)

Therefore

Z = 75000 XE+ 50000 X1

Z is to be maximized under the following constraints or conditions :-

2XE+ 3X1  ≤ 10 (Availability constraint of raw material A)

3XE + 2X1 ≤ 15 (Availability constraint of raw material B)

X1 – XE ≤ 2 (Demand constraint – Demand of interior paints daily cannot exceed more than 2 tons that of exterior paint)

X1 ≤ 3 (Demand of interior paint cannot exceed 3 tons every day)

Also,

X1  ≥ 0 (Non-negativity constraint of interior paints)

XE ≥ 0 (Non-negativity constraint of exterior paints)

The complete mathematical model for MIS Steadfast Ltd problem maybe written as follows :-

Determine the tons of interior and exterior paints, X1 and. XEto be produced in order to maximize Z = 75000 XE + 50000 X, (objective function) under the constraints (conditions) of

2XE+ 3X1  ≤ 10

3XE + 2X1 ≤  15

X1- XE ≤ 2

X1 ≤  3

XE ≤ 0

X1. ≥ 0

Let us verify if the above model satisfies the conditions of Linearity as we are using LP method to solve the problem, Linearity demands that Proportionality and Additivity must be satisfied.

1. Proportionality. It requires that usage of resources is directly proportional to that value of the variables, in this case XEand X1. Suppose MIS Steadfast have .a promotion policy that they will sell external paint at Rs 60,000 per ton after sales of exterior paintis more than 2 tons per day, then the equation of the objective function will no longer be true as each ton of external paint produced does not bring a revenue of Rs 75,000/- Actually for XE ≤ 2 tons, the revenue will be Rs 75000 per ton and Rs 60,0001- per ton for XE ≥ 2 tons. This situation does not satisfy the condition of direct proportionality with XE.

2. Additivity demands that the objective function (Z) must be the direct sum of the individual contribution of different variables. For example, in case of two competing products like tea and coffee where an increase in sale of coffee adversely affects the sale of tea, these two products do not satisfy the conditions of additivity.

Since the mathematical model written above, satisfies both the properties of proportionality and additivity, we can use LP method to solve the problem.

Example 2.3.A chemical manufacturer produces two types of chemicals X and Y. Each typeof chemical is manufactured by a two-step process that involves machines A and B. The processing time for each unit of the two products 011 machines A and B are as follows :-

Product

Machine A (Hours/unit)

Machine B (Hours/unit)

X

2

3

Y

4

2

For a period of one month, the availability of machine A is 160 hours and availability of machine B is 120 hours. The manufacturer has found out from the market that maximum sale price of chemical X can be Rs 400 per unit and that of chemical Y can be Rs 500 per unit. Also, maximum of 20 units of chemical X can be sold in the market per month and 25 units of chemical B.

Problem.How much of chemical X and chemical Yshould be produced so that the profits can be maximized?

Sol. Let X – be the quantity of chemical X to be produced.

Y – be the quantity of chemical Y to be produced.

Z = 400X + 500Y

Z is to be maximized under the following constraints or conditions :-

2X + 4 Y  ≤ 160 (Availability constraint of machine A)

3X + 2Y ≤ 120 (Availability constraint of machine B)

X ≤20 (Marketing constraint of product X)

Y ≤ 25 (Marketing constraint of product B)

X ≥ 0 (Non-negativity constraint of product X)

Y ≥ 0 (Non-negativity constraint of product Y)

The complete mathematical model for the problem may be written as follows :-

Determine the quantity of product X and Y to be produced in order to maximize

Z = 400X + 500Y (objective function) under the constraint of

2X +4Y  ≤ 160

3X + 2Y  ≤ 120

X ≤ 20

Y ≤ 25

X  ≥ 0

Y  ≥ 0

Example 2.4.MIS Plastics Ltd. produces two major products tables and chairs. Each table can be sold in the market at a profit of Rs 50 and chair can be sold at a profit of Rs 60. The requirement of man hours and machine hours for manufacturing of one table and one chair is as given below:

Product

Manhours (Hours per unit)

Machine (Hours per unit)

Table

10

1.5

Chair

6

1.0

 

Total number of man hours available for this activity during the year is 1200 and machine is available only for 800 hours. The company carried out a survey to find out the maximum demand for their table and chairs. It was revealed that they can sell only 100 tables and 160 chairs in one year.

Problem.How many tables and chairs should MIS Plastics Ltd. produce to maximize their profit?

Sol. Let X be the number of tables to be produced.

Y be the number of chair to be produced.

Z = profit earned (objective function which is to be maximized)

Therefore Z = 50X + 60Y

Z is to be maximized under the following conditions

10X + 6Y ≤ 1200 (Availability constraint of man hours)

1· 5X + Y ≤ 800 (Availability constraint of machines hours)

X ≤ 100 (Demand constraint)

Y ≤  160 (Demand constraint)

X ≥ 0 (Non-negativity constraint of tables)

Y ≥ 0 (Non-negativity constraint of chairs)

Example 2.5.An oil refinery uses blending process to produce gasoline in a typical manufacturing process. Crude A and B are mixed to produce gasoline G1and gasoline G2. The inputs and outputs of the process are as follows :-

Availability of Crude A is only 200 tons and B 300 tons. Market demand of Gasoline G1is 150 tons and Gasoline G2is 120 tons. Profit by using process 1is Rs 200 per ton and by using process 2 is Rs 250 per ton. What is the optimal mix of two blending processes so that the refinery can maximize its profits?

Sol. Let X be the number of tons to be produced by process 1

Y be the number of tons to be produced by process 2

Z = Profit earned (Objective function which is to be maximized)

Therefore Z = 200X + 250Y

Z is to be maximized under the followingconditions

5X + 6Y ≤ 200

2X + 8Y ≤ 300

6X+ 5Y ≤  150

8X + 7Y   ≤ 120

X ≥ 0

Y ≥ 0

Example 2.6.Vitamin C and Vitamin Eare found in two different fruits F1 and F2. One unit of fruit F1 Contains 3units of vitamin C and 2 units of vitamin E. Similarly, one unit of fruit F2 contains 2 units of vitamin C in it and 2 units of vitamin E in it. A patient needs minimum of 30 units of vitaminC and 20 units of vitamin E. Also one unit offruit F1 costs Rs 20 and oneunit of fruit F2 costs Rs 25. The problem, the hospitals faces is to find such units of ‘fruit FI and F!, which should be supplied to the patients at minimum cost.

Sol. Let X be the number of units of fruit F1

Y be the number of unitsof fruit F2

Z = Minimum cost (Objective function which is to be minimized)

Therefore Z = 20X + 25Y

Z is to be minimized subject to the following constraints :-

3X + 2Y  ≥ 30 (Minimum requirement of vitamin C)

2X + 2Y  ≥ 20 (Minimum requirement of vitamin E)

X  ≥ 0

Y≥ 0

Note – In this example a unit means different measures like grams and caloric values etc.

Example 2.7.Manufacturing company XYZ Lid manufactures two different types of products, refrigerators and washing machines. Boththese products have to be processed through two machines, Machine A and Machine B. Machine A is available for 200 hours and machine B is available for 100 hours. The requirement of time on these machines is as follows:-

Refrigerator

Washing machine

Machine A

10

6

Machine B

5

4

 

The company makes a profit ofRs 800 on sale of one refrigerator and Rs 500 on sale of one washing machine. What quantities of refrigerators and washing machines should company XYZ Ltd. produce to maximize its profits?

Sol. Let X be the number of refrigerators to be manufactured.

and Y be the number of washing machines to be manufactured.

Z = Profit earned (Objective function which is to be maximized)

Therefore      Z = 800X + 50Y

Z is to be maximized subject to the following constraints :-

10X + 6y ≤ 200 (Availability of machine A)

5X + 4Y ≤ 100 (Availability of machine B)

X  ≥ 0

Y ≥  0

Example 2.8.A manufacturer can manufacture two different types of products, FRP sheets and FRP bath tubs. Each unit of FRP sheets of a particular size needs 5 kg of raw material A and 2 kg of raw material B. Each unit of FRP bath tubs needs 7 kg of raw material A and I kg of raw material B. Availability of raw material A in the market is 500 kg and that of raw material B 100 kg. Each FRP sheet contributes profit of Rs 100 and each FRP tub contributes profit of Rs 400. What is the most suitable product mix for the manufacturer to maximizeprofits?

Sol. Let X be the number of FRP sheets to be produced

and Y be the number of FRP tubs to be produced

Z = Profit earned (Objective function which is to be maximized)

Therefore Z = 100X + 400Y

Z is to be maximized subject to the following constraints

5X + 2Y  ≤ 500

7X+ Y ≤ 100

X ≥ 0

Y ≥ 0

Example 2.9.A company manufactures three types of electrical products, electric iron, fan and toaster. All the three products have to be processed on two machines A and B. The processing time required by each product on both the machines is as given below:-

Electric Iron

Fan

Toaster

Machine A

2

3

2

Machine B

1

2

3

 

Machine A is available only for 200 hours and machine B is available for 160 hours. The firm should not manufacture more than 400 electric irons, more than 500 fails and more than 200 toasters. All electric iron gives a profit of Rs 110, a fall of Rs 150 and a toaster of Rs 80. What product mix would you recommend to the company so that its profits are maximized?

Sol. Let X1 be the number of electric irons to be manufactured.

X1be the number of fans to be manufactured

X3 be the number of toasters to be manufactured

Z = profits generated (Objective functions which is to be maximized)

Z is to be maximized under the following constraints :-

Z = 110X1+ 150X2+ 80X3

2X1 + 3X2+ 2X3 ≤ 200

X1 + 2X2 + 3X3 ≤ 160

Also, X1, X2, X3 ≥ 0

Example 2.10.A manufacturer of furniture makes twoproducts, chairs and tables. Processing of these products is done on two machines A and B. A chair requires two hours on machine A and 6 hours Oil machine B. A table requires 5 hours on machine A and notime onmachine B. There are 16 hours of time per day available on machine A and 30 hourson machine B. Profit gained by the manufacturer from a
chair and table is Rs 2 and Rs 10 respectively. What should be daily production of the two products?

Sol. Let X be the number of chairs to be manufactured

and Y be the number of tables to be manufactured

Let Z = Profit earned (objective function which is to be maximized)

Z is to be maximized under the following constraints

Z=2X+ 10Y

2X + 5Y   ≤ 16 (Availability constraint of machine A)

6X  ≤ 30 (Availability constraint of machine B)

X  ≥ 0

Y ≥ 0

Example 2.11.Sandeep Electric company produces two products, motors and fans which are produced and sold onmonthly basis. The monthly production of motors cannot exceed 100 and that of fans 150 because of the limitation of the facilities of production. Motor requires 12 man-hours of labour per month and fanrequires 12 man hours of labour per month. The company has a total of 50 employees. Profit margin on sale ofRs 110 and on sale of fan Rs 90. Formulate a LP problem.

Sol. Let X be the number of motors produced

and Y be the number of fans produced

Z = profit earned (objective function)

Z = 110X + 90Y

X ≤ 100 (Constraint of production of motors)

Y  ≤ 150 (Constraint of production of fans)

30X + 12Y ≤  50 (Constraint of availability of man-hours)

X ≥ 0

Y ≥ 0

Example 2.12.A carpenter has 90, 80 and 50 running feet respectively of teak. plywood and rosewood. Product A requires 2, 1 and 1 running feet of teak, plywood and rosewood respectively. Product B requires 1, 2 and 1 running feet of teak, plywood and rosewood respectively. If A would sell/or Rs 48 and B would sell for Rs 40 per unit, how much of each should he make and sell in order to obtain the
maximum gross income out of his stock of wood ? Give a mathematical formulation to this linear programming problem.

Sol. Let X be the number of products A produced.

and Y be the number of products B produced.

Let Z = gross income (objective function)

Z is to be maximized under the following constraints

Z =48X +40Y

2X + Y ≤ 90 (Constraint of availability of teak)

X + 2Y ≤ 80 (Constraint of availability of plywood)

X + Y ≤  50 (Constraint of availability of rosewood)

X ≥ 0

Y  ≥ 0

Example 2.13.A retailer deals in two items only, item A and item B. He has Rs 50,000 to invest and a space to store at the most 60 pieces. An item of A costs himRs 2500 and B costs him 500. Net profit to him all items A is Rs 500 and on item B is Rs 150. If he can sell all the items that he purchases, how much should he invest his amount to have maximum profit? Give a mathematical formulation to the linear programming problem.

Sol. Let X be the number of items A the retailer should be purchase

and Y be the number of items B the retailer should purchase

Z = profit earned (objective function)

Z = 500X + 150Y

X + Y  ≤ 60 (Constraint of total capacity)

2500X + 500 Y ≤  50,000 (Constraint of total investment)

X  ≥ 0

Y ≥ 0

 

Example 2.14.Company produces two types of products A and B. Product B is superior quality and product B is lower quality. Profit all the two types of products are Rs 30 and Rs 40 respectively. The data about resources required and availability of resources are given below:

Requirement of Resources

Product A

Product B

Capacity available (per month)

Raw material (Kg)

60

120

12000

Machine hours (per piece)

8

5

630

Assembly

3

4

500

 

How should the company manufacture the two types of products in order to get the maximum overall profits? Formulate the problem as a LPP.

Sol. Let X be the quantity of product A manufactured

and Y be the quantity of product B manufactured.

Z = overall profits (Objective function)

Zis to be maximized under the following conditions

Z = 30X + 40Y

60x + 120Y ≤ 12000 (Constraint ofraw material)

8X + 5Y ≤ 630 (Constraint of machine hours)

3X + 4Y ≤500 (Constraint of assembly)

X ≥0

Y ≥0

Example 2.15.A machine is producing either product A or B. Itcan produce product A by using 2 units of chemicals and 1unit of a compound and can produce B by using 1unit of chemical and 2 units of compound. Only 800 units of chemical and 100 units of compound are available. The profits available per unit of A and Bare Rs 30 and Rs 20 respectively. Find the optimum allocation of units between A and B tomaximize the total profit. Give a mathematical formulation to the LP problem.

Sol. Let X be the quantity of product A produced

andY be the quantity of product B produced

Z = Total profits (Objective function)

Z is to be maximized under the following constraints.

Z = 30X + 20Y

2X + Y ≤ 800 (Constraint of availability of chemical)

X + 2Y ≤1000 (Constraint of availability of compound)

X≥0

Y≥0

Example 2.16.ABC Ltd manufacturers two products A and B. For manufacturing product A, a machine has tobe used for 3 hours and the operator has toput in4 hours. For manufacturing product B the machine use is for 4 hours and the operator works only for 2 hours. In one week. the company can use amaximum of 60 hours of machine time and the worker is available for 80 hours. The contribution of profit by product A is Rs 40 and by product B is Rs 50. Formulate the problem as a LP model andfind what quantities should ABC Lid manufacture of product A and B tobe able tomaximize their profits. Assume that all quantities of A and B produced are sold without any problem.

Sol. Let X and Y be the quantities of product A and B respectively to be manufactured, The data of the problem can be put in the form of a table as shown below.

Decision Variable
Quantities to be manufactured

Product

Time Required
(Hours)

Product/Unit
(Rs)

Machine

Operator

X

A

2

4

40

Y

B

4

2

50

Max time available (Hours)

60

78

 

Objective function

Maximise Z = 40X + 50Y

Constraints

(i) 2X + 4Y ≤  60

(ii) 4X + 2Y ≤  78 Converting the inequalities into equalities for solving these

(iii) X, Y ≥ 0

Now X and Y can be solved to have maximum of Z.

X= 16

Y=7

Z=40 ×16+50 ×7=640+350=Rs990

ABC should manufacturer 16 number of component A and 7 of component B.

Example 2.17.An ancillary unit of HMT tractors produces two types of tractor parts A and B. All the parts that are manufactured by the ancillary units are purchased by HMT. The manufacturing cost of part A is Rs 2400 and of part B it is Rs 3200. HMT purchases part A for Rs 3200 and part Bfor Rs 3600. The company’s present production capacity is limited because of the following three constraints.

(a)             Budget constraints – cash available at the beginning of year Rs 10,50,000

(b)             Machine time available 4200 hours/year

(c)             Manufactures available/year 5600.

Part A needs 8 hours of machine time and part B needs only 4 hours. Also, the man-hours requiredfor part A are 8 hours and for part B 8 hours, find out what quantities of these should be produced by the ancillary unit to maximize its profits,

Sol. The problem can be formulated as an LP problem with constraints as follows. The given data can be put in the form of a table for ease of understanding let X1be the quantity of part A to be manufactured and X, be the quantity of part B to be manufactured

Decision Variable Quantities to be manufactured

Part

Machine Time (Hours)

Operator Time (Hours)

Cost Price Rs

Selling Price (Rs)

X1

A

8

20

2400

3200

X2

B

4

8

3200

3600

Total availability

4200

5600

Budget availability

=Rs 25,60,000

 

Objective Function

Maximize Z = 800X1+ 400X2

2400 X1 + 3200 X2≤4,60,000 (Constraints of Budget)

8X1+ 4X2≤ 4200 (Constraint of machine time)

20XI + 8X2≤5600 (Constraint of man hours)

X1, X2≥ 0 (non-negativity constraints)

These inequalities can be converted into equations and solved.

3200 X1+ 3600 X2= 25,60,000

8X1 + 4X2= 4200

90X1+ 8X2= 5600

X1 = 200

X2= 650

It can be seen that the cost of quantity of parts manufactured remains within the total budget.

Also Z = Rs 4,20,000.

Example 2.18.A company supplying three types of parts to an automatic manufacturing company, purchases castings of three parts from a nearby foundry and performs three types of operators before selling these and cost per hour of these machines is given in the table below

Machine

Capacity/hour

Capacity/hour

Capacity/hour

Cost/hour (Rs.)

Cutting

20

60

25

150

Drilling

40

20

40

100

Polishing

50

50

20

200

The cost ofthe castings for A, Rs 120, for B Rs 200, for C Rs 400 and the selling price of these parts is Rs 200, Rs 350 and Rs 500 respectively. All the parts that are processed by the company can be sold, What quantity of various parts should the company process for selling in order to maximize their profits?

Sol. Let X1, X2 and X3 be the number of parts the company should process.

Example 2.19.ABC Lid is assembling two products P1 and P2, The cost of assembling one unit of products P1 and P2 is Rs 200 and Rs 240 respectively. The availability of work station for two products is limited to 60 hours and the two products spend 6 hours and 2 hours respectively all tile work station. The products can be sold for Rs 280 and Rs 320 respectively. Total mall-hours available are 400 and P1 requires 2 man-hours and P2 requires 4 man-hours, Formulate the problem as a LPP.

Sol. Let X1and X2be the number of products P1and P2to be assembled.

Maximise Z = 80X1+ 80X2 with the following constraints

6X1+ 2X2≤ 60 (Workstation constraints)

2X1 + 4X2≤400 (man-hours constraints)

X1,X2≥0

These inequalities can be converted into equations and solved for X1and X2.

Example 2.20.A person has 12,00,000 which he wants to investin different types of investment opportunities. He has consulted investment consultant to advise him on this who has given him the following advice.

(a)             Government sector, PoswfficeiNSC etc upto 40%

(b)             Share X and Share Y both put together Rs 2,00.000.

(c)             Mutual funds A and B. At least 25%

The rate of return which he gets from these investments are

(a) Government sectors 0 ·1

(b) Share X 0·2

(c) Share Y0.25

(d) Mutual fund A 0 ·15

(e) Mutual fund B 0 ·12

Formulate the above as a LP problem.

Sol. LetX1= Government sector investment

X2= Share X

X3= Share Y

X4= Mutual Fund A

X5= Mutual Fund B

Objective function is to be maximized = Z = 0.1 X, + 0.2 X2 + 0.25 X3 + 0.15 X4 + 0.12 X5 the following constraints.

X1 + X2 + X3+ X4+ X5≤ 12,00,000 (total investment constraint)

X≤4,80,000

X2+ X3≤ 2,00,000

X4 + X5≥3,00,000

X1, X2, X3, X4, X5≥ 0

These inequalities can be converted into equations and solved for X1, X2 , X3 , X4 and X5and hence for objective function Z.

Example 2.21.A firm manufacturing three products A, B and C. Time to manufacture product A is twice that for B and thrice that for C and they are to be produced in the ratio 2: 3 : 4. The relevant data is given below. If the whole labour is engaged in manufacturing product A. 2000 units of the product call be produced. There is a demand for at least 200. 300 and 400 units of the product A. B. C and the profit earned per unit is Rs 100, Rs 70 and Rs 50 respectively. Formulate the problem as a linear programming problem.

Raw Material

Requirement per unit of product

Total availability (Kg)

A

B

C

P

6

5

9

4000

Q

4

8

6

5000

Maximize Z=100A+70B+50C (Objective function) subject to the following constraints

6A + 58 + 9C≤ 4000 (Constraint of raw material P)

4A + 88 +6C ≤ 5000 (Constraint of raw material Q)

Example 2.22.A patient wants to decide the constituents of diet which his daily
requirement of proteins, fats and carbohydrate’s at the minimum cost. The choice is to be made from three different types offoods. The yields per unit of these foods are as follows:

Food Type

Yield per unit

Cost per unit (Rs.)

Proteins Fats Carbohydrates

1

2

2

4

40

2

4

2

4

50

3

8

10

8

60

Minimum requirement

1200

400

800

Formulate linear programming model for the problem.

Sol. Let A, Band C be the number of units of food type 1,2,3, and respectively.

Maximise Z = 40A + SOB + 60C (Objective function) subject to the following constraints)

2A + 4B + 8C ≥ 1200 (Constraint of Proteins)

2A + 2B + 10C ≥400 (Constraints of Fats)

4A + 4B + 8C ≥ 800 (Constraints of Carbohydrates)

A≥0

B ≥ 0 (Non-negativity constraints)

C≥0

 

Example 2.23.A firm manufactures three products A, Band C. The profits are Rs3. Rs2 and Rs 4 respectively. The firm has two machines and the required processing time in minutes for each machine on each product if given below:

Product

A

B

C

Machine

C

4

3

5

D

2

2

4

 

Machine C and D have 2000 and 2500 machine-minutes respectively. The firm must manufacture 100 A’s, 200 B’s and 50 C’s but 110t more than 150 A’s, Set lip all LP model to maximize the profit.

Sol. Maximize Z: 3A + 2B + 4C

Subject to 4A + 38 + 5C ≥ 2000

2A + 28 + 4C ≥ 2500

A ≥ 100

B ≥ 200

A ≤ 150

C ≤  50

A, B, C ≥ 0 .

Example 2.24.A [inn can produce three types of cloth A, Band C. Three kinds of wool are required for it. say, red wool. green and blue wool. One unit length of type A cloth needs 2 yards of red wool and 3 yards of blue wool, one unit length of type B cloth needs 3 yards of red wool. 2 yards of green wool and 2 yards of blue wool and one unit length of type C cloth needs 5 yards of green wool and 4 yards of blue wool. The firm has a stock of only 8 yards of red wool. 10 yards of green wool and 15 yards of blue wool. It is assumed that the income obtained from one unit of type A cloth is Rs3. of type B cloth is Rs5 and that of type C cloth is Rs4. Formulate the problem as linear programming problem,

Sol. Maximize Z: 3A + 58 + 4C (Objective function) subject to the following constraints

2A + 38 ≤ 8 (Constraints of availability of red wool)

2A + 5B ≤ 10 (Constraints of availability of green wool)

3A + 28 + 4C ≤ 15 (Constraints of availability of blue wool)

A, 8, C ≥ 0 (Non-negativity constraints)

Example 2.25.Orient Paper Mills produces two grades of paper X and Y. Because of raw material restrictions not more than 400 tons of grade X and no/more than 300 tons of grade Y can be produce a ton of week. There are 160 production hours in a week. It requires 0·2 and 0.4 hours to produce a ton of products X and Y respectively. with corresponding profit of Rs 20 and Rs 50 per ton. Formulate a linear programming problem to optimize the product mix for maximum profit.

Sol. Maximize Z = 20X + 50Y (Objective function)

Subject to the constraints

0·2X + 0·4Y ≤ 160 (Constraints of production hours)

X ≤ 400 (Constraints of raw material)

Y ≤ 300 (Constraints of raw material)

X ≥ 0

Y ≥ 0 (Non-negative constraints)