The procedure for mathematical formulation of a linear programming problem consists of the following steps.

**Mathematical
formulation of a linear programming problem:**

The procedure for mathematical formulation of a linear programming problem consists of the following steps.

(i) Identify
the decision variables.

(ii) Identify
the objective function to be maximized or minimized and express it as a
linear function of decision variables.

(iii) Identify
the set of constraint conditions and express them as linear inequalities or
equations in terms of the decision variables.

**Example 10.1**

A
furniture dealer deals only two items viz., tables and chairs. He has to invest
Rs.10,000/- and a space to store atmost 60 pieces. A table cost him Rs.500/–
and a chair Rs.200/–. He can sell all the items that he buys. He is getting a
profit of Rs.50 per table and Rs.15 per chair. Formulate this problem as an
LPP, so as to maximize the profit.

*Solution:*

**(i) Variables: **Let** ***x*_{1}** **and** ***x*_{2}** **denote the number of tables and chairs
respectively.

**(i) Objective function:**

Profit on
*x*_{1 }tables = 50 *x*_{1}

Profit on
*x*_{2 }chairs = 15 *x*_{2}

Total
profit = 50 *x*_{1}+15*x*_{2}

Let *Z* = 50 *x*_{1}+15 *x*_{2}
, which is the objective function.

Since the total profit is to be
maximized, we have to maximize *Z*=50*x*_{1}+15*x*_{2}

**(iii) Constraints:**

The dealer has a space to store
atmost 60 pieces

*x*_{1}+*x*_{2}* *≤ 60

The cost of *x*_{1} tables = Rs.500 *x*_{1}

The cost of *x*_{2} tables = Rs. 200 *x*_{2}

Total cost = 500 *x*_{1} + 200 *x*_{2} ,which cannot be more than 10000

500 *x*_{1} + 200 *x*_{2}
≤ 10000

* 5x*_{1}+ 2*x*_{2} ≤ 100

**(iv) Non-negative restrictions: **Since the
number of tables and chairs cannot be** **negative,
we have *x*_{1} ≥0, *x*_{2} ≥0

Thus, the mathematical
formulation of the LPP is

Maximize *Z*=50 *x*_{1}+15*x*_{2}

Subject to the constrains

*x*_{1}+*x*_{2}* *≤60

5*x*_{1}+2*x*_{2}≤100

*x*_{1},* x*_{2}≥0

**Example 10.2**

A company
is producing three products *P*_{1},
*P*_{2} and *P*_{3}, with profit contribution
of Rs.20,Rs.25 and Rs.15 per unit respectively. The resource requirements per
unit of each of the products and total availability are given below.

Formulate
the above as a linear programming model.

*Solution:*

**(i) Variables: **Let** ***x*_{1}** **,**
***x*_{2}** **and** ***x*_{3}** **be the number of units of products** ***P*_{1},** ***P*_{2}** **and** ***P*_{3}** **to**
**be produced.

**(ii) Objective function: **Profit on** ***x*_{1}** **units of the product** ***P*_{1}** **= 20** ***x*_{1}

Profit on *x*_{2} units of the product *P*_{2} = 25 *x*_{2}

Profit on *x*_{3} units of the product *P*_{3} = 15 *x*_{3}

Total profit = 20 *x*_{1} +25 *x*_{2} +15 *x*_{3}

Since the total profit is to be
maximized, we have to maximize *Z*= 20 *x*_{1}+25 *x*_{2} +15 *x*_{3}

**Constraints:**

6*x*_{1} +3 *x*_{2} +12 *x*_{3}
≤ 200

2 *x*_{1} +5 *x*_{2}
+4 *x*_{3} ≤ 350

*x*_{1}* *+2* x*_{2}* *+* x*_{3}* *≤ 100

Non-negative restrictions: Since
the number of units of the products A,B and C cannot be negative, we have *x*_{1}, *x*_{2}, *x*_{3} ≥ 0

Thus, we have the following
linear programming model.

Maximize *Z* = 20 *x*_{1} +25 *x*_{2} +15 x_{3}

Subject
to

6 *x*_{1} +3 *x*_{2} +12 *x*_{3}≤
200

2*x*_{1}+5*x*_{2}
+4 *x*_{3} ≤ 350

*x*_{1}+2* x*_{2}* *+* x*_{3}≤ 100

*x*_{1},* x*_{2},* x*_{3}* *≥* *0

**Example 10.3**

A
dietician wishes to mix two types of food F_{1} and F_{2} in
such a way that the vitamin contents of the mixture contains atleast 6units of
vitamin A and 9 units of vitamin B. Food F_{1} costs Rs.50 per kg and F_{2}
costs Rs 70 per kg. Food F_{1} contains 4 units per kg of vitamin A and
6 units per kg of vitamin B while food F_{2} contains 5 units per kg of
vitamin A and 3 units per kg of vitamin B. Formulate the above problem as a
linear programming problem to minimize the cost of mixture.

*Solution:*

**(i) Variables: **Let the mixture contains** ***x*_{1}** **kg of food F_{1}** **and** ***x*_{2}** **kg of food F_{2}

**(ii) Objective function:**

cost of *x*_{1} kg of food F_{1} =
50 *x*_{1}

cost of *x*_{2} kg of food F_{2} = 70*x*_{2}

The cost is to be minimized

Therefore minimize Z= 50 *x*_{1}+70*x*_{2}

**(iii)
Constraints:**

We make
the following table from the given data

* 4x*_{1}+5*x*_{2}* *≥6 (since the mixture contains ‘atleast 6’ units of vitamin A ,we
have the* *inequality of the type ≥ )

6*x*_{1}+3*x*_{2}* *≥9(since the mixture contains ‘atleast 9’ units of vitamin B ,we
have the* *inequality of the type ≥ )

**(iv) Non-negative restrictions:**

Since the number of kgs of
vitamin A and vitamin B are non-negative , we have *x*_{1}, *x*_{2}* *≥0

Thus, we
have the following linear programming model

Minimize
Z = 50 *x*_{1} +70*x*_{2}

subject
to 4 *x*_{1}+5*x*_{2 }≥6

6 *x*_{1}+3*x*_{2 }≥9

and *x*_{1}, *x*_{2}≥0

**Example 10.4**

A soft
drink company has two bottling plants *C*_{1}
and *C*_{2}. Each plant
produces three different soft drinks S_{1},S_{2} and S_{3}.
The production of the two plants in number of bottles per day are:

A market
survey indicates that during the month of April there will be a demand for
24000 bottles of S_{1}, 16000 bottles of S_{2} and 48000
bottles of S_{3}. The operating costs, per day, of running plants C_{1}
and C_{2} are respectively Rs.600 and Rs.400. How many days should the
firm run each plant in April so that the production cost is minimized while
still meeting the market demand? Formulate the above as a linear programming
model.

*Solution:*

(i) **Variables:** Let *x*_{1}
be the number of days required to run plant C_{1} and x_{2} be
the number of days required to run plant C_{2}

Objective
function: Minimize *Z* = 600 *x*_{1} + 400 *x*_{2}

**(ii) Constraints: **3000** ***x*_{1}** **+ 1000** ***x*_{2}** **≥ 24000(since there is a demand of 24000** **bottles of drink A, production should
not be less than 24000)

*1000x*_{1} + 1000 *x*_{2} ≥ 16000

*2000x*_{1}+ 6000 *x*_{2} ≥ 48000

(iii) **Non-negative restrictions: **Since be the
number of days required of a firm are** **non-negative,
we have *x*_{1}, *x*_{2} ≥0

Thus we
have the following LP model.

Minimize
Z = 600 *x*_{1} + 400 *x*_{2}

subject
to 3000 *x*_{1} + 1000 *x*_{2} ≥ 24000

1000 *x*_{1} + 1000 *x*_{2} ≥ 16000

2000 *x*_{1} + 6000 *x*_{2} ≥ 48000

and *x*_{1}, *x*_{2} ≥0

Tags : Linear programming problem , 11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research

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11th Business Mathematics and Statistics(EMS) : Chapter 10 : Operations Research : Mathematical formulation of a linear programming problem | Linear programming problem

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