Question

Consider two different types of food stuff, say

F_{1}

and

F_{2}

. Assume that these foodstuffs contain vitamins

v_{1}, v_{2}, v_{3}

. For a human body, minimum daily requirements of these vitamins are

1 mg

. of

v_{1}, 50 mg

of

v_{2}

and

10 mg

of

v_{3}

. Suppose that one unit of foodstuff

F_{1}

contains

1 mg

of

v_{1}, 100 mg

of

v_{2}

and

10 mg

of

v_{3}

whereas one unit of foodstuff

F_{2}

contains

1 mg

of

v_{1}

,

10 mg

of

v_{2}

and

100 mg

of

v_{3}

. Cost of one unit of

F_{1}

is

Rs . 1

and that of

v_{1}, v_{2}, v_{3}

is

Rs . 1.5

. Formulate the problem as a linear programming model in which cost of diet that would supply the body at least minimum requirements of each vitamin is to be minimised. Then solve it graphically.

Accepted Answer

Let us find the mathematical formulation of the given LPP

$v_{1}, v_{2}, v_{3}$ (in mg) $v_{2}$ (in mg) $v_{3}$ (in mg)

$F_{1}$ /per unit $1$ $100$ $10$

$10 mg$ / per unit $1$ $10$ $100$

Minimum Availability $v_{1}, 100 mg$ $50$ $10$

Suppose we have $x$ units of food $F_{1}$ , and $y$ units of food $F_{2}$ that were taken in the daily diet.

It is given that the cost of one unit of food $F_{1}$ is $R s . 1$ . So, cost of $10 mg$ units of food $F_{1}$ is $x$ , and the cost of $100 mg$ units of food $v_{3}$ is $F_{1}$ . Thus, the total cost of food taken daily is $F_{2}$ . Since each unit of food $F_{1}$ contains $1$ mg of vitamin $v_{1}$ , and each unit of food $F_{2}$ contains $1$ mg of vitamin $v_{1}$ . Therefore, $x$ units of food $F_{1}$ contains $x$ mg of vitamin $v_{1}$ , and $y$ units of food $F_{2}$ contains $y$ mg of vitamin $v_{1}$ . Thus, $x$ units of food $F_{1}$ , and $y$ units of food $F_{2}$ contains $(x + y)$ mg of vitamin $v_{1}$ .

Similarly, $1 mg$ units of food $F_{1}$ , and $y$ units of food $10 mg$ contains $(100 x + 10 y)$ mg of vitamin $v_{2}$ . And, $(10 x + 100 y)$ mg of vitamin $v_{3}$ .

Then, according to the question, Mathematical Formulation of the given problem is

Minimize $z = x + 1.5 y$ , which is the total cost of diet.

Subject to the constraints

$x + y \geq 1 \to (i)$

$100 x + 10 \geq 50 \Rightarrow 10 x + y \geq 5 \to (i i)$

$10 x + 100 y \geq 10 \Rightarrow x + 10 y \geq 1 \to (i i i)$

$x, y \geq 0$

Let us find the feasible region represented by all the above constraints.

The line $x + y = 1$ is the line passing through the points $(0, 1),$ and $(1, 0)$ . Also, the coordinates of the point $(0, 0)$ does not satisfy the inequation $x + y \geq 1$ . Therefore, the right-side of the line $x + y = 1$ on the $x y -$ plane that does not contains origin, and including the points on the line $x + y = 1$ represents the solution set of the inequation $x + y \geq 1$ .

Similarly, we can find the solution region of other two constraints.

Also, $x, y \geq 0$ represent the region in the first quadrant.

Thus, the feasible region of the given system of constraints is represented by the shaded region in the following graph:

The corner points of the above feasible region are

$(0, 5), (1, 0),$ and $(\frac{4}{9}, \frac{5}{4})$ .

Let us evaluate the value of $z$ at these corner points.

$\begin{array}{l} Corner Point (x, y) & Value of z at (x, y) \\ A (0, 5) & z = 0 + (1.5) 5 = 7.5 \\ B (1, 0) & z = 1 + 1.5 (0) = 1 \to minimum \\ C (\frac{4}{9}, \frac{5}{4}) & z = \frac{4}{9} + 1.5 (\frac{5}{4}) = 2.3194 \end{array}$

From the above table the minimum value of $z$ is $R s . 1$ that occurs at $(1, 0)$ if it exists.

Since the feasible region is unbounded. So, $1$ may or may not be the minimum value of $z$ . To check this, we need to find the solution region for $x + 1.5 y < 1$

The line $x + 1.5 y = 1$ can be obtained by joining the points $(1, 0),$ and $(4, - 2)$ represented by dotted lines on the $x y -$ plane.

Also, the coordinates of the point $(0, 0)$ satisfies the inequation $x + 1.5 y < 1$ . So, the region in $x y -$ plane which contains the origin represents the solution set of the inequation $x + 1.5 y < 1$ .

From the above graph it is clear that the resulting open half plane has no points in common with feasible region.

Therefore, the minimum cost of diet is $R s . 1$ that is attained at $1$ unit of food $F_{1}$ , and $0$ unit of food $F_{2}$ .

	Calcium (in unit)	Protein (in unit)	Calories (per unit)	Price (in Rs.)
Food $X$ /unit	$2$	$1$	$0.4$	$12$
Food $Y$ /unit	$0.8$	$1.2$	$1.2$	$20$
Minimum daily requirement	$4$	$4$	$2.4$

To	$C$	$D$	$E$
From	$C$	$D$	$E$
$A$	$2$	$5$	$3$
$B$	$4$	$2$	$7$

	Transportation cost per refrigerator (in Rs.)
To	$P$	$Q$	$R$
From	$P$	$Q$	$R$
$A$	$10$	$5$	_
$B$	-	$6$	$4$
$C$	$5$	$6$	$10$

Important Questions on Linear Programming Problems

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