AMU MCA Linear Programming Previous Year Questions (PYQs) – Page 1 of 2
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Showing 11 questions
The following LPP
Maximize $z = x_1 + x_2$
Subject to
$x_1 + x_2 \le 1$
$-3x_1 + x_2 \ge 3$
$x_1, x_2 \ge 0$
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Solution
From $-3x_1 + x_2 \ge 3$
$x_2 \ge 3 + 3x_1$
But from $x_1 + x_2 \le 1$
$x_2 \le 1 - x_1$
So we need:
$3 + 3x_1 \le 1 - x_1$
$4x_1 \le -2$
$x_1 \le -\frac{1}{2}$
This contradicts $x_1 \ge 0$
Hence problem is infeasible.
he following LPP:
Maximize $Z = 6x_1 - 2x_2$
Subject to:
$2x_1 - x_2 \le 2$
$x_1 \le 3$
$x_1, x_2 \ge 0$
has optimal value as
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Solution
The following LPP:
Minimize $Z = x - y$
Subject to:
$2x + 3y \le 6$
$0 \le x \le 4$
$0 \le y \le 3$
then the number of extreme points on the feasible region and the number of basic feasible solutions are
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Solution
The minimum value of $z=2x_1+3x_2$ subject to the constraints
$2x_1+7x_2\geq22,\ x_1+x_2\geq6,\ 5x_1+x_2\geq10$ and $x_1,x_2\geq0$ is
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Solution
Optimal value of the following LPP:
Max $z = 2x_1 + 3x_2$
Subject to
$6x_1 + 5x_2 \le 25$
$x_1 + 3x_2 \le 10$
$x_1, x_2 \ge 0$
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Solution
Corner points of feasible region are checked.
Maximum value of $z$ occurs at intersection of
$6x_1 + 5x_2 = 25$ and $x_1 + 3x_2 = 10$.
Substituting gives $x_1 = 2.5$, $x_2 = 3$.
$z = 2(2.5) + 3(3) = 5 + 9 = 13.5$.
Max $z = 6x_1 - x_2$
Subject to
$2x_1 - x_2 \le 2$
$x_1 \le 3$
$x_1, x_2 \ge 0$
The above LPP has:
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Solution
The feasible region allows $z$ to increase indefinitely, hence the objective function is unbounded.
When 40% of a number is added to 42, the result is the number itself. The number is:
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Solution
Let the number $= x$
$0.4x + 42 = x$
$x - 0.4x = 42$
$0.6x = 42$
$x = 70$
The population of a town increases by 5% annually. If it is 15,435 now, its population 2 years ago was:
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Solution
Present population $= 15435$
$P(1.05)^2 = 15435$
$P \times 1.1025 = 15435$
$P = \frac{15435}{1.1025} = 14000$
An $n$-tuple $(x_1,x_2,\dots,x_n)$ which satisfies all the constraints of a linear programming problem and for which the objective function is maximum is called
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Solution
Maximum value satisfying constraints ⇒ optimal solution
Final Answer: $\boxed{\text{optimal solution}}$
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