Tag: linear programming problem

Questions Related to linear programming problem

In linear programming context, sensitivity analysis is a technique to

  1. Allocate resources optimally.

  2. Minimize cost of operations.

  3. Spell out relation between primal and dual.

  4. Determine how optimal solution to LPP changes in response to problem inputs.


Correct Option: D
Explanation:

A sensitivity analysis is performed to determine the sensitivity of the solution to changes in parameters.
Option D is correct.

Choose the wrong statement:

  1. In order that dual to an LPP may be written, it is necessary that it has at least as many constraints as the number of variables.

  2. The dual represents an alternate formulation of LPP with decision variables being implicit values.

  3. The optimal values of the dual variables can be obtained by inspecting the optimal tableau of the primal problem as well.

  4. Sensitivity analysis is carried out having reference to the optimal tableau alone.


Correct Option: A
Explanation:

In order to write LPP, it is not necessary that it has at least as many constraints as the number of variables.

The number of constraints allowed in a linear program is which of the following?

  1. Less than 5

  2. Less than 72

  3. Less than 512

  4. Less than 1,024

  5. Unlimited


Correct Option: E
Explanation:

there is no limit on constraints allowed in linear programming.
so the number of constraints is unlimited.

Which of the following is an essential condition in a situation for linear programming to be useful?

  1. Linear constraints

  2. Bottlenecks in the objective function

  3. Non-homogeneity

  4. Uncertainty

  5. None of the above


Correct Option: A
Explanation:

For linear programming, the constraints must be linear.

Choose the most correct of the following statements relating to primal-dual linear programming problems:

  1. Shadow prices of resources in the primal are optimal values of the dual variables.

  2. The optimal values of the objective functions of primal and dual are the same.

  3. If the primal problem has unbounded solution, the dual problem would have infeasibility.

  4. All of the above.


Correct Option: D
Explanation:

From the primal-dual relationship,

The shadow prices of resources in the primal are optimal values of the dual variables.

If one of the problems has an optimal feasible solution then the other problem also has an optimal feasible solution. The optimal objective function value is same for both primal and dual problems.

If one problem has an unbounded solution then the other problem is infeasible.

Apply linear programming to this problem. A firm wants to determine how many units of each of two products (products D and E) they should produce to make the most money. The profit in the manufacture of a unit of product D is $100 and the profit in the manufacture of a unit of product E is $87. The firm is limited by its total available labor hours and total available machine hours. The total labor hours per week are 4,000. Product D takes 5 hours per unit of labor and product E takes 7 hours per unit. The total machine hours are 5,000 per week. Product D takes 9 hours per unit of machine time and product E takes 3 hours per unit. Which of the following is one of the constraints for this linear program?

  1. $5 D + 7 E≤ 5,000$

  2. $9 D + 3 E ≥4,000$

  3. $5 D + 7 E = 4,000$

  4. $5 D + 9 E ≤5,000$

  5. $9 D + 3 E ≤5,000$


Correct Option: E
Explanation:

Given, product D takes 5 hours per unit of labour, and
product E takes 7 hours per unit of labour.
Therefore, to produce D units of product D takes $5D$ hours and
to produce E units of product E takes $7E$ hours 
Given, total labour hours per week are $4000$ hours.
Hence, $5D+7E\leq 4000$

Given, product D takes 9 hours per unit of machine time, and
product E takes 3 hours per unit of machine time.
Therefore, to produce D units of product D takes $9D$ hours and
to produce E units of product E takes $3E$ hours 
Given, total machine hours per week are $5000$ hours.
Hence, $9D+3E\leq 5000$

To write the dual; it should be ensured that  
I. All the primal variables are non-negative.
II. All the bi values are non-negative.
III. All the constraints are $≤$ type if it is maximization problem and $≥$ type if it is a minimization problem.

  1. I and II

  2. II and III

  3. I and III

  4. I, II and III


Correct Option: C
Explanation:

To write the dual, then all the primal variables must be non-negative.

All the constraints are $\leq$ type if it ia maximization problem and $\geq$ type if it is a minimization problem.

If $x=\log _{2^2}2+\log _{2^3}2^2+\log _{2^4}2^3......+\log _{2^{n+1}}2^n+$, then the minimum value of $x$ will be-

  1. $\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  2. $n\left(\dfrac 1{n+1}\right)^{\tfrac 1n}$

  3. $\left(\dfrac n{n+1}\right)^{\tfrac 1n}$

  4. None of the above.


Correct Option: B
Explanation:
given $x={log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}$

we know that $A.M. \geq G.M.$

$\implies \dfrac{log _{2^2}2+log _{2^3}{2^2}+log _{2^4}{2^3}+...+log _{2^{n+1}}2^n}{n}\geq \sqrt[n]{log _{2^2}2*log _{2^3}{2^2}*log _{2^4}{2^3}*...*log _{2^{n+1}}{2^n}}$

$\implies \dfrac{x}{n}\geq \sqrt[n]{\dfrac{log2}{log2^2}*\dfrac{log2^2}{log2^3}*\dfrac{log2^3}{log2^4}*...*\dfrac{log2^n}{log2^{n+1}}}$

$\implies x\geq n\sqrt[n]{\dfrac{log2}{log2^{n+1}}}$

$\implies x\geq n({\dfrac{log2}{(n+1)log2}})^{\dfrac 1n}$

$\implies x\geq n({\dfrac{1}{n+1}})^{\dfrac 1n}$

therefore the minimum value of $x$ is $ n({\dfrac{1}{n+1}})^{\dfrac 1n}$

If $a,b >0$, $a+b=1$, then the least value of $(1+\dfrac 1a)(1+\dfrac 1b)$, is

  1. $3$

  2. $6$

  3. $9$

  4. $12$


Correct Option: C
Explanation:
Given, $a+b=1$
we know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b}{2}\geq \sqrt{ab}$

$\implies \dfrac{1}{2}\geq \sqrt{ab}$

$\implies \sqrt{ab}\leq \dfrac{1}{2}$

squaring on both sides

$\implies ab \leq \dfrac{1}{4}$  --------------(1)

Similarly

$\implies \dfrac{1+a+1+b}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+(a+b)}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{2+1}{2}\geq \sqrt{(1+a)(1+b)}$

$\implies \dfrac{3}{2}\geq \sqrt{(1+a)(1+b)}$

squaring on both sides

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{4}{9}$  ---------------(2)

multiplying (1) and (2) we get

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{4}*\dfrac{4}{9}$

$\implies \dfrac{ab}{(1+a)(1+b)}\leq \dfrac{1}{9}$

$\implies \dfrac{1}{(1+a)(1+b)}\leq \dfrac{1}{9ab}$

$\implies \dfrac{(1+a)(1+b)}{ab}\geq 9$

$\implies \dfrac{(1+a)}{a}*\dfrac{(1+b)}{b}\geq 9$

$\implies (1+\dfrac{1}{a})(1+\dfrac{1}{b})\geq 9$

Therefore, the minimum value of $ (1+\dfrac{1}{a})(1+\dfrac{1}{b})$ is $ 9$

If $l,m,n$ be three positive roots of the equation $x^3-ax^2+bx+48=0$, then the minimum value of $\dfrac 1l +\dfrac 2m+\dfrac 3n$ is

  1. $1$

  2. $2$

  3. $\dfrac {-3}{2}$

  4. $\dfrac 52$


Correct Option: C
Explanation:

we Know that, $A.M.\geq G.M.$

$\implies \dfrac{a+b+c}{3}\geq \sqrt[3]{abc}$

let $a=\dfrac{1}{l}, b=\dfrac{2}{m}, c=\dfrac{3}{n}$

Therefore,

$\dfrac{1}{3}(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq \sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$


$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1\times2\times3}{lmn})}$

Given, the roots of the polynomial $x^3-ax^2+bx+48=0$ are $l,m,n$
Therefore, the product of the roots $lmn=-(\dfrac{48}{1})=-48$

Substituting $lmn=-48$ in the above equation

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{6}{-48})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(\dfrac{1}{-8})}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times\sqrt[3]{(-\dfrac{1}{2})^3}$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq 3\times(-\dfrac{1}{2})$

$(\dfrac{1}{l}+\dfrac{2}{m}+\dfrac{3}{n})\geq (-\dfrac{3}{2})$

therefore, the minimum value is $-\dfrac{3}{2}$