Question
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.

Answer

  1. All of the above.

Solution:

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.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

From mean value theorem $f(b) - f(a) = $ $(b - a)f'({x_1});$   $a < {x_1} < b$ if $f(x) = {1 \over x}$, then ${x_1} = $
Let $f(x)=x^5+2 x^3+3 x+1, x \in R$, and $g(x)$ be a function such that $g(f(x))=x$ for all $x \in R$. Then $\frac{g(7)}{g^{\prime}(7)}$ is equal to :
If $y=y(x)$ is the solution of the differential equation $x \frac{d y}{d x}+2 y=x e^{x}, y(1)=0$ then the local maximum value of the function $z(x)=x^{2} y(x)-e^{x}$, $x \in R$ is
If $\text{x}=\text{a}\cos\ \text{nt}-\text{b}\sin\ \text{nt}$ then $\frac{\text{d}^2\text{x}}{\text{dt}^2}$ is:
  1. n2x
  2. -n2x
  3. -nx
  4. nx
The integral $\int\left(\frac{x}{x \sin x+\cos x}\right)^{2} d x$ is equal to

(where $C$ is a constant of integration)

The identity element for the binary operation × defined on Q - {0} as $\text{a}\times\text{b}=\frac{\text{ab}}{2}\ \forall$ a, b $\in$ Q - {0} is:
  1. 1
  2. 0
  3. 2
  4. None of these.
If $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors,then the greatest value of $\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|$ is:
  1. $2$
  2. $2\sqrt{2}$
  3. $4$
  4. $\text{None of these}$
Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three units vectors such that $\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0} .$ If $\lambda=\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \cdot \overrightarrow{\mathrm{a}} $ and $\overrightarrow{\mathrm{d}}=\overrightarrow{\mathrm{a}} \times \overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}}+\overrightarrow{\mathrm{c}} \times \overrightarrow{\mathrm{a}},$ then the ordered pair $(\lambda, {\mathrm{\vec d}})$ is equal to 
The point at which the maximum value of $Z=3 x+2 y$ subject to the constraints $x+2 y \leq 2, x \geq 0, y \geq 0$ is $.....$
Let $g: R \rightarrow R$ be a non constant twice differentiable such that $g^{\prime}\left(\frac{1}{2}\right)=g^{\prime}\left(\frac{3}{2}\right)$. If a real valued function $f$ is defined as $\mathrm{f}(\mathrm{x})=\frac{1}{2}[\mathrm{~g}(\mathrm{x})+\mathrm{g}(2-\mathrm{x})]$, then