Uncertain random quadratic bottleneck assignment problem

Abstract

Uncertain random expected value simulation is one of the most important techniques to solve uncertain random optimization problems where random variables coexist with uncertain variables. A general simulation algorithm is designed to estimate the uncertain random expected value. Furthermore, an uncertain random quadratic bottleneck assignment problem is proposed and an uncertain random expected value model is presented. An algorithm is designed to find a lower bound of the uncertain random quadratic bottleneck assignment problem.

Appendix

Example 1

Assume that $\eta _{1}\sim U(1, 2)$ and $\eta _{2}\sim U(2, 4)$ are independent random variables, and assume that $\tau _{1}\sim \L (10, 20)$ and $\tau _{2}\sim \L (3, 6)$ are independent uncertain variables . Then, $\xi =\displaystyle \frac{\eta _{1} \vee \tau _{1}}{\eta _{2}\wedge \tau _{2}}$ is an uncertain random variable. Next, we calculate the expected value of $\xi$.

The inverse uncertainty distribution of $\tau _{1}$ is $\Phi ^{ -1}_{1}(\alpha )= (1-\alpha )10 +\alpha 20$ and the inverse uncertainty distribution of $\tau _{2}$ is $\Phi ^{ -1}_{2}(1-\alpha )=\alpha 3 +(1-\alpha )6$. The probability distributions of $\eta _{1}$ and $\eta _{2}$ are

$$\begin{aligned} \Psi _1(y_1) = \left\{ \begin{array}{ll} 0 &{}\quad if\; y_1 \le 1,\\ {y_1-1} &{}\quad if \; 1 \le y_1 \le 2,\\ 1 &{}\quad if \; y_1 \ge 2, \end{array} \right. \end{aligned}$$

and

$$\begin{aligned} \Psi _2(y_2) = \left\{ \begin{array}{ll} 0 &{}\quad if\; y_2 \le 2,\\ \displaystyle {\frac{y_2-2}{2}} &{} \quad if \; 2 \le y_2 \le 4,\\ 1 &{}\quad if \; y_2 \ge 4. \end{array} \right. \end{aligned}$$

The expected value of $\xi$ is formulated as

$$\begin{aligned} \begin{aligned} E[\xi ]&=\frac{1}{2}\int _{2}^{4}\int _1^{2}\int _0^1f(y_1,y_2, \Phi ^{ -1}_{1}(\alpha ),\\&\quad \Phi ^{ -1}_{2}(1-\alpha ))\mathrm{d}\alpha \mathrm{d}\Psi _1(y_1)\mathrm{d}\Psi _2(y_2)\\&=\frac{1}{2}\int _{2}^{4}\int _1^2\int _0^1 \frac{y_1\vee (10+10\alpha ) }{ y_2\wedge (6-3\alpha )}\mathrm{d}\alpha \mathrm{d}y_1 \mathrm{d}y_2.\\ \end{aligned} \end{aligned}$$

For simplicity, we set $N=100$ and $K=100$. Let random variable $\eta _{1}$ take values in $\{y_1|y_1=1+0.01\cdot i, \; \mathrm {for}\;$i$=1, 2,\ldots , 100 \}$ and random variable $\eta _{2}$ take values in $\{y_2|y_2=2+0.02\cdot j, \; \mathrm {for}\;$j=$1, 2,\ldots , 100 \}$. $\alpha$ take values in $\{\alpha |\alpha =0.01\cdot k, \; \mathrm {for}\;$k=$1, 2,\ldots , 100 \}$.

Then, we get

$$\begin{aligned} \begin{aligned} E[\xi ]=\frac{1}{2}\sum \limits _{i=1}^{100}\sum \limits _{j=1}^{100} \sum \limits _{k=1}^{100}\frac{(1+0.01\cdot i)\vee [(10+10(0.01\cdot k)]}{(2+0.02\cdot j)\wedge [6-3(0.01\cdot k)]}\cdot 0.01\cdot 0.01\cdot 0.02.\\ \end{aligned} \end{aligned}$$

The value of $f(y_1,y_2,\Phi ^{ -1}_{1}(\alpha ),\Phi ^{ -1}_{2}(1-\alpha ))$ is listed in the Table 3.

Table 3 Value of $f(y_1,y_2,\Phi ^{ -1}_{1}(\alpha ),\Phi ^{ -1}_{2}(1-\alpha ))$

We approximate the value of the integral using numerical integral technique and get $E(\xi )=5.24$. From this example, we can see that the algorithm is easy to implement.