Reliability of a flow network. The journal Networksperiodically publishes studies on the reliability of flownetworks. For example, Networks (September 2007) provided applications in mobile ad hoc and sensor networks.Consider a similar network with four activities, called arcs.The probability distribution of the capacity x for each ofthe four arcs is provided in the following table.

Source: Adapted from J. Lin, “On Reliability Evaluation of CapacitatedFlow Network in Terms of Minimal Path sets.” Networks, Vol. 25, No. 3, May 1995 (Table 1).

a. Verify that the properties of discrete probability distributions are satisfied for each arc capacity distribution.

b. Find the probability that the capacity for Arc 1 willexceed 1.

c. Repeat part b for each of the remaining three arcs.

d. Compute for each arc and interpret the results.

e. Compute for each arc and interpret the results.

Properties of discrete probability distribution:

1) for all values of x,$\mathrm{P}\left(x\right)⩾0$

2)$\sum \mathrm{P}\left(x\right)=1$

The probability distribution of the capacity for Arc 1 is given below:

From the given table, it is clear that $\mathrm{P}\left(x\right)⩾0$, for all values of x.

Now

$\begin{array}{rcl}\sum \mathrm{P}\left(x\right)& =& 0.05+0.10+0.25+0.60\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 1.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 2.

The probability distribution of the capacity for Arc 2 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)⩾0$, for all values of x.

Now

$\begin{array}{rcl}\sum \mathrm{P}\left(x\right)& =& 0.10+0.30+0.60+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 2.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 3

The probability distribution of the capacity for Arc 3 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)⩾0$, for all values of x.

Now

$\begin{array}{rcl}\sum \mathrm{P}\left(x\right)& =& 0.05+0.25+0.70+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 3.

Verifying Properties of the discrete probability distribution are satisfied for capacity for arc 4

The probability distribution of the capacity for Arc 4 is given below:

From the table, it is clear that $\mathrm{P}\left(x\right)⩾0$for all values of x.

Now

$\begin{array}{rcl}\sum \mathrm{P}\left(x\right)& =& 0.90+0.10+0+0\\ & =& 1\\ & & \end{array}$

Hence, the properties of the discrete probability distribution are satisfied for capacity for arc 4.

To obtain the probability that the capacity for arc 1 exceeds 1, it means to find$\mathrm{P}\left(x>1\right)$

$\begin{array}{rcl}P\left(x>1\right)& =& 1-\mathrm{P}\left(x⩽1\right)\\ & =& 1-\left(0.05+0.10\right)\\ & =& 1-0.15\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.85$

Hence the probability that the capacity for arc 1 exceeds 1 is 0.85.

To obtain the probability that the capacity for arc 2 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x⩽1\right)\\ & =& 1-\left(0.10+0.30\right)\\ & =& 1-0.40\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.60$

Therefore, the probability that the capacity for arc 2 exceeds 1 is 0.60.

Calculating the probability that the capacity for arc 3 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x⩽1\right)\\ & =& 1-\left(0.05+0.25\right)\\ & =& 1-0.30\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0.70$

Therefore, the probability that the capacity for arc 3 exceeds 1 is 0.70.

Calculating the probability that the capacity for arc 4 exceeds 1.

$\begin{array}{rcl}P\left(x>1\right)& =& 1<\mathrm{P}\left(x⩽1\right)\\ & =& 1-\left(0.90+0.10\right)\\ & =& 1-1\\ & & \end{array}$

$\mathrm{P}\left(x>1\right)=0$

Therefore, the probability that the capacity for arc 4 exceeds 1 is 0

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0\times 0.05+1\times 0.10\times 2\times 0.25+3\times 0.60\\ & =& 2.4\\ & & \end{array}$

Thus, the mean capacity for arc 1 is 2.4.

Calculating the mean for arc 2

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0\times 0.10+1\times 0.30\times 2\times 0.60+3\times 0\\ & =& 1.5\\ & & \end{array}$

Thus the mean capacity for arc 2 is 21.5

Calculating the mean for arc 3

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0\times 0.05+1\times 0.25\times 2\times 0.70+3\times 0\\ & =& 1.65\\ & & \end{array}$

Thus,the mean capacity for arc 3 is$\begin{array}{rcl}& & 1.65\end{array}$

Calculating the mean for arc 3

$\begin{array}{rcl}E\left(x\right)& =& x\mathrm{P}\left(x\right)\\ & =& 0\times 0.90+1\times 0.10\times 2\times 0+3\times 0\\ & =& 0.10\\ & & \end{array}$

Thus,the mean capacity for arc 4 is$\begin{array}{rcl}& & 0.10\end{array}$

${\sigma }^2=E\left[{\left(x-\mu \right)}^2\right]$

${\sigma }^2={\left(x-\mu \right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{\sigma }^2& =& {\left(0-2.4\right)}^{2}0.05+{\left(1-2.4\right)}^{2}0.10+{\left(2-2.4\right)}^{2}0.25+{\left(3-2.4\right)}^{2}0.60\\ & =& 0.288+0.196+0.04+0.216\\ & =& 0.74\\ & & \end{array}$

$\begin{array}{rcl}\sigma & =& \sqrt{0.74}\\ & =& 0.86\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}\mu \pm 2\sigma & =& 2.4\pm 2\times 0.86\\ & =& 2.4\pm 1.72\\ & =& \left(2.4-1.72,2.4+1.72\right)\\ & & \end{array}$

$\mu \pm 2\sigma =\left(0.68,4.12\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${\sigma }^2=E\left[{\left(x-\mu \right)}^2\right]$

${\sigma }^2={\left(x-\mu \right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{\sigma }^2& =& {\left(0-1.5\right)}^{2}0.10+{\left(1-1.5\right)}^{2}0.30+{\left(2-1.5\right)}^{2}0.60+{\left(3-1.5\right)}^{2}0\\ & =& 0.225+0.075+0.15+0\\ & =& 0.45\\ & & \end{array}$

$\begin{array}{rcl}\sigma & =& \sqrt{0.45}\\ & =& 0.67\\ & & \end{array}$

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}\mu \pm 2\sigma & =& 1.5\pm 2\times 0.67\\ & =& 1.5\pm 1.34\\ & =& \left(1.5-1.34,1.5+1.34\right)\\ & & \end{array}$

$\mu \pm 2\sigma =\left(0.16,2.84\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${\sigma }^2=E\left[{\left(x-\mu \right)}^2\right]$

${\sigma }^2={\left(x-\mu \right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{\sigma }^2& =& {\left(0-1.65\right)}^{2}0.05+{\left(1-1.65\right)}^{2}0.25+{\left(2-1.65\right)}^{2}0.70+{\left(3-1.65\right)}^{2}0\\ & =& 0.1316+0.1056+0.0857+0\\ & =& 0.327\\ & & \end{array}$

$\begin{array}{rcl}\sigma & =& \sqrt{0.327}\\ & =& 0.57\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}\mu \pm 2\sigma & =& 1.65\pm 2\times 0.57\\ & =& 1.65\pm 1.14\\ & =& \left(1.65-1.14,1.65+1.14\right)\\ & & \end{array}$

$\mu \pm 2\sigma =\left(0.51,2.79\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean

Calculating the standard deviation for arc 2.

${\sigma }^2=E\left[{\left(x-\mu \right)}^2\right]$

${\sigma }^2={\left(x-\mu \right)}^2\mathrm{P}\left(x\right)$

$\begin{array}{rcl}{\sigma }^2& =& {\left(0-0.1\right)}^{2}0.90+{\left(1-0.1\right)}^{2}0.10+{\left(2-0.1\right)}^{2}0+{\left(3-0.1\right)}^{2}0\\ & =& 0.009+0.081+0+0\\ & =& 0.09\\ & & \end{array}$

$\begin{array}{rcl}\sigma & =& \sqrt{0.327}\\ & =& 0.3\\ & & \end{array}$

Interpretation of result:

The two standard deviations from the mean are calculated as:

$\begin{array}{rcl}\mu \pm 2\sigma & =& 0.1\pm 2\times 0.3\\ & =& 0.1\pm 0.6\\ & =& \left(0.1-0.6,0.1+0.6\right)\\ & & \end{array}$

$\mu \pm 2\sigma =\left(-0.5,0.7\right)$

Hence it can be concluded that most of the observations will fall within two standard deviations from the mean