91Ó°ÊÓ

Average time utilization $\left(\mathrm{P}\right)=\frac{\mathrm{λ}}{\mathrm{μ}}$Where,

P = Average time utilization

μ = Rate of service delivery

λ = Rate of students' arrival

Now, putting these values in the equation for obtaining the following results:

$P=\frac{4}{6}$= 0.66 or 66.67% of the time

Hence, the average time utilization of the clerk is 0.667 or 66.67% of the time.

${\mathrm{L}}_{\mathrm{q}}=\frac{{\mathrm{λ}}^2}{\mathrm{μ}×\left(\mathrm{μ}-\mathrm{λ}\right)}$

Where

$\mathrm{μ}$=Rate of service delivery

$\mathrm{λ}$= Rate of students' arrival

Now, substitute the above values in the equation for obtaining the following results:

$$\begin{gathered} {\mathrm{L}}_{\mathrm{q}}=\frac{\left(4^2\right)}{6×\left(6-4\right)} \\ =1.333\mathrm{students} \end{gathered}$$

Hence, the average number of students in the system, excluding the graduate student is1.333

Average time in the system $\left({\mathrm{W}}_{\mathrm{s}}\right)$= $\frac{{\mathrm{L}}_{\mathrm{s}}}{\mathrm{λ}}$

$=\frac{{\displaystyle \frac{\mathrm{λ}}{\mathrm{μ}-\mathrm{λ}}}}{\mathrm{λ}}$

Here, $\mathrm{μ}$= 6 and $\mathrm{λ}$= 4 Substitute the above values in the equation for obtaining the following result:

role="math" localid="1650877079217" $$\begin{gathered} {\mathrm{W}}_{\mathrm{s}}=\frac{{\displaystyle \frac{4}{\left(6-4\right)}}}{4} \\ =0.5\mathrm{hour}\mathrm{or}30\min . \end{gathered}$$

Hence, the average time spent in the system is 30 min.

The probability of four or more students in the system is

P = (1 probability of fewer than 4 students in the system)

$\mathrm{P}=1-\left({\mathrm{P}}_0+{\mathrm{P}}_1+{\mathrm{P}}_2+{\mathrm{P}}_3\right)$

${\mathrm{P}}_{\mathrm{n}}$(Probability of n systems in the system) =$\left(1-\frac{\mathrm{λ}}{\mathrm{μ}}\right){\left(\frac{\mathrm{λ}}{\mathrm{μ}}\right)}^n$

μ = Rate of service delivery

2 = Rate of customers' arrival

Therefore, at

n =0,

$$\begin{gathered} {\mathrm{P}}_0=0.3333 \\ {\mathrm{P}}_1=0.2222 \\ {\mathrm{P}}_2=0.1481 \end{gathered}$$

At n= 3, ${\mathrm{P}}_3$=0.0987

Now, substitute the above values in equation (4) to obtain the following results:

P=1-(0.3333+0.2222+0.1481+0.0987)

P=0.1975

Hence, the probability of four or more students in the system is

${\mathrm{L}}_{\mathrm{S}}=\frac{\mathrm{λ}}{\mathrm{μ}-\mathrm{λ}}$

Where,

${\mathrm{L}}_{\mathrm{s}}$= Average line length

$\mathrm{λ}$= Rate of students' arrival

$\mathrm{μ}$= Rate of service delivery

Here? = 6 per hour and μ = 6 per hour now, substitute the above values in the equation to obtain the following results:

$$\begin{gathered} {\mathrm{L}}_{\mathrm{S}}=\frac{6}{6-6} \\ =\frac{6}{0} \\ =\left(∞\right) \end{gathered}$$

Hence, it can be concluded that at? = 6 per hr. the arrival rate would be equivalent to the service rate.