Question

On a railway station reservation counter, average 10 customers per hour visit for reservation. The probability that a customer should not spend more than 12 minutes in the system is 8%. What will be the service rate?

• A. 22.6 customers/hour
• B. 24 customers/hour
• C. 21.5 customers/hour
• D. 29 customers/hour

Answer 1

The correct option is A 22.6 customers/hour

$P(W_s\ge T)=e^{-T/W_s}$

$0.08=e^{-12/W_s}$

$ln\left(0.08\right)=-\frac{12}{W_s}$

$W_s=-\frac{12}{ln\left(0.08\right)}=\frac{-12}{-2.52573}=\frac{-12}{-2.526}$

$=4.75min$

$\text{We know that,}$

$W_s=\left(\frac{\rho }{1-\rho }\right)\times \frac{1}{\lambda }$

$4.75=\left(\frac{\rho }{1-\rho }\right)\times \frac{1}{1/6}$

$\frac{4.75}{6}=\left(\frac{\rho }{1-\rho }\right)$

$\frac{1}{\rho -1}=\frac{6}{4.75}$

$\frac{1}{\rho }=2.263$

$\rho =0.442$

$\frac{\lambda }{\mu }=0.442$

$\mu =\frac{10}{0.442}=22.62\text{customer/hour}$