Question

Consider a single server queuing model with Poisson arrivals $(\lambda =4/hour)$ and exponential service $(\mu =4/hour.)$ The number in the system is restricted to a maximum of 10. The probability that a person who comes in leaves without joining the queue is

• A. 1/2
• B. 1/10
• C. 1/11
• D. 1/9

Answer 1

The correct option is C 1/11

$\text{Given data:}$

$\lambda =4\text{per hour,}\mu =4\text{per hour}$

$\text{Traffic intensity,}\rho =\frac{\lambda }{\mu }=\frac{4}{4}=1$

We know that,

${\sum }_{n=0}^{10}{P}_n=1$

$\therefore P_0+P_1+P_3+....+P_{10}=1$

$P_0+\rho P_0+{\rho }^2{P}_0+{\rho }^3{P}_0+...+{\rho }^{10}{P}_0=1$

$P_0(1+\rho +{\rho }^2+...+{\rho }^{10})=1$

$P_0(1+1+...+1)=1$

$\therefore P_0=\frac{1}{11}$

Probability that a person who comes in leaves without joining the queue i.e.

$P_{11}={\rho }^{11}{P}_0=(1{)}^{11}\times \frac{1}{11}=\frac{1}{11}$