Question

Consider a single server queuing model with Poisson arrivals (λ = 4/hour) and exponential service (μ = 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

1. 1/11
2. 1/10
3. 1/9
4. ½

Answer 1

Correct Answer - Option 1 : 1/11

Concept:

Traffic intensity \(ρ \; = \;\frac{\lambda }{\mu}\; \)

\(\mathop \sum \limits_{n\; = \;0}^{n} {P_n}\; = \;1\)

The probability that there are n customer in the system Pn = P₀ × ρⁿ

where, P₀ = probability of zero customer in queue

Calculation:

Given: 

λ = 4 per hour

μ = 4 per hour

Traffic intensity \(ρ \; = \;\frac{\lambda }{\mu}\; = \;\frac{4}{4}\; = \;1\)

We know that,

\(\mathop \sum \limits_{n\; = \;0}^{10} {P_n}\; = \;1\)

∴ P₀ + P₁ + P₃ + … + P₁₀ = 1

P₀ + ρP₀ + ρ²P₀ + ρ³P₀ + … + ρ¹⁰P₀ = 1

P₀(1 + ρ + ρ² + …. + ρ¹⁰) = 1

P₀(1 + 1 + …. + 1) = 1

\({P_0}\; = \;\frac{1}{{11}}\)

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

\({P_{11}}\; = \;{ρ ^{11}}{P_0}\; = \;{\left( 1 \right)^{11}} × \frac{1}{{11}}\; = \;\frac{1}{{11}}\)