In a three-channel queuing system, the arrival rate is 1 unit per time period and the service rate is 0.6 unit per time period. Determaine the probability that an arriving unit does not have to wait.
To determine the probability that an arriving unit does not have to wait in a three-channel queuing system, we can use the concept of the M/M/C queuing model. In this model, "M" represents a memory-less arrival process (Poisson process), "M" represents a memory-less service time distribution (exponential distribution), and "C" represents the number of channels or servers in the system.
The formula to calculate the probability "P0" that an arriving unit does not have to wait is given as:
P0 = ((λ/μ)^C) / (C! * Σ((λ/μ)^n/n!, n=0 to C))
Where:
- λ is the arrival rate (1 unit per time period).
- μ is the service rate (0.6 units per time period).
- C is the number of channels or servers in the system.
Let's calculate the probability using this formula:
C = 3 (Three-channel queuing system)
λ = 1 unit per time period
μ = 0.6 units per time period
Using the formula:
P0 = ((λ/μ)^C) / (C! * Σ((λ/μ)^n/n!, n=0 to C))
= ((1/0.6)^3) / (3! * Σ((1/0.6)^n/n!, n=0 to 3))
Calculating further:
P0 = (1.667^3) / (6 * (1 + (1/0.6) + (1/0.6^2) + (1/0.6^3)))
Using a calculator, we can compute:
P0 ≈ 0.234
Therefore, the probability that an arriving unit does not have to wait in a three-channel queuing system is approximately 0.234 or 23.4%.