Question

The sandwich maker of the EE-Cole-Eye Sandwich Truck was just fired (for a reason described below) and EE-Cole-Eye needs to find a replacement. They want to make sure that the replacement is fast enough so that customers don't have to wait too long on line. Customers arrive according to a Poisson process with arrival rate 2 customers/min and sandwich makers take an exponentially distributed length of time to make a sandwich, with mean time service time of 29 seconds. There are no limits to the number of customers that wait for sandwiches. Consider the queue of customers and analyze it assuming it is in steady state.

What is the probability there are 20 customers or fewer in the system?

Answer 1

The probability of having 20 customers or fewer in the system is 0.972.

Step-by-step explanation:

To find the probability of having 20 customers or fewer in the system, we need to calculate the steady-state probability distribution of the number of customers in the system. This can be done using Little's formula, which relates the average number of customers in the system (L), the average arrival rate (λ), and the average service rate (μ):

L = λW

Here, W is the average time a customer spends in the system, which is equal to the sum of the average time spent waiting in line (Wq) and the average service time (Ws):

W = Wq + Ws

Using Little's formula, we can find L:

L = λ(Wq + Ws)

To calculate Wq, we use the formula for the expected value of a queue with Poisson arrivals and exponentially distributed service times:

Wq = (ρ^2 + ρ)/(2(1 - ρ))

Here, ρ is the traffic intensity, which is equal to the product of the arrival rate and the service rate:

ρ = λ/μ

Substituting values, we get:

ρ = (2)/(1/31) = 66.67%

Wq = (0.6667^2 + 0.6667)/(2(1 - 0.6667)) = 0.5333 seconds

To calculate Ws, we use the formula for the expected value of an exponentially distributed random variable:

Ws = 1/μ = 1/0.3438 = 2.9444 seconds

Now that we have all necessary values, we can calculate L:

L = (2)/(1/31) * (0.5333 + 2.9444) = 15.8889 customers

The steady-state probability distribution for this queue can be found using a technique called matrix algebra. However, since we only need to find P{X ≤ 20}, where X is the number of customers in the system, we can use a simpler method called convolution. This involves finding P{X = k} for k = 0, 1, ..., and then calculating P{X ≤ k} by summing up all probabilities up to k:

P{X = k} = P{X_{waiting} = k} * P{X_{service} = k} + P{X_{waiting} = k - 1} * P{X_{service} = k + 1} - P{X_{waiting} = k - 1} * P{X_{service} = k}, where X_{waiting} and X_{service} are random variables representing the number of customers waiting in line and in service, respectively. Using this formula, we can calculate P{X ≤ k}:

P{X ≤ k} = sum(P{X = j}, j = 0..k) = P{X_{waiting} <= k} * P{X_{service} <= k} + P{X_{waiting} <= k - 1} * P{X_{service} <= k + 1} - P{X_{waiting} <= k - 1} * P{X_{service} <= k}, where X_{waiting} and X_{service} are random variables representing the number of customers waiting in line and in service, respectively. Using this formula, we can calculate P{X <= k}:

P{X <= k} = sum(P{X = j}, j = 0..k) = P{X_{waiting} <= k} * P{X_{service} <= k} + P{X_{waiting} <= k - 1} * P{X_{service} <= k + 1} - P{X_{waiting} <= k - 1} * P{X_{service} <= k}, where X_{waiting}, X_{service}, X are random variables representing the number of customers waiting in line, in service, and in total, respectively. Using this formula with our specific parameters (λ=2,μ=0.3438), we can calculate P{X <= k}:

P{X <= k}: [0.5333; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0; 0]' [1-exp(-2*53); exp(-2*53)-exp(-2*54); exp(-2*54)-exp(-2*55); exp(-2*55)-exp(-2*56); exp(-2*56)-exp(-2*57); exp(-2*57)-exp(-2*58); exp(-2*58)-exp(-2*59); exp(-2*59)-exp(-2*60); exp(-2*60)-exp(-2*61); exp(-2*61)-exp(-2*62); exp(-2*62)-exp(-2*63); exp(-2*63)-exp(-2*64); exp(-2*64)-exp(-2*65); exp(-2*65)-exp(-2*66); exp(-2*66)-exp(-2*67); exp(-2*67)-exp(-2*68); exp(-2*68)-exp(-2*69); exp(-