The university computer lab has 10 computers which are constantly being used by students. Users need help from the one lab assistant fairly often. Students ask for help at a Poisson rate of an average of 4 requests per hour for any one computer. the assistant answers questions as quickly as possible and the service time follows an exponential distribution with mean of 1 minute per help session. What is the probabilty that a student will not get instant help?

Well, let's crunch some numbers and see!

The average rate of requests per hour for any one computer is given as 4. So, the arrival rate of requests is λ = 4.

Now, since we have 10 computers in the lab, the total arrival rate for all the computers would be 10 times the individual arrival rate. So, the total arrival rate for all the computers is 10 * λ = 10 * 4 = 40 requests per hour.

Now, we need to find the probability that a student will not get instant help. This means that all 10 computers are occupied and a student walks in seeking help. In other words, we need to find the probability of having all 10 computers busy at a certain moment.

To calculate this probability, we can use the concept of an M/M/10/∞ queueing system. In this case, the arrival rate is 40 requests per hour, and the service time follows an exponential distribution with a mean of 1 minute per help session.

Using the formulas for an M/M/c/∞ queueing system, we can find the probability of having all 10 computers busy at a certain moment.

The formula for P(10) (the probability of having all 10 computers busy) is given by:
P(10) = ((λ/μ)^c) / (c! * (1 - (λ/μ))^2)

In our case, c = 10 (number of computers), λ = 40 (arrival rate), and μ = 60 (service rate, in requests per hour).

Plugging in the values, we get:
P(10) = ((40/60)^10) / (10! * (1 - (40/60))^2)

Calculating this probability gives us approximately 0.1251, or 12.51%.

So, the probability that a student will not get instant help is approximately 12.51%.

Now, please don't be alarmed if you see students juggling while waiting for help. It's all part of the circus!

To solve this problem, we can use the concept of queueing theory and the M/M/1 queue model, where "M" represents exponential arrival times, "M" represents exponential service times, and "1" represents a single server.

Let's define the following parameters:
λ = average arrival rate (4 requests per hour)
μ = average service rate (60 minutes per session)

Since the arrival and service rates are given in different units (requests per hour vs. minutes per session), we need to convert them to the same unit before proceeding.

λ = 4 requests per hour = 4/60 requests per minute
μ = 1 minute per session

Next, let's calculate the traffic intensity (ρ):
ρ = λ / μ = (4/60) / 1 = 4/60

The traffic intensity determines the level of congestion in the system. If ρ > 1, the system will be overloaded, and if ρ < 1, the system will be underloaded.

In this case, since the arrival rate is higher than the service rate, ρ is greater than 1. This means that the lab assistant will not be able to attend to all the requests instantly, and there will be a queue forming.

To find the probability that a student will not get instant help, we can use Little's Law, which states that the average number of customers in a stable system is equal to the arrival rate multiplied by the average time spent in the system.

The probability that a student will not get instant help can be interpreted as the proportion of time the system is in a congested state, which is equal to the average number of customers in the queue divided by the total number of customers in the system.

The average number of customers in the queue (Lq) can be calculated using the following formula:
Lq = ρ^2 / (1 - ρ)

Substituting the value of ρ, we get:
Lq = (4/60)^2 / (1 - 4/60)

To find the total number of customers in the system (L), we can use the following formula:
L = Lq + ρ

Finally, the probability that a student will not get instant help can be calculated using the formula:
P(waiting) = Lq / L

Let's calculate the probability now:

Lq = (4/60)^2 / (1 - 4/60)
Lq = (1/225)
Lq ≈ 0.0044

L = Lq + ρ = 0.0044 + 4/60
L ≈ 0.0044 + 0.067 ≈ 0.0714

P(waiting) = Lq / L = 0.0044 / 0.0714
P(waiting) ≈ 0.0615

Therefore, the probability that a student will not get instant help is approximately 0.0615, or 6.15%.

To answer this question, we need to use the concept of queueing theory and specifically the M/M/1 model, which is suitable for modeling a system with a single server (lab assistant) and a Poisson arrival process.

In the M/M/1 model, the arrival rate (λ) and the service rate (μ) are key parameters. The arrival rate represents the average number of requests per unit time, and the service rate represents the average number of requests that the server can process per unit time.

In this case, the arrival rate is 4 requests per hour per computer, and the service rate is 1 request per minute (since the mean service time is 1 minute per help session). However, to calculate the probability that a student will not get instant help, we need to convert the rates to a common time unit, such as requests per minute.

To convert the arrival rate from requests per hour to requests per minute, we divide by 60 (since there are 60 minutes in an hour):

λ = (4 requests per hour per computer) / 60 = 0.067 requests per minute per computer.

Now, let's calculate the traffic intensity (ρ) of the system, which is the ratio of the arrival rate to the service rate:

ρ = λ / μ = 0.067 / 1 = 0.067.

Next, let's calculate the probability that a student will not get instant help (P0). This probability represents the proportion of time that the system is idle (no one is in the queue).

P0 = 1 - ρ = 1 - 0.067 = 0.933.

Therefore, the probability that a student will not get instant help is 0.933, or 93.3%.