3. A computer processes two kinds of requests. One is an online query, and the other is a request to run an application program. The queries arrive according to a Poisson process at a rate of 90 per hour. The run requests arrive according to an independent Poisson process at a rate of 10 per hour. Answer the following questions:

a. What is the expectation of the total number of requests (online and run) to arrive between 8:00 am and 10:00 am?

b. If 60 queries arrive between 9:00 am and 9:30 am, what is the expectation of the number of run requests for that same time period?

c. If 25 queries arrive between 10:00 am and 10:30 am, what is the expectation of the total number of queries for 10:00 am to 11:00 am?

d. If 8 run requests arrive between 1:00 pm and 1:30 pm, what is the probability that no requests will arrive between 1:30 pm and 1:40 pm?

e. If the last run request was 20 minutes ago and the last query was 5 minutes ago, what is the probability that no requests (run or query) will occur in the next two minutes?

4. You observed a small business’s telephone calls over a two-hour period. During that time, 22 calls came in. However, during that same period, the time during which all telephone lines were busy was 30 minutes.

a. Determine the maximum-likelihood estimate of the arrival rate. Express your answer in calls per hour, rounded to the nearest tenth of a call.

b. Determine a 90% confidence interval for the true, but unknown, arrival rate, assuming the arrivals have a Poisson distribution. Express the lower and upper bounds.

5. A small parking lot has room for four cars. The cars arrive according to a Poisson process with a mean rate of 7 per hour. If an arriving driver sees the lot is full, he or she will not wait but will park at another, less convenient lot. The time cars spend parked in the lot is an exponential random variable with a mean of 30 minutes.

a. Decide on an appropriate interpretation of N(t) and draw a rate diagram for this lot below. Be sure to clearly identify all rates.

b. Determine the long-run steady state distribution for this queueing system. State your answers rounded to three decimal places. You may use MS Excel or MatLab, but not QTS, for this part. If you do this by hand, show your calculations here. If you use a computer, be sure to state your answer here and include the file with your submission.

c. Determine the average number of cars in the lot, expressed to the nearest hundredth of a car.

d. Determine the expectation of the number of cars that will enter this lot per hour, expressed to the nearest hundredth of a car.

a. The total number of requests (online and run) to arrive between 8:00 am and 10:00 am can be found by using the Poisson distribution formula.

The expected number of queries arriving in 2 hours can be calculated as shown below:E(X) = λ * tE(X) = 90 * 2 = 180E(Y) = λ * tE(Y) = 10 * 2 = 20E(X + Y) = E(X) + E(Y)E(X + Y) = 180 + 20E(X + Y) = 200Therefore, the expectation of the total number of requests to arrive between 8:00 am and 10:00 am is 200.b. If 60 queries arrive between 9:00 am and 9:30 am, the number of run requests follows the Poisson distribution with a rate of λ = 10 per hour. The expected number of run requests can be calculated as shown below:E(Y | X = 60) = λ * tE(Y | X = 60) = 10 * 0.5E(Y | X = 60) = 5.

Therefore, the expectation of the number of run requests for that same time period is 5.c. If 25 queries arrive between 10:00 am and 10:30 am, then the total number of queries arriving between 10:00 am and 11:00 am follows a Poisson distribution with a rate of 90 per hour. The expected number of queries can be calculated as shown below:E(X | X = 25) = λ * tE(X | X = 25) = 90 * 1E(X | X = 25) = 90.

Therefore, the expectation of the total number of queries for 10:00 am to 11:00 am is 90 + 25 = 115.d. If 8 run requests arrive between 1:00 pm and 1:30 pm, then the number of requests arriving in the next 10 minutes follows a Poisson distribution with a rate of λ = 10 / 6 per 10 minutes. The probability that no requests will arrive between 1:30 pm and 1:40 pm can be calculated as shown below:P(X = 0) = e^(-λt) * (λt)^x / x!P(X = 0) = e^(-10/6 * 1/6) * (10/6 * 1/6)^0 / 0!P(X = 0) = 0.824.

Therefore, the probability that no requests will arrive between 1:30 pm and 1:40 pm is 0.824.e. If the last run request was 20 minutes ago and the last query was 5 minutes ago, then the probability that no requests (run or query) will occur in the next two minutes can be calculated as shown below:P(X = 0) = e^(-λt) * (λt)^x / x!P(X = 0) = e^(-(90/60 + 10/60) * 2) * ((90/60 + 10/60) * 2)^0 / 0!P(X = 0) = 0.018Therefore, the probability that no requests (run or query) will occur in the next two minutes is 0.018.

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