{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

finalF03Sols

finalF03Sols - IEOR 6711 Stochastic Models I Fall 2003...

This preview shows pages 1–3. Sign up to view the full content.

IEOR 6711: Stochastic Models I, Fall 2003, Professor Whitt Solutions to Final Exam: Thursday, December 18. Below are six questions with several parts. Do as much as you can. Show your work. 1. Two-Pump Gas Station (12 points) Potential customers arrive at a full-service, two-pump gas station according to a Poisson process at a rate of 40 cars per hour. There are two service attendants to help customers, one for each pump. If the two pumps are busy, then arriving customers wait in a single queue, to be served in the order of arrival by the first available pump. However, customers will not enter the station to wait if there are already two customers waiting, in addition to the two in service. Suppose that the amount of time required to service a car is exponentially distributed with a mean of three minutes. ———————————————————————– This problem is a minor variant of Problem 15 in Chapter 6, on page 393, of the easier Ross book, Introduction to Probability Models . To answer the questions posed here, we need to find the steady-state distribution of the CTMC. In general, the steady-state distribution can be found by solving the equation αQ = 0, where α is the steady-state probability vector. This CTMC is a birth-and-death process, so we can find the steady-state solution directly, by solving the local balance equations α k λ k = α k +1 μ k +1 for all k . But, first, we need to read carefully to make sure we get the model right. We need to recognize that there are five states: k for 0 k 4. We thus can write down the steady-state probabilities in terms of an unknown x , and then normalize to find x : α 0 = 1 * x, α 1 = λ 0 μ 1 * x, α 2 = λ 0 μ 1 λ 1 μ 2 * x , α 3 = λ 0 μ 1 λ 1 μ 2 λ 2 μ 3 * x , α 4 = λ 0 μ 1 λ 1 μ 2 λ 2 μ 3 λ 3 μ 4 * x . A main issue is to be careful and get the rates right, using the same units. We could either use cars per hour or cars per minute, but we should be consistent. We should realize that a mean service time of 3 minutes means one service per 3 minutes or 20 services per hour. So, here, inserting the birth rates and death rates, we get α 0 = 1 * x, α 1 = 40 20 * x, α 2 = 40 20 40 40 * x , α 3 = 40 20 40 40 40 40 * x , α 4 = 40 20 40 40 40 40 40 40 * x or α 0 = 1 * x, α 1 = 2 * x, α 2 = 2 * x , α 3 = 2 * x , α 4 = 2 * x , so that α 0 = 1 / 9 , α 1 = α 2 = α 3 = α 4 = 2 / 9 . ———————————————————————–

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
(a) In the long-run, what rate are cars served? ———————————————————————– Cars are served at rate 20 * α 1 + 40 * ( α 2 + α 3 + α 4 ) = 280 / 9 = 31 . 111 . . . per hour . Alternatively, we could multiply the arrival rate times 1 - α 4 getting 40 × (7 / 9) = 280 / 9, as above. ———————————————————————– (b) In the long-run, what proportion of potential customers fail to be served?
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 11

finalF03Sols - IEOR 6711 Stochastic Models I Fall 2003...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online