EE444HW1.sln - EE 444 INTRODUCTION TO COMPUTER NETWORKS...

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

View Full Document Right Arrow Icon
EE 444 INTRODUCTION TO COMPUTER NETWORKS 2009 Spring Homework #1 with Answers Deadline: March 13 2009 @ 23:59 (No late submission is allowed) Posted on March 06 2009. Question # 1 Consider a car dealer that is also responsible for repairing the cars they sell in case they fail. There are a total of M cars that have been sold by the dealer and the dealer does not sell new cars any more due to the global economic recession. However, the repair service must be continuously provided. The cars fail after an average of F time units whenever they are repaired. The repair service is provided by the dealer by k repairmen. Each repairman is capable of repairing a broken car in an average of P time units. Assuming that any car does not die out, obtain upper and lower bounds ( in terms of F , M , P and k ) on the number of car failures per unit time and on the average time between repairs of the same car. The bounds must be as tight as possible. A.1. We will use Little’s theorem to come up with the desired bounds. There are M cars in a cycle of three possible stages: 1. working without any problem 2. waiting to be repaired 3. being repaired We define the system as a combination of these three stages. Let T NP , T W , T P be the times spent in the first, second and third stage respectively. Then, the time spent in the system T S is T S = T NP + T W + T P Average of T S is E(T S ) = F + P + E(T W ) Note that average waiting time is not given. We will bound it. Two extreme cases are i. zero cars in the waiting stage ii. M - k cars in the waiting stage and the repair process of k cars has just started. Working w/o problem Repairman 1 Waiting to be repaired Repairman 2 Repairman k λ f
Background image of page 1

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

View Full DocumentRight Arrow Icon
In (i), the waiting time T W is zero. In case (ii), T W is (M/k -1)P as k cars are served at a time.
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/09/2010 for the course EE ee444 taught by Professor Akan during the Spring '10 term at Middle East Technical University.

Page1 / 4

EE444HW1.sln - EE 444 INTRODUCTION TO COMPUTER NETWORKS...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online