# CS 5480/6480: Computer Networks Spring 2012 Homework 1 Solutions Due by 9:00 AM MT on January 31 st 2012

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1 CS 5480/6480: Computer Networks Spring 2012 Homework 1 Solutions Due by 9:00 AM MT on January 31 st 2012 Important: No cheating will be tolerated. No extension. CS 5480 total points = 32 CS 6480 total points = 44 Question 1 (Packet Switching versus Circuit Switching) 5 points: Suppose 40 users share a 1Mbps link. Also suppose that each user alternates (independently of the other users) between periods of activity, when the user generates data at a constant rate of 100 Kbps and periods of inactivity when the user generates no data. Suppose further that the user is active (independent of the other users) only 10 percent of the time. 1. (3.5 points) When circuit switching is used to allocate resources on the shared link, 10 users can be supported. When packet switching is used, what is the probability that 11 or more users are active simultaneously? (You must show the steps for full credit.) The probability of r users being active simultaneously, P(r) = 40 C r * (0.1) r * (0.9) 40-r The probability <= 10 users being active simultaneously = P(0) + P(1) + + P(10) ~ The probability of 11 or more users being active simultaneously = = (1.5 points) Do you expect the result of part 1 to change when all the 40 users synchronize in their use of the shared link (i.e. all the 40 users generate data or are idle exactly at the same time)? Explain with arguments. No calculations are required for this part of the question. When 40 users are active at the same time none of the users is likely to get the desired 100 Kbps. This example shows that packet switching is useful when users are active independent of each other. However, when they are active in a correlated manner, packet switching does not help and one might need to revert to circuit switching to allow at least 10 users to get the desired bandwidth. Question 2 (Internet Delay Components) 3 points: The figure below shows a router and two links which are part of a bigger network.

6 a. Suppose that u s (u s + u u N )/N. Specify a distribution scheme that has a distribution time of F/u s. b. Suppose that u s (u s + u u N )/N. Specify a distribution scheme that has a distribution time of NF/(u s + u u N ). c. Conclude that the minimum distribution time is in general given by max{f/u s, NF/(u s + u u N )} a. Define u = u 1 + u u N. By assumption u s <= (u s + u)/n Equation 1 Divide the file into N parts, with the i th part having size (u i /u)f. The server transmits the i th part to peer i at rate r i = (u i /u)u s. Note that r 1 + r r N = u s, so that the aggregate server rate does not exceed the link rate of the server. Also have each peer i forward the bits it receives to each of the N-1 peers at rate r i. The aggregate forwarding rate by peer i is (N-1)r i. We have (N-1)r i = (N-1)(u s u i )/u <= u i, where the last inequality follows from Equation 1. Thus the aggregate forwarding rate of peer i is less than its link rate u i. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in F/u s. b. Again define u = u 1 + u u N. By assumption u s >= (u s + u)/n Equation 2 Let r i = u i /(N-1) and r N+1 = (u s u/(n-1))/n In this distribution scheme, the file is broken into N+1 parts. The server sends bits from the i th part to the i th peer (i = 1,., N) at rate r i. Each peer i forwards the bits arriving at rate r i to each of the other N-1 peers. Additionally, the server sends bits from the (N+1) st part at rate r N+1 to each of the N peers. The peers do not forward the bits from the (N+1) st part. The aggregate send rate of the server is r r N + N r N+1 = u/(n-1) + u s u/(n-1) = u s

7 Thus, the server s send rate does not exceed its link rate. The aggregate send rate of peer i is (N-1)r i = u i Thus, each peer s send rate does not exceed its link rate. In this distribution scheme, peer i receives bits at an aggregate rate of Thus each peer receives the file in NF/(u s +u). (For simplicity, we neglected to specify the size of the file part for i = 1,., N+1. We now provide that here. Let Δ = (u s +u)/n be the distribution time. For i = 1,, N, the i th file part is F i = r i Δ bits. The (N+1) st file part is F N+1 = r N+1 Δ bits. It is straightforward to show that F F N+1 = F.) c. We know from section 2.6 that Combining this with (a) and (b) gives the desired result. (c) (2 points) Consider an overlay network with N active peers, with each pair of peers having an active TCP connection. Additionally, suppose that the TCP connections pass through a total of M routers. How many nodes and edges are there in the corresponding overlay network? There are N nodes in the overlay network. There are N(N-1)/2 edges. Question 7 (required for CS6480, extra credit for CS5480) 12 points: Read the following paper Internet Indirection Infrastructure by I. Stoica, D. Adkins, S. Zhuang, S. Shenker and S. Surana, ACM Sigcomm Conference, August ( Answer the following questions that are based on this paper: 1. (3 points) Consider a model in which data sent from two senders is added together before being sent to a receiver. Show and describe how this could be implemented using the i3 architecture.

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