Computer Networks. Homework 1 Due Date June 11, 2013

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1 Computer Networks Homework 1 Due Date June 11, 2013 Problem 1 Consider a packet-switched network of N nodes connected by the following topologies: Star: one central node (hub) and all other nodes are attached to the hub. Loop: each node connects to two other nodes to form a closed loop. Fully connected: Each node if directly connected to all other nodes 1. For each case above, give the average number of hops between stations. 2. Prove that for any fixed set of node locations, the minimum possible total length of a tree interconnecting all nodes is less than the minimum possible length of any ring interconnecting all nodes. Problem 1 Solution 1. The number of hops is one less than the number of nodes visited. For a star topology, the fixed number of hops is 2. For a loop topology, the furthest distance from a source node is situated at N/2 hops if the number of nodes in the loop is even, and at (N-1)/2 hops if the number of nodes in the loop is odd. o Even Case: Notice that there are 2 nodes located at i hops from the source, for 1<= i <= N/2-1, and only one node, the furthest from the source, located at hops from the source. Therefore, the average number of hops is: 2 (N 1) N 2! 1 i!1 i + N 2 = N 2 4(N 1) o Odd Case: In this case, there are exactly 2 nodes located at i hops from the source, for 1<= i <= (N - 1)/2. Therefore, the average number of hops is: 2 (N 1) N 2! 1 i!1 N + 1 i = 4 For a fully connected topology, each node is directly connected to all other nodes. Therefore, the number of hops is simply Since a tree is defined to be any geometry which interconnects all nodes so that there is a single path between any two nodes but not loops or circuits, a tree can be formed from any ring configuration by deleting one link. Thus, for any ring configuration of a fixed set of node

2 locations, a tree configuration, with shorter total length than the length of the ring, that interconnects all these nodes can be found by removing one link from the original ring. Problem 2 Two musicians located in two different cities are connected by a network that transmits the sound in a cable at the speed of light of 2.3 x 10 8 meters/second. The two musicians want to hear themselves playing music in real-time as if they were 10 meters apart of each other in the same room. Given that the speed of sound is approximately 330 meters/second, find the maximum possible distance between the musicians in order for them to interact in real-time. Problem 2 Solution The first step is to find the delay for the sound when the musicians are 10 meters apart: t 10 = 10/330 = milliseconds The maximum distance is the time required for a real-time 'experience' times the cable speed: Problem 3 d = (2.3 x 10 8 ) x (30.30 x 10-3 ) = 6,969,000 meters = 6969 kilometers The propagation delay is the time that is required for the energy of a signal to propagate from one point to another. a. Find the propagation delay for a signal traversing the following networks at the speed of light in cable (2.3 x 10 8 meters/second): A circuit board 10 cm A room 10 m A building 100 m A metropolitan area 100 km A continent 5000 km Up and down to a geostationary satellite 2 x km Solution a. To find the propagation delay, divide distance by the speed of light in cable. Thus we have: a circuit board t prop = x seconds a room t prop = x 10-8 seconds a building t prop = x 10-7 seconds a metropolitan area t prop = x 10-4 seconds a continent t prop = seconds up and down to a geostationary satellite t prop = seconds b. How many bits are in transit during the propagation delay in the above cases, if bits are entering the above networks at the following transmission speeds: 10,000 bits/second; 1 megabit/second; 100 megabits/second; 10 gigabits/second?

3 Solution The number of bits in transit is obtained by multiplying the transmission rate R by the propagation delay: Distance (m) 10 Kbps Gbps x x x x x x x10 9 Problem 4 Consider the following networks and assume that a signal traversing the following networks at the speed of light in cable (2.3 x 10 8 meters/second): A circuit board 10 cm A room 10 m A building 100 m A metropolitan area 100 km A continent 5000 km Up and down to a geostationary satellite 2 x km How long does it take to send an L-byte file and to receive a 1-byte acknowledgment back? Let L=1, 10 3, 10 6, and 10 9 bytes? Solution The total time required to send a file and receive an acknowledgment of its receipt is given by: t total = L message /R + L ack /R +2* t prop = L message /R + L ack /R + 2*d/c, where (1) L message is the message length in bits, L ack is the acknowledgment length in bits, R is the transmission bit rate, d is the distance traversed, and The parameter c is the speed of light. Equation (1) shows that there are two main factors that determine total delay: 1. Message and ACK transmission time, which depends on the message length and the transmission bit rate; 2. Propagation delay, which depends solely on distance. When the propagation delay is small, message and ACK transmission times determine the total delay. On the other hand, when the bit rate becomes very large, the propagation delay provides a delay component that cannot be reduced no matter how fast the transmission rate becomes. The tables below show the two main components of the total delay in microseconds.

4 Distance Table 1 Message Length = 1 Byte Distance Table 2 Message Length = 1000 Bytes

5 8.00E E Distance E E E E E E E E E E E E Table 3 Message Length = 10 6 Bytes Distance E E E E E E E E E E E E+06 Table 4 Message Length = 10 9 Bytes

6 Problem 5 Assume that writing a program of N lines of code costs N 2 units. If the program is written in a modular form, however, a module of m lines costs a 2 + m 2 to develop. The cost a 2 is caused by the need to follow the specifications of the modular construction. 1. Compute the number of modules, k, which minimizes the cost of writing a program of N lines assuming that each module has N/k lines? 2. For what value of N is it preferable to decompose a program of N lines into modules rather than to write it as a single program? Solution 5 1. Assume that the program can be decomposed into k modules where the size of each module N/k lines. The cost of writing these modules is: C k = k a 2 + N K 2 = ka 2 + N2 k 2. Note that the function C(k) increases to infinity for k going to zero and for k going to infinity. Thus, we expect that there is a minimum value. To find the value of k that corresponds to that minimum, we calculate the derivative of C(k) with respect to k and we set it to zero. This gives: dc(k) dk = a2 - N2 = 0 k2 k = N a C k = N a a2 + N2 N a = 2Na We can find the values of N such as C * < N 2 (i.e, it is preferable to decompose the program into modules). This is the case when 2Na < N 2, i.e., N > 2a.

NETWORK PROBLEM SET Due Date 6/28/2014

NETWORK PROBLEM SET Due Date 6/28/2014 Problem 1 Consider a packet-switched network of N nodes connected by the following topologies: Star: one central node (hub) and all other nodes are attached to the