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hw6 - Note that K3 will be the key stream for encryption...

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1 Homework Assignment 7 ELEC315 Q1. A CSMA/CD cable LAN has the following parameters: Channel capacity: 5 Mbps Cable length: 1 Km Signal speed: 2x10^8 m/sec Packet length: 600 bits (a) Use the formula given in the class notes, compute the maximum efficiency of the channel? (b) Repeat part (a) if we increase the channel capacity to 30 Mpbs Q2. Two 10 Mbps LAN segments are linked together a bridge and there are 100 stations on each segments. Ignore the collisions in our calculations below. (a) Find the total capacity that can be achieved in the system. (b) Repeat (a) if we divide users into 5 groups and put each group on one of the 5 segments of a 5-port bridge. Q3. Assume all the packet forwarding tables in the following Ethernet switches are empty. (a) The first packet is sent from station X to Station Y. How many packets will be in total from all switches? (b) The second packet is sent from Z to X, how many packets are sent out by Switch 4? Switch 1 Switch 2 Switch 3 Switch 4 X Y Z

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2 (4) The random sequence generator of a stream cipher is given below.
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Unformatted text preview: Note that K3 will be the key stream for encryption. cipher text = (data ) XOR (key). The random sequence generator’s initial state is the XOR result of two four-bit vectors: IV (initial vector) and KV (key vector). That is, Initial State = IV xor KV IV will be selected by the transmitter and will be sent to the receiver with the cipher text as shown below: IV (four bits, unencrypted) + cipher text KV, also a four-bit vector, is the shared key between the sender and the receiver. KV must be kept secrete. Otherwise, the encryption algorithm will be compromised. (a)What is the problem if we only use IV, but not KV to set the initial state? (b) What is the problem if we only use KV to set the initial state? (c) Assume IV= [1110] and at time = 0, KV = [1101]. Determine the cipher text for the following plain text (the rightmost bit is sent in first for encryption at t=0) 1010 0111 D Q D Q D Q D Q XOR K0 K1 K2 K3...
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hw6 - Note that K3 will be the key stream for encryption...

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