CS 742 Computer Communication Networks Final Exam - Name ...

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5. (7 pts.) Using the divisor polynomial x 4 + x + 1 for CRC, what frame will be transmitted for the data M = 110011001? ns: M(x) = 110011001, C(x) = 10011, r = 4 1101101 -------------- 10011 )1100110010000 10011 ------ 10101 10011 ------- 11000 10011 ------ 10111 10011 ------- 10000 10011 -------- 1100 So the transmission frame T(x) is 1100110011100 6. (8 pts.) Suppose that 2 n stations use the adaptive tree walk protocol to arbitrate access to a shared channel. At a certain instant 2 stations become ready. What are the minimum and maximum number of bit slots needed to walk the tree if 2 n is much larger than 1? (a) The minimum number happens when two stations are in two different subtrees from the root. It will be 3 bit slots. (b) The maximum number happens when two stations are in the same deepst subtree. Traversing from the root node, there are n collision nodes and n + 1 nodes without collision. So the maximum number is n + n + 1 = 2n + 1. 7. (6 pts.) Suppose a router has built up the routing table as shown in the following table. The router can deliver packets directly over interfaces eth0 and eth1, or it can forward packets to other routers in the table. Destination Netmask Gateway 156.26.10.0 255.255.255.128 eth0 156.26.10.128 255.255.255.128 eth1 156.26.11.0 255.255.255.192 R1 R2 Describe what the router does with a packet addressed to each of the following destinations: (a) 156.26.10.239 - deliver packets directly through eth1 (b) 156.26.11.208 - forward to R2 (c) 156.26.12.27 - forward to R2 8. (9 pts.) For the network 168.78.16.0/20, answer the following questions: (a) How many IP addresses can be allocated? (b) What is the last IP address? (c) What is the netmask? Ans: (a) 2 32−20 = 2 12 = 4096. (b) 16 × 256 = 4096, 16 + 16 − 1 = 31 ⇒ 168.78.31.255. (c) The host ranges from 168.78.16.0 to 168.78.31.255. The first, second, and forth part of the netmask are 255, 255, and 0, respectively. The third part of the netmask 256 − 2 24−20 = 240. Hence, the netmask is 255.255.240.0.
9. (12 pts.) An organization has a class C network 192.168.1 and wants to form subnets for 3 departments, with hosts as follows: A B C 72 hosts 46 hosts 52 hosts (a) There are 170 hosts in all. Give a possible arrangement of network and subnet masks to make this possible. (b) Suggest what the organization might do if the department C grows to 70 hosts. Ans: (a) Department Network Netmask IP Addresses A 192.168.1.0/25 255.255.255.128 128 B 192.168.1.128/26 255.255.255.192 64 C 192.168.1.192/26 255.255.255.224 64 (b) These are possible choices: either assign multiple subnets to single department, abandon subnets and buy a bridge, or let B and C use the same subnet. Here is a possible deployment giving A two subnets, of sizes 64 and 32; every other department gets a sigle subnet of size of the next highest power of 2: Department Network Netmask IP Addresses A 192.168.1.0/26 255.255.255.192 64 A 192.168.1.64/27 255.255.255.224 32 B 192.168.1.96/26 255.255.255.192 64 C 192.168.1.160/27 255.255.255.224 32 C 192.168.1.192/26 255.255.255.192 64 10. (12 pts.) For True or False questions, if it is false, explain why. (a) Each host computer or router can only be assigned one Internet (IP) address. (b) The address 156.26.255.255 is a valid host IP address. (c) The ARP is used to obtain an IP address. (d) For wireless communication, error-detecting can be used to make the transmission reliable. Ans: (a) False, a host with two Ethernet cards should have two IP addresses, also routers always have multiple IP addresses assigned (b) False, the address 156.26.255.255 is used to broadcast a message to all hosts on the network 156.26.0.0. (c) False, the ARP is used to obtain the hardware address. (d) False, error-correcting should be used to make the transmission reliable. 11. (8 pts.) Encrypt the following plaintext using a transposition cipher based on the key LEOPARD. plaintext = H A V E A G R E A T W I N T E R B R E A K L E O P A R D 4 3 5 6 1 7 2 H A V E A G R E A T W I N T E R B R E A K ciphertext = A I E R T K A A R H E E V T B E W R G N A
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Page 3: 3. (9 pts.) Recall the Nyquist’s

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