Computer Networking: A Top-Down Approach, 7th Edition
计算机网络自顶向下 第七版
Solutions to Review Questions and Problems
Chapter 6 Review Questions
1. The transportation mode, e.g., car, bus, train, car.
2. Although each link guarantees that an IP datagram sent over the link will be received at the other end of the link without errors, it is not guaranteed that IP datagrams will arrive at the ultimate destination in the proper order. With IP, datagrams in the same TCP connection can take different routes in the network, and therefore arrive out of order. TCP is still needed to provide the receiving end of the application the byte stream in the correct order. Also, IP can lose packets due to routing loops or equipment failures.
3. Framing: there is also framing in IP and TCP; link access; reliable delivery: there is also reliable delivery in TCP; flow control: there is also flow control in TCP; error detection: there is also error detection in IP and TCP; error correction; full duplex: TCP is also full duplex.
4. There will be a collision in the sense that while a node is transmitting it will start to receive a packet from the other node.
5. Slotted Aloha: 1, 2 and 4 (slotted ALOHA is only partially decentralized, since it requires the clocks in all nodes to be synchronized). Token ring: 1, 2, 3, 4.
6. After the 5th collision, the adapter chooses from {0, 1, 2,…, 31}. The probability that it chooses 4 is 1/32. It waits 204.8 microseconds.
7. In polling, a discussion leader allows only one participant to talk at a time, with each participant getting a chance to talk in a round-robin fashion. For token ring, there isn’t a discussion leader, but there is wine glass that the participants take turns holding. A participant is only allowed to talk if the participant is holding the wine glass.
8. When a node transmits a frame, the node has to wait for the frame to propagate around the entire ring before the node can release the token. Thus, if L/R is small as compared to tprop, then the protocol will be inefficient.
9. 248 MAC addresses; 232 IPv4 addresses; 2128 IPv6 addresses.
10. C’s adapter will process the frames, but the adapter will not pass the datagrams up the protocol stack. If the LAN broadcast address is used, then C’s adapter will both process the frames and pass the datagrams up the protocol stack.
11. An ARP query is sent in a broadcast frame because the querying host does not which adapter address corresponds to the IP address in question. For the response, the sending node knows the adapter address to which the response should be sent, so there is no need to send a broadcast frame (which would have to be processed by all the other nodes on the LAN).
12. No it is not possible. Each LAN has its own distinct set of adapters attached to it, with each adapter having a unique LAN address.
13. The three Ethernet technologies have identical frame structures.
14. 2 (the internal subnet and the external internet)
15. In 802.1Q there is a 12- bit VLAN identifier. Thus 212 = 4,096 VLANs can be supported.
16. We can string the N switches together. The first and last switch would use one port for trunking; the middle N-2 switches would use two ports. So the total number of ports is 2+ 2(N-2) = 2N-2 ports.
Chapter 6 Problems
Problem 1
1 1 1 0 1 0 1 1 0 0 1 0 0 1 0 1 1 0 1 1 1 1 0 0 0
Problem 2
Suppose we begin with the initial two-dimensional parity matrix:
0 0 0 0 1 1 1 1 0 1 0 1 1 0 1 0
With a bit error in row 2, column 3, the parity of row 2 and column 3 is now wrong in the matrix below:
0 0 0 0 1 1 0 1 0 1 0 1 1 0 1 0
Now suppose there is a bit error in row 2, column 2 and column 3. The parity of row 2 is now correct! The parity of columns 2 and 3 is wrong, but we can't detect in which rows the error occurred!
0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0
The above example shows that a double bit error can be detected (if not corrected).
Problem 3
01001100 01101001 + 01101110 01101011 ------------------------------ 10111010 11010100 + 00100000 01001100 ------------------------------ 11011011 00100000 + 01100001 01111001 -----------------------------
00111100 10011010 (overflow, then wrap around) + 01100101 01110010 ------------------------------ 10100010 00001100
The one's complement of the sum is 01011101 11110011
Problem 4
a) To compute the Internet checksum, we add up the values at 16-bit quantities:
00000001 00000010 00000011 00000100 00000101 00000110 00000111 00001000 00001001 00001010 ------------------------- 00011001 00011110
The one's complement of the sum is 11100110 11100001.
b) To compute the Internet checksum, we add up the values at 16-bit quantities:
01000010 01000011 01000100 01000101 01000110 01000111 01001000 01001001 01001010 01001011 ------------------------- 10011111 10100100
The one's complement of the sum is 01100000 01011011
c) To compute the Internet checksum, we add up the values at 16-bit quantities:
01100010 01100011 01100100 01100101 01100110 01100111 01101000 01101001 01101010 01101011
------------------------- 00000000 00000101
The one's complement of the sum is 11111111 11111010.
Problem 5
If we divide 10011 into 1010101010 0000, we get 1011011100, with a remainder of R=0100. Note that, G=10011 is CRC-4-ITU standard.
Problem 6
a) we get 1000110000, with a remainder of R=0000. b) we get 0101010101, with a remainder of R=1111. c) we get 1011010111, with a remainder of R=1001.
Problem 7
a) Without loss of generality, suppose ith bit is flipped, where 0<= i <= d+r-1 and assume that the least significant bit is 0th bit.
A single bit error means that the received data is K=D*2r XOR R + 2i. It is clear that if we divide K by G, then the reminder is not zero. In general, if G contains at least two 1’s, then a single bit error can always be detected.
b) The key insight here is that G can be divided by 11 (binary number), but any number of odd-number of 1’s cannot be divided by 11. Thus, a sequence (not necessarily contiguous) of odd-number bit errors cannot be divided by 11, thus it cannot be divided by G.
Problem 8
a)
N?1E(p)?Np(1?p)
N?1N?2E'(p)?N(1?p)?Np(N?1)(1?p) N?2?N(1?p)((1?p)?p(N?1))
E'(p)?0?p*?1N
b)
计算机网络自顶向下 第七版 第六章答案 - 图文
