1N)11N?11N?1NE(p*)?N(1?)?(1?)?1NNN1?N
111lim(1?)?1lim(1?)N?N??NNe N?? Thus
1limE(p*)?N??e
(1?Problem 9
E(p)?Np(1?p)2(N?1)
E'(p)?N(1?p)2(N?2)?Np2(N?1)(1?p)2(N?3)
?N(1?p)2(N?3)((1?p)?p2(N?1))
E'(p)?0?p*?1 2N?1
E(p*)?N1(1?)2(N?1) 2N?12N?1111 ??2e2e
N??limE(p*)?
Problem 10
a) A’s average throughput is given by pA(1-pB). Total efficiency is pA(1-pB) + pB(1-pA).
b) A’s throughput is pA(1-pB)=2pB(1-pB)= 2pB- 2(pB)2. B’s throughput is pB(1-pA)=pB(1-2pB)= pB- 2(pB)2. Clearly, A’s throughput is not twice as large as B’s.
In order to make pA(1-pB)= 2 pB(1-pA), we need that pA= 2 – (pA / pB).
c) A’s throughput is 2p(1-p)N-1, and any other node has throughput p(1-p)N-2(1-2p).
Problem 11
a) (1 – p(A))4 p(A)
where, p(A) = probability that A succeeds in a slot p(A) = p(A transmits and B does not and C does not and D does not)
= p(A transmits) p(B does not transmit) p(C does not transmit) p(D does not transmit) = p(1 – p) (1 – p)(1-p) = p(1 – p)3
Hence, p(A succeeds for first time in slot 5)
= (1 – p(A))4 p(A) = (1 – p(1 – p)3)4 p(1 – p)3
b) p(A succeeds in slot 4) = p(1-p)3 p(B succeeds in slot 4) = p(1-p)3 p(C succeeds in slot 4) = p(1-p)3 p(D succeeds in slot 4) = p(1-p)3
p(either A or B or C or D succeeds in slot 4) = 4 p(1-p)3 (because these events are mutually exclusive)
c) p(some node succeeds in a slot) = 4 p(1-p)3 p(no node succeeds in a slot) = 1 - 4 p(1-p)3
Hence, p(first success occurs in slot 3) = p(no node succeeds in first 2 slots) p(some node succeeds in 3rd slot) = (1 - 4 p(1-p)3)2 4 p(1-p)3
d) efficiency = p(success in a slot) =4 p(1-p)3
Problem 12
Problem 13
The length of a polling round is
N(Q/R?dpoll).
The number of bits transmitted in a polling round is NQ. The maximum throughput therefore is
NQ?N(Q/R?dpoll)
1?R dpollRQProblem 14
a), b) See figure below. E C 192.168.1.001 A 192.168.2.001 192.168.3.001 00-00-00-00-00-00 44-44-44-44-44-44 77-77-77-77-77-77 Router 1 Router 2 LALALA N N N 192.168.2.002 192.168.2.003 192.168.1.002 192.168.3.002 55-55-55-55-22-22-22-22-22-22 33-33-33-33-33-33 88-88-88-88-88-88 192.168.2.004 192.168.1.003 B D 66-66-66-66-192.168.3.003 F 11-11-11-11-11-11 99-99-99-99-99-99 c)
1. Forwarding table in E determines that the datagram should be routed to interface 192.168.3.002.
2. The adapter in E creates and Ethernet packet with Ethernet destination address 88-88-88-88-88-88.
3. Router 2 receives the packet and extracts the datagram. The forwarding table in this router indicates that the datagram is to be routed to 198.162.2.002.
4. Router 2 then sends the Ethernet packet with the destination address of 33-33-33-33-33-33 and source address of 55-55-55-55-55-55 via its interface with IP address of 198.162.2.003.
5. The process continues until the packet has reached Host B.
a) ARP in E must now determine the MAC address of 198.162.3.002. Host E sends out an ARP query packet within a broadcast Ethernet frame. Router 2 receives the query packet and sends to Host E an ARP response packet. This ARP response packet is carried by an Ethernet frame with Ethernet destination address 77-77-77-77-77-77.
Problem 15
a) No. E can check the subnet prefix of Host F’s IP address, and then learn that F is on the same LAN. Thus, E will not send the packet to the default router R1. Ethernet frame from E to F: Source IP = E’s IP address Destination IP = F’s IP address Source MAC = E’s MAC address
Destination MAC = F’s MAC address
b) No, because they are not on the same LAN. E can find this out by checking B’s IP address.
Ethernet frame from E to R1: Source IP = E’s IP address Destination IP = B’s IP address Source MAC = E’s MAC address
Destination MAC = The MAC address of R1’s interface connecting to Subnet 3.
c) Switch S1 will broadcast the Ethernet frame via both its interfaces as the received ARP frame’s destination address is a broadcast address. And it learns that A
resides on Subnet 1 which is connected to S1 at the interface connecting to Subnet 1. And, S1 will update its forwarding table to include an entry for Host A.
Yes, router R1 also receives this ARP request message, but R1 won’t forward the message to Subnet 3.
B won’t send ARP query message asking for A’s MAC address, as this address can be obtained from A’s query message.
计算机网络自顶向下 第七版 第六章答案 - 图文
