chapter5.txt

网络书的答案

Chapter 5 

Review Questions

1．	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.

2．	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.

3．	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.

4．	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．	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.

7．	248 MAC addresses; 232 IPv4 addresses; 2128  IPv6 addresses.

8．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.

9．	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).

11．	The three Ethernet technologies have identical frame structures.

12．	20 million transitions per second.

13．	After the 5th collision, the adapter chooses from {0, 1, 2,…, 31}. The probability that 
it chooses 4 is 1/32. It waits 4 x 512 x 0.1 = 204.8 microseconds.


Problems

1. 

The rightmost column and bottom row are for parity bits 000100011.

1 0 1 0 0
1 0 1 0 0
1 0 1 0 0
1 0 1 1 1
0 0 0 1 1

5. 

a)
   
 
       

 

b)
 
              

Thus
 
6. 

   		          
 
   

 



 

 


8. 

The length of a polling round is
 .

The number of bits transmitted in a polling round is  . The maximum throughput 
therefore is
 

9. 

a), b), c) See figure below.

 


d) 
1.	Forwarding table in   determines that the datagram should be routed to interface 
111.111.111.002.
2.	Host   uses ARP to determine the LAN address for 111.111.111.002, namely 
22-22-22-22-22.
3.	The adapter in   creates and Ethernet packet with Ethernet destination address 
22-22-22-22-22-22.
4.	The first router receives the packet and extracts the datagram. The forwarding 
table in this router indicates that the datagram is to be routed to 122.222.222.003.
5.	The first router then uses ARP to obtain the associated Ethernet address, namely 
55-55-55-55-55-55.
6.	The process continues until the packet has reached Host  .

e)
ARP in   must now determine the LAN address of 111.111.111.002. Host   sends out 
an ARP query packet within a broadcast Ethernet frame. The first router receives the 
query packet and sends to Host   an ARP response packet. This ARP response packet is 
carried by an Ethernet frame with Ethernet destination address 00-00-00-00-00-00. 

. 

10. 

Wait for 51,200 bit times. For 10 Mbps, this wait is

 msec

For 100 Mbps, the wait is 512  sec.

11. 

At     transmits. At  ,   would finish transmitting. In the worst case,   
begins transmitting at time  . At time   's first bit arrives at 
 . Because  ,   aborts before completing the transmission of the packet, as it 
is supposed to do.

Thus   cannot finish transmitting before it detects that  transmitted. This implies that if 
  does not detect the presence of a host, then no other host begins transmitting while   
is transmitting.











12. 


Time,  
 Event
0
A and B begin transmission
225
A and  B detect collision
225+48=273
A and B finish transmitting jam signal
273+225 = 498
B's last bit arrives at A, A detects an idle channel
(for KA=0)
498+96=594
A starts transmitting after A sense idle channel for 
96 bit times


273+512 = 785
B returns to Step2 , B detects an idle channel (for 
KB=1)
594+225=819
A's transmission reaches B
785+96=881

At 819, B detects an not idle channel while it waits 
for 96 bit times (from 785 to 881) before 
retransmition 



Because  's retransmission reaches   before  's scheduled retransmission time,   
refrains from transmitting while   retransmits. Thus   and   do not collide. Thus the 
factor 512 appearing in the exponential backoff algorithm is sufficiently large.

13. 

We want   or, equivalently,  .   m/sec 
and  bits bits/sec sec. Solving for   we obtain   
meters. For the 100 Mbps Ethernet standard, the maximum distance between two hosts is
200 m. 

For transmitting station   to detect whether any other station transmitted during  's 
interval,   must be greater than  m m/sec sec. Because 
 ,   will detect  's signal before the end of its transmission.

15. 

a)

 

                                             

b)
?	At time  , both   and   transmit.
?	At time  ,   detects a collision.
?	At time   last bit of  's aborted transmission arrives at  .
?	At time   first bit of  's retransmission arrives at  .
?	At time    's packet is completely 
delivered at  .

c)  

17.