1. Design and describe an application-level protocol to be used between an automatic teller machine and a bank’s centralized computer. You protocol should allow a user’s card and password to be verified, the account balance (which is maintained at the centralized computer) to be queried, and an account withdrawal to be made (that is, money disbursed to the user). Your protocol entities should be able to handle the
all-too-common case in which there is not enough money in the account to cover the withdrawal. Specify your protocol by listing the messages exchanged and the action taken by the automatic teller machine or the bank’s centralized computer on transmission and receipt of messages. Sketch the operation of your protocol for the case of a simple withdrawal with no errors, using a diagram similar to that in Figure 1.2. Explicitly state the assumption made by your protocol about underlying end-to-end transport service. Answers:
Messages from ATM machine to Server
Msg name purpose
-------- -------
HELO
machine
ATM card transmits user ID to Server
PASSWD
WITHDRAWL
BYE user all done
Messages from Server to ATM machine (display)
Msg name purpose
-------- -------
PASSWD Ask user for PIN (password)
OK last requested operation (PASSWD, WITHDRAWL) OK ERR last requested operation (PASSWD, WITHDRAWL) in
ERROR
AMOUNT
BYE user done, display welcome screen at ATM Correct operation:
client server
HELO (userid) --------------> (check if valid userid)
<------------- PASSWD
PASSWD
<------------- OK (password is OK)
BALANCE -------------->
<------------- AMOUNT
WITHDRAWL
withdrawl
<------------- OK
ATM dispenses $
BYE -------------->
<------------- BYE
In situation when there's not enough money:
HELO (userid) --------------> (check if valid userid)
<------------- PASSWD
PASSWD
<------------- OK (password is OK)
BALANCE -------------->
<------------- AMOUNT
WITHDRAWL
error msg displayed
no $ given out
BYE -------------->
<------------- BYE
Problem 5 ==================================================== Consider sending a packet of F bits over a path of Q links. Each link transmits at R bps. The network is lightly loaded so that there are no queuing delays. Propagation delay is negligible.
a.Suppose the network is a packet-switched virtual circuit network. Denote the VC
set-up time by ts seconds. Suppose the sending layers add a total of h bits of header to each packet. How long does it take to send the file from source to destination?
b. Suppose the network is a packet-switched datagram network and a connectionless service is used. Now suppose each packet has 2h bits of header. How long does it take to send the file?
c. Finally, suppose that the network is a circuit-switched network. Further suppose that the transmission rate of the circuit between source and destination is R bps. Assuming ts set-up time and h bits of header appended to the entire file, how long does it take to send the file?
Answer
a.Time cost = ts+(F+h)*Q/R
b.Time cost = (F+2h)*Q/R
c.Time cost = ts +(F+h)*Q/R
Problem6 ====================================================
This elementary problem begins to explore propagation delay and transmission delay, two central concepts in data networking. Consider two hosts, Host A and B, connected by a single link of rate R bps. Suppose that the two hosts are separated by m meters, and suppose the propagation speed along the link is s meters/sec. Host
A is to send a packet of size L bits to Host B.
a. Express the propagation delay, dprop in terms of m and s.
b. Determine the transmission time of the packet, dtrans in terms of L and R.
c. Ignoring processing and queuing delays, obtain an expression for the end-to-end delay.
d. Suppose Host A begins to transmit the packet at time t = 0. At time t = dtrans, where is the last bit of the packet?
e. Suppose dprop is greater than dtrans. At time t = dtrans, where is the first bit of the packet?
f. Suppose dprop is less than dtrans. At time t = dtrans, where is the first bit of the packet?
g. Suppose s = 2.5*108, L = 100 bits, and R = 28 kbps. Find the distance m so that dprop equals dtrans.
Answer
a.dprop = m/s;
b.dtrans = L/R
c.The end to end delay = dprop + dtrans = m/s +L/R;
d.The last bit has just left Host A;
e.The first bit is on the link and with a distance L*s/R from the host A;
f.The first bit has left the link and arrived at Host B.
g.m = L*s/R= 100bits *2.5*108 /(28kbps)=893km
Problem7====================================================
In this problem we consider sending voice from Host A to Host B over a packet-switched network (for example, Internet phone). Host A converts on-the-fly analog voice to a digital 64- Kbps bit stream. Host A then groups the bits into 48-byte packets. There is one link between host A and B; its transmission rate is 1 Mbps and its propagation delay is 2 mesc. As soon as Host A gathers a packet, it sends it to Host B. As soon as Host B receives an entire packet, it converts the packet’s bits to an analog signal. How much time elapses from the time a bit is created (from the original analog signal at A) until a bit is decoded (as part of the analog signal at B)?
Answer
The processing time on host A = 48Byte/64Kbps=6mesc
The decoding time on host B =The processing time on host A =6mesc
The transmission time = 48Byte/1Mbps= 0.384mesc
Time cost from the time bit is created to the time a bit is decoded =
6mesc+0.384mesc+2mesc = 8.384 mesc
Problem 10 ================================================== Consider the queuing delay in a router buffer (preceding an outbound link). Suppose all packets are L bits, the transmission rate is R bps, and that N packets arrive to the buffer every LN/R seconds. Find the average queuing delay of a packet.
Answer
It takes LN/R seconds to transmit N packets ,and when the last packet is transmitted over ,the buffer is empty ,at the same time another N packets arrive.
The queuing delay for the first packet is 0;
The queuing delay for the second packet is L/R;
The queuing delay for the third packet is 2L/R; ......
The queuing delay for the Nth packet is (N-1)L/R;
Then the average queuing delay of a packet is
(0+L/R+2L/R+......+(N-1)L/R)/N=(N-1)L/(2R)
Problem 14==================================================
Suppose two hosts, A and B, are separated by 10,000 kilometers and are connected by a direct link of R = 1 Mbps. Suppose the propagation speed over the link is 2.5* 8
10meters/sec.
a.Calculate the bandwidth-delay product, R* prop
t
b.Consider sending a file of 400,000 bits from Host A to Host B. Suppose the file is
sent continuously as one big message. What is the maximum number of bits that will be in the link at any given time?
c.Provide an interpretation of the bandwidth-delay product.
d.What is the width (in meters) of a bit in the link? Is it longer than a football field?
e.Derive a general expression for the width of a bit in terms of the propagation speed s,
the transmission rate R, and the length of the link m.
Answer
a.R*tprop = 1Mbps *10000km/2.5*
8
10meters/sec=40000bits
b. dprop=10000km/2.5*
8
10meters/sec=0.04sec, dtrans = 400000bits/1Mbps=0.4sec
Considering dtrans is far more than dprop,so we know that when first bit arrives at Host B the transmission is not over. So the maximum number of bits that will be in the link at any given time is 0.04sec *1Mbps = 40000bits
c.The bandwidth-delay product means the maximum number of bits that is in the link at given time.
d.The width (in meters) of a bit in the link =10000km/40000=250m.
e. Assume that the length of the packet is L,and we use w to represent width of a bit.
If(m/s <= L/R) then w =m/(R*tprop)=m/(R*m/s)=s/R
If (m/s>L/R) then w =(L*s/R)/L=s/R
In a nutshell, the w =s/R
Problem17================================================== Refer again to problem 14.
a.How long does it take to send the file, assuming it is sent continuously?
b.Suppose now the file is broken up into 10 packets with each packet containing
40,000bits. Suppose that each packet is acknowledged by the receiver and the
transmission time of an acknowledgment packet is negligible. Finally, assume that the sender cannot send a packet until the preceding one is acknowledged. How long does it take to send the file?
https://www.doczj.com/doc/2913545987.html,pare the results from (a) and (b).
Answer
a.Assume t is the time it takes to send the file when it is sent continuously.
Then t = tprop + dtrans= 0.04sec+0.4sec=0.44sec.
b.Assume T is the time it takes to send the file .
T=10*(40000bit/1Mbps +0.04sec +0.04sec)=1.2sec.
c.When the file is broken up into several packets ,it takes more time to send it than in a whole packet,the time cost including the but it's safer to use smaller packets .
Problem 20==================================================
In modern packet-switched networks, the source host segments long, application-layer messages (for examples, an image or a music file) into smaller packets and sends the packets into the network. The receiver then reassembles the packets back into the original message. We refer to this process as message segmentation. Figure 1.21 illustrates the end-to-end transport of a message with and without message segmentation. Consider a message that is 7.5*10e6 bits long that is to be sent from source to destination in
Figure1.21. Suppose each link in the figure is 1.5Mbps. Ignore propagation, queuing, and processing delays
a. Consider sending the message from source to destination without message segmentation. How long does it take to move the message from source to destination without message segmentation? How long does it take to move the message from the source host to the first packet switch, what is the total time to move the message from source host to destination host?
b. Now suppose that the message is segmented into 5,000 packets, with each packet being 1,500 bits long. How long does it take to move the first packet from the source host to the first packet switch? When the first packet is being sent from the first switch to the second switch, the second packet is being sent from the source host to the first switch. At what time will the second packet be fully received at the first switch?
c. How long does it take to move the file from source host to destination host when message segmentation is used? Compare this result with your answer in a part (a) and comment.
d. Discuss the drawbacks of message segmentation.
Answer
a.Time to send message from source host to first packet switch =
7.5*10e6bits/1.5Mbps=5sec. So the total time to move message from source host to destination host =5sec *3 =15sec.
b.Time to send first packet from source host to first packet switch =
1500bits/1.5Mbps=1msec;It takes 1msec for the second packet to arrive at the first switch,so at the time 2msec the second packet is fully received by the first switch.
https://www.doczj.com/doc/2913545987.html,ing segmentation ,the time to send the file from source to destination
=1msec*3+4999*1msec=5002msec=5.002s Compare this number with 15sec ,we learn that segmentation will decrease the delay by 60%.
d.The drawback of segmentation is: these packets must be in the right order when they arrived at the destination. And also that the total length of all the packets is larger. Problem 22================================================== Consider sending a large file of F bits from Host A to Host B. There are two links (and one switch) between A and B, and the links are uncongested (that is, no queuing delays). Host A segments the file into segments of S bits each and adds 40 bits of header to each segment, forming packets of L = 40 + S bits. Each link has a transmission rate of R bps. Find the value of S that minimizes the delay of moving the packet from Host A to Host B. Neglect propagation delay.
Answer
The time that send first packet from A to B =L*2/R sec=(40+S)*2/R.Then after each L/R time , B received a packet.
So the total time cost to send the file from A to B
T=2*L/R+((F/S-1)*L/R=(F/S+1)(40+S)/R
To minimizes the delay of moving the packet from A to B , we have dT/dS=0, so we get S =sqrt(40F).
Problem in chapter 1 slides PPT 95.==================================
A router’s outgoing bandwidth is 100 kbps.
Arrival packet’s number of bits has expo. distr. With mean number of 1 kbits
Poisson arrival process, λ = 80 packets/sec
a.How many packets in router expected by a new arrival?
b.What is the expected waiting time in queue for a new arrival?
c.What is the expected access delay (response time)?
d.What is the prob. that the server is idle?
e.What is P( N > 5 )?
f.Suppose you can increase router bandwidth, what is the minimum bandwidth to support av
g. access delay of 20ms?
Answer
A.μ=100kbps/(1kbits/packet)=100packet/sec, then ρ = λ/μ= 80/100=0.8
Pn=ρ^n*P0,P0= 1-ρ=0.2
So packets in server when a new arrival comes =∑∞
=0*n Pn n =ρ/(1-ρ)=4 B.The average waiting time =ρ
ρ-*1R L = 1/100 *(0.8/(1-0.8))=0.04sec=40msec C.The access delay = average waiting time +outgoing time
=0.04sec+1/μ=0.05sec=50msec
D.P(server idle)=P(0)=1-ρ=1-0.8=0.2
E.P(N>5)=1-P(N<5)=1-(P0+P1+P2+P3+P4)
Pn=ρ^n*P0,P0= 1-ρ=0.2
Hence,P(N>5)=1-0.67232=0.32768
F. We can conclude that access delay =average waiting time +outgoing time =ρρ-*1R L +R
L =20msec And we have ρ=λ/μ=L λ/R Then we can figure out that the minimum R is 130kbps to support avg. access delay of 20ms
Computer networks Homework Assignment 2
1. When a user requests a Web page that consists of some text and two images. For this page, the client will send one request message and receive three response messages.False
2. Two distinct Web pages(for example,https://www.doczj.com/doc/2913545987.html,/reaearch.html and
https://www.doczj.com/doc/2913545987.html,/students.html)can be send over the same persistent connection.True
6.================================================================
Suppose within your Web browser you click on a link to obtain a Web page. The IP address for the associated URL is not cached in your local host, so a DNS look-up is necessary to obtain the IP address. Suppose that n DNS servers are visited before your host receives the OP address form DNS; the successive visits incur an RTT of 1RTT ,…n RTT .Further suppose that the Web page associated with the link contains exactly one object, consisting of a small amount of HTML text. Let 0RTT denote the RTT between the local host and the server containing the object. Assuming zero transmission time of the object, how much time elapses form when the client clicks on the link until
the client receives the objects?
Time elapses form when the client clicks on the link until the client receives the objects=1RTT +.....+n RTT +2*0RTT
7.================================================================
Referring to Problem 6, suppose the HTML file references three very small objects on the same server. Neglecting transmission times, how much time elapses with
a. Non-persistent HTTP with no parallel TCP connections?
1RTT +.....+n RTT +2*0RTT +3*20RTT =1RTT +.....+n RTT +80RTT
b. Non-persistent HTTP with parallel connections?
1RTT +.....+n RTT +2*0RTT +20RTT =1RTT +.....+n RTT +40RTT
c. Persistent HTTP with pipelining?
1RTT +.....+n RTT +20RTT +0RTT =1RTT +.....+n RTT +30RTT
9.================================================================
Consider Figure 2.11,for which there is an institutional network connected to the Internet. Suppose that the average object size is 900,000 bits and that the average request rate from institution ’s browsers to the origin servers is 1.5 requests per second. Also suppose that the amount of time it takes from when the router on the Internet side of the access link forwards an HTTP request until it receives the response in two seconds on average(see Section 2.2.6). Model the total average response time as the sum of the average access delay (that is, the delay from Internet router to institution router)and the average Internet delay. For the average access delay, use /(1)β?-?, where ? is the average time required to send an object over the access link and β is the arrival rate of objects to the access link. a. Find the total average response time.
The object size L=900,000bits, the access link has a rate R of 15Mbps;
Hence ?=L/R=900,000bits/15Mbps=0.06s;
The arrival rate of objects to the access link β equals the request rate from institution ’s browsers to the origin servers ,thus β=1.5request/sec
So the average response time = average access delay + average Internet delay =/(1)β?-?+2 =0.06s/(1-0.06*1.5)+2=0.066s+2s=2.066s;
b. Now suppose a cache is installed in the institutional LAN. Suppose the hit rate is 0.4. Find the total response time.
Since we have a cache of hit rate 0.4 installed in the institutional LAN ,then only 60%requests needed to be send outside, the left 40% will be satisfied all most immediately, let's say zero compared with the access delay.
Then the average response time for the 60%= 0.06s/0.06s/(1-0.06*1.5*0.6)+2=2.0634s Thus the total average response time =0.6*2.0634s+0.4* 0=1.238s
A) The time to transmit an object of size L over a link or rate R is L/R . The average time
is the average size of the object divided by R:
Δ= (900,000 bits)/(1,500,000 bits/sec) = .6 sec
The traffic intensity on the link is (1.5 requests/sec)(.6 msec/request) = .9. Thus, the average access delay is (.6 sec)/(1 - .9) = 6 seconds. The total average response time is therefore 6 sec + 2 sec = 8 sec.
B) The traffic intensity on the access link is reduced by 40% since the 40% of the requests are satisfied within the institutional network. Thus the average access delay is (.6 sec)/[1 – (.6)(.9)] = 1.2 seconds. The response time is approximately zero if the request is satisfied by the cache (which happens with probability .4); the average response time is 1.2 sec + 2 sec = 3.2 sec for cache misses (which happens 60% of the time). So the average response time is (.4)(0 sec) + (.6)(3.2 sec) = 1.92 seconds. Thus the average response time is reduced from 8 sec to 1.92 sec.
20.================================================================ Consider query flooding in P2P file sharing, as discussed in Section 2.6. Suppose that each peer is connected to at most N neighbors in the overlay network.. Also suppose that the node-count field is initially set to K.Suppose Alice makes a query. Find an upper bound on the number of query messages that are sent into the overlay network.
The upper bound on the number of query messages that are sent into the overlay network happens when the query flooding access all different nodes .The query flooding process builds a tree with each nodes have N-1 child nodes and have K layers.Then the number of query messages equals to the number of edges in this tree.
On the first layer , in which each node have a node-count field of K-1, number of edges N,
on the 2th layer , the number of edges goes to (N-1)*N.
3th layer ------N*(N-1)*(N-1) Kth layer-----N*(N-1)1-K
The total edges = N+(N-1)*N+N*(N-1)*(N-1)+.......+N*(N-1)1-K=N*
21
)1 (
--
-
N
N K
Problem4===================================================== Consider our motivation for correcting protocol rtd2.1. Show that the receiver, shown in the figure on the following page, when operating with the sender shown in Figure3.11, can lead the sender and receiver to enter into a deadlock state, where each is waiting for an event that will never occur.
Suppose that the sender is in state "wait for the ack or nak 0", and the receiver is in state "wait for 1 from below". If the ack0 packet is corrupted when it arrives at sender, the sender will retransmit the packet0, but when this packet0 arrives at the receiver ,the receiver will send a nack0 to the sender. The sender receives the nack0 and will resend the packet0, then it will repeat these steps above. Thus they enter into a deadlock state. Problem 11=====================================================
Consider a reliable data transfer protocol that uses only negative acknowledgments. Suppose the sender sends data only infrequently. Would a NAK-only protocol be preferable to a protocol that uses ACKs? Why? Now suppose the sender has a lot of data to send and the end-to-end connection experiences few losses. IN this second case, would a NAK-only protocol be preferable to a protocol that uses ACKs? Why?
In a NAK-only protocol, the receiver get a packet x without corruption will do nothing
later. If receiver get a corrupted one ,it will send a nak. Then the sender will know that the former packets is correctly received. At the receiver side, if packet receives packet x then packet x+2, the receiver will realize that packet x+1 is missed,then nak(x+1) will be sent. If there is a long delay between packet x and x+1, then there will be a long time for packet x to resend. If the nak is lost or corrupted , things get more complicated.
But if data is being sent often, then recovery will happen quickly.And if error occurs infrequently, the nak will be occasionally sent and ack never be sent. The backward channel has only a small amount of packets . In this point of view ,it's somehow better than the acks protocol.
Problem 14=====================================================
In the generic SR protocol that we studied in Section 3.4.4, the sender transmits a message as soon as it is available (if it is in the window) without waiting for an acknowledgment. Suppose now that we want an SR protocol that sends messages tow at a time. That is, the sender will send a pair of messages and will send the next pair of messages only when it knows that both messages in the first pair have been received correctly.
Suppose that the channel may lose messages but will not corrupt or reorder messages. Design an error-control protocol for the unidirectional reliable transfer of messages. Give an FSM description of the sender and receiver. Describe the format of the packets sent between sender and receiver, and vice versa. If you use any procedure calls other than those in Section3.4, clearly state their actions. Give an example(a timeline trace of sender and receiver) showing how your protocol recovers from a lost packet.
The sender The Receiver
Problem 16=====================================================
Consider the GBN protocol with a sender window size of 3 and a sequence number range of 1,024. Suppose that at time t, the next in- order packet that the receiver is expecting has a sequence number of k. Assume that the medium doesn’t reorder messages. Answer the following questions:
a. what are the possible sets of sequence numbers inside the sender’s window at time t? Justify your answer.
b. what are all possible values of the ACK field in all possible messages currently propagating back to the sender at time t? Justify your answer.
a.The receiver has acked k-1 and all the proceeding packets .If all the acks has been received by the sender ,then the set is {k,k+1, k+2}. If all the acks are lost, then the set is {k-3, k-2,k-1}
b.The receiver is expecting the packet k, so the acks of the proceeding packets must be sent to the receiver. If all the former 3 acks hasn't arrived at the sender ,then values of ack includes k-1,k-2,k-3; since packet k-1 has been sent , the ack of packet k-4 is no doubt that has got the sender so there are no acks less than k-3 in the currently propagating back to the sender ,neither acks more than or equal to k.
Thus the possible values is {k-3,k-2,k-1}
Problem 18=====================================================
Consider the GBN and SR protocols.
Suppose the sequence number space
is of size k. What is the largest
allowable sender window that will
avoid the occurrence of problems
such as that in Figure 3.27 for each
of these protocols?
Suppose the window size is N
The GBN protocol: k>N The
SR protocol: k>=2N
In order to avoid the scenario of
Figure 3.26, we want to avoid
having the leading edge of the
receiver's window (i.e., the one with the “highest” sequence number) wrap around in the sequence number space and overlap with the trailing edge (the one with the "lowest" sequence number in the sender's window). That is, the sequence number space must be large enough to fit the entire receiver window and the entire sender window without this overlap condition. So - we need to determine how large a range of sequence numbers can be covered at any given time by the receiver and sender windows (see Problem 13).
Suppose that the lowest-sequence number that the receiver is waiting for is packet m. In this case, it's window is [m,m+w-1] and it has received (and ACKed) packet m-1 and the w-1 packets before that, where w is the size of the window. If none of those w ACKs have been yet received by the sender, then ACK messages with values of [m-w,m-1] may still be propagating back. If no ACKs with these ACK numbers have been received by the sender, then the sender's window would be [m-w,m-1].
Thus, the lower edge of the sender's window is m-w, and the leading edge of the receivers window is m+w-1. In order for the leading edge of the receiver's window to not overlap with the trailing edge of the sender's window, the sequence number space must thus be big enough to accommodate 2w sequence numbers. That is, the sequence number space must be at least twice as large as the window size, w
k2
Problem 19=====================================================
Answer true or false to the following questions and briefly justify your answer:
a. With the SR protocol, it is possible for the sender to receive an ACK for a Pkt that falls outside of its current window.
b. With GBN, it is possible for the sender to receive an ACK for a Pkt that falls outside
of its current window.
c. The alternating-bit protocol is the same as the SR protocol with a sender and receiver window size of 1
d. The alternating-bit protocol is the same as the GBN protocol with a sender and receiver window size of 1
A.True. When the below circumstances occurred , at the time the sender get the ack1 for the second time , the window has slide to next three packets and the packet 1-3 has been outside of the current window.
B. True . The same reason as a. C true;
D true;In this case ,a cumulative ACK is just an ordinary ACK in this situation, since it can only refer to the single packet within the window.
Problem 21=====================================================Consider the TCP procedure for estimating RTT. Suppose that x = 0.1. Let SampleRTT1 be the most recent sample RTT, let SampleRTT2 be the next most recent sample RTT, etc.
(a) For a given TCP connection, suppose 4 acknowledgements have been returned with corresponding sample RTTs Sample RTT4 , SampleRTT3 , SampleRTT2 , and SampleRTT1 . Express Estimated RTT in terms of the four sample RTTs.
(b) Generalize your formula for n sample round-trip times.
(c) For the formula in part (b) let n approach infinity. Comment on why this averaging procedure is called an exponential moving average.
A.Estimated RTT(1)=sampleRTT1
EstimatedRTT(2)=(1-x)*Estimated RTT(1)+x*sampleRTT2
=(1-x)*sampleRTT1+x*sampleRTT2
Estimated RTT(3)=(1-x)*Estimated RTT(2)+x*sampleRTT3
=(1-x)*((1-x)*sampleRTT1+x*sampleRTT2)+x*sampleRTT3
=(1-x)^2*sampleRTT1+(1-x)*x*sampleRTT2+x*sampleRTT3
Estimated RTT(4)=(1-x)*Estimated RTT(3)+x*sampleRTT4
=(1-x)^3*sampleRTT1+(1-x)^2*x*sampleRTT2
+(1-x)*x*sampleRTT3+x*sampleRTT4
=0.9^3*sampleRTT1+0.9^2*0.1*sampleRTT2+0.9*0.1*sampleRTT3+0.1*sampleRTT4
B.Estimated RTT(n)=1
12*)1(*)1(*sampleRTT x sampleRTT x x n i i
n n i --=-+-∑ C. If a = 0.1 and n -> ∞, Estimated RTT(n) ≈sampleRTTi n
i i n ∑=-29.0. Since little value is
given to past samples, the average changes and remains recent. I would assume that this is why it is considered the exponential moving average.
Denote )(n TT EstimatedR
for the estimate after the n th sample. 1
)1(SampleRTT TT EstimatedR = 21)2()1(SampleRTT x xSampleRTT TT EstimatedR -+=
1)3(xSampleRTT TT EstimatedR =]
)1()[1(32SampleRTT x xSampleRTT x -+-+ 21)1(xSampleRTT x xSampleRTT -+=3
2)1(SampleRTT x -+ )3(1)4()1(TT EstimatedR x xSampleRTT TT EstimatedR -+=
21)1(x S a m p l e R T T x x S a m p l e R T T -+= 4
332)1()1(SampleRTT x xSampleRTT x -+-+ b)j
n j j n SampleRTT x x TT EstimatedR ∑-=-=1
1)()1(n n SampleRTT x )1(-+
c)∑∞=∞--=1)()1(1j j j SampleRTT x x x TT EstimatedR ∑∞==1
9.91j j j SampleRTT Problem 28=====================================================
Refer to Figure3.53 , which illustrates the convergence of TCP ’s AIMD algorithm. Suppose that instead of a multiplicative decrease, TCP decreased the window size by a constant amount. Would the resulting AIAD converge to an equal share algorithm? Justify your answer using a diagram similar to Figure3.53.
The answer is no.The resulting AIAD will
not converge to an equal share algorithm.In
the end, connection1 will have the full
bandwidth.
Problem
31==================================
===================
Recall the idealized model for the
steady-state dynamics of TCP. In the period
of time from when the connection ’s window size varies from W/(2*RTT) to W/RTT, only one packet is lost.
a. Show that the loss rat is equal to L= loss rate = 1/(3 2w /8 +3w/4).
b. Use the result above to show that if a connection has loss rate L, then its average bandwidth is approximately given by = 1.22*MSS/(RTT*
L ).
a.The number of packets sent during this period of time
sum= W/2 + (W/2+1)+...+W
Since there are (W/2 +1)RTT then
sum=(W/2+W)/2 *(W/2+1)=3w 2/8 +3w/4
Hence the loss rate L=1/sum=1/(3w 2/8 +3w/4)
b.The average bandwidth =(W/2+W)*MSS/(2*RTT)=RTT MSS W 4*3 Since W W 43832>>( W is very large ) ,hence 23/8W L ≈, then
L W 38≈. Thus the average bandwidth RTT MSS L ?=3843
L RTT MSS ??=22.1
Problem 38=====================================================
In this problem we fill in some of the detail s in the derivation of latency in Section3.7.2. a. Derive the formula Q =
2log (1/(/))RTT S R + + 1 b. Use the identity 11
2P k k -=∑ = 2P - 1 to derive the formula Latency = 2RTT + O/R + P[RTT + S/R] – (2P - 1)*S/R
A. Q is the number of times the server idles if the object were of infinite size.
Time server idles due to slow start. If the object were of infinite size, then:
The first window contains two segments, the second window contains four segments, and the third window contains eight segments. More generally, the q th window contains 2q
segments. As the figure shows that, the
server stalls after transmitting the first
window. We now consider the stall time
after transmitting the q th window.The
time from the start of transmission of the q
th window until the reception of
acknowledgment of the first segment in
the windows is S/R+RTT, the transmission
time of the q th window is 2q *S/R.
Then the stall time ST,defined as the
difference between these two quantities ,is
idle time after the q th window
=S/R+RTT - 2q *S/R.
Q=max{q: S/R+RTT - 2q *S/R>=0}
=max{q: 2q <=1+RTT/(S/R)}
=max{q: q<=log 2(1+RTT/(S/R))} = 2log (1/(/))RTT S R +
+ 1
B. P is the number of times TCP idles at server K is the number of windows that cover the object. P= min{Q,K-1} S/R+RTT = time from when server starts to send segment until server receives ack .21-k S/R=time to transmit the kth window. S/R+RTT - 21-k *S/R= idle time after the kth window
Delay=O/R+2RTT+∑=P p p idleTime 1=O/R+2RTT+∑=--+P p K R
S RTT R S 11]2[
=O/R+2RTT+P(R S +RTT) - (2P -1)R
S
=======================================================================
8.Consider a datagram network using 8-bit host address. Suppose a router uses longest prefix matching and has the following forwarding table;
For each of the four interfaces, give the associated range of destination
host addresses and the number and the number of addresses in the range.
Interface 0: range 00000000~00111111
Interface 1: range 01000000~01111111
Interface 2: range 10000000~10111111
Interface 3: range 11000000~11111111
The number of addresses in each range =26=64
=======================================================================
16.Suppose datagr am’s are limited to 1,500 bytes (including header) between source Host
A and destination Host B. Assuming a 20-byte IP header, how many datagram’s would be required to send an MP3 file consisting of 4 million bytes?
The payload of a packet is limited to (1500-20)Byte=1480B in the datagram.Suppose we use the TCP protocol to carry the file, then every datagram can carry 1480B-20B
=1460B .So for a mp3 file of 4 million bytes ,274014601046=??
?????=, then we need a number of 2740 datagrams.
=======================================================================
17. Consider the network setup in Fig 4.22 (the figure below). Suppose that the ISP instead assigns the router the address 126.13.89.67 and that the network address of the home network is 192.168/16.
a. Assign addresses to all interfaces in the home network.
b. Suppose each host has two ongoing TCP connections, all to port 80 at host
128.119.40.86. Provide the six corresponding entries in the NAT translation table.
A. Homework network addresses: 192.168.0.1, 192.168.0.2, 192.168.0.3 with the router interface being 192.168.0.4
21. Consider the network in P22 (shown below). Using Dijkstra’s algorithm, and showing your work using a table for each of a, b and c, do the following:
z
x
y
w v t u s 2
14
4 6
1 3
1
3 1 9 1
4 2 1 For node x. step
N' D(w), p(w) D(v),p(v ) D(y),p(y)
D(z),p(z ) D(t),p(t) D(u),p(u)
D(s),p(s) 0
x 1,x 3,x 6,x ∞ ∞ ∞ ∞ 1
xw 2,w 6,x ∞ ∞ 4,w ∞ 2
xwv 3,v ∞ 11,v 3,w ∞ 3
xwvy 17,y 7,y 3,w ∞ 4
xwvyu 17,y 5,u 7,u 5
xwvyut 7,t 6,t 6 xwvyuts 7,t
26. Consider the three node topology shown in Figure 4.27. Rather than having the link costs shown in Figure 4.2-4, the link costs are c(X,Y)=5, c(Y,Z)=6, c(Z,X)=2. Compute the distance tables after the initialization step and after each iteration of a synchronous version of the distance vector algorithm (as we did in our earlier discussion of Fig4.27.)
28. Consider the following network. ISP B provides national backbone service to regional ISP A. ISP C provides national backbone service to regional ISP D. Each ISP consists of one AS. Band C peer with each other in two places using BGP. Consider traffic going from A to D. B would prefer to hand that traffic over to C on the West Coast (so that C would have to absorb the cost of caring the traffic cross-country), while C
would prefer to get the traffic via, its East Coast
peering point with B (so that B would have cared
the traffic across the country). What BGP
mechanism might C use, so that B would hand over
A-to-D traffic at its East Coast peering point? To
answer this question, you will need to dig into the
BGP specification.
To make B hand over A-to-D traffic at its East Coast peering point is to only advertise its route to D via its east coast peering point with C.
31. Consider the two basic approaches identified towards achieving multicast: unicast emulation and network-layer-multicast. Consider a single sender and 32 receivers. Suppose the sender is connected to the receiver through a binary tree of routers. What is the cost of sending a multicast packet in the case of unicast emulation and network-layer multicast for this topology? Here, each time a packet (or copy of a
packet) is sent over a single link, it incurs a unit of "cost". What topology for interconnecting the sender, receivers, and routers will bring the cost of unicast emulation and true network-layer-multicast as far apart as possible.? You can choose as many routers as you'd like.
The binary tree has 6 floors and each floor respectively has 1, 2, 4, 8, 16, 32 nodes.
For unicast emulation :cost =32 * unicast cost = 32 * 5=160
For network-layer broadcast : cost = all the edges = N-1=(1+2+4+8+16+32)-1=62
33. Reverse path forwarding. Consider the topology and link costs shown in Figure 4.41 and suppose that node E is the broadcast source. Using arrows like those shown in Figure 4.41, indicate links over which packets will be forwarded using RPF, and links over which packets will not be forwarded, given that node E is the source.
35.In Section 4.5.1 we studied Dijkstra’s link-state routing algorithm for computing the unicast paths that are individually the least-cost paths from the source to all destinations. The union of these paths might be thought of as forming a least-unicast-cost path tree(or a shortest unicast path tree, if all link costs are identical). By constructing a counterexample, show that the least-cost path tree is not always the same as a minimum spanning tree.
In this figure , assume A is the source node ,then compute the least-cost paths from the A to other nodes, then form a least-unicast-cost path tree which includes edges:AC,AB,CD with a least unicast cost 10, however the MST of the graph is consisted of edges
CD,BD,AB .The MST has a least cost 7.
Problems 9 =================================================== Consider three LANs interconnected by two routers, as shown in the diagram below.
a) Redraw the diagram to include adapters. b) Assign IP addresses to all of the interfaces. For LAN 1 use addresses of the form 111.111.111.xxx ;
for LAN 2 uses addresses of the form 122.222.222.xxx ;
and for LAN 3 use addresses of the form 133.333.333.xxx .
c) Assign LAN addresses to all of the adapters.
d) Consider sending an IP datagram from host A to host F. Suppose all the ARP tables are up-to-date. Enumerate all the steps as done for the single-router example in section
5.4.2.
e) Repeat (d), now assuming that the ARP table in the sending host is empty (and the other tables are up-to- date).
Problems 12=================================================== Suppose nodes A and B are on the same 10 Mbps Ethernet segment, and the propagation delay between the two nodes is 225 bit times. Suppose A and B send frames at the same time, the frames collide, and then A and B choose different values of K in the CSMA/CD algorithm. Assuming no other nodes are active, can the retransmissions from A and B collide? For our purposes, it suffices to work out the following example. Suppose A and B begin transmission at t=0 bit times. They both detect collisions at t=225 bit times. They finish transmitting jam signal at t= 225+48= 273 bit times. Suppose Ka =0 and Kb =1. At what time does B schedule its retransmission? At what time does A begin transmission?
(Note, the nodes must wait for an idle channel after returning to Step 2- see protocol.) At what time does A's signal reach B? Does B refrain from transmitting at its scheduled time?
Problem 13=================================================== Consider a 100Mbps 100BT Ethernet. In order to have an efficiency of .50, what should be the maximum distance between a node and the hub? Assume a frame length of 64 bytes and that there are no repeaters. Does this maximum distance also ensure that a transmitting node A will be able to detect whether any other node transmitted while A was transmitting? Why or why not? How does your maximum distance compare to the actual 100 Mbps standard?
Problem 14 ===================================================
In this problem you will derive the efficiency of a CSMA/CD-like multiple access protocol. In this protocol, time is slotted and all adapters are synchronized to the slots. Unlike slotted ALOHA, however, the length of a slot (in seconds) is much less than a frame time (the time to transmit a frame). Let S be the length of a slot. Suppose all frames are of constant length L = k R S, where R is the transmission rate of the channel and k is a large integer. Suppose there are N nodes, each with an infinite number of frames to send. We also assume that tprop < S, so that all nodes can detect a collision before the end of a slot time. The protocol is as follows:
If for a given slot, no node has possession of the channel, all nodes contend for the channel; in particular, each node transmits in the slot with probability p. If
exactly one node transmits in the slot, that node takes possession of the channel
for the subsequent k-1 slots and transmits its entire frame.
计算机网络习题及答案 This manuscript was revised on November 28, 2020
计算机网络习题及答案 第一章计算机网络的基本概念 一、选择题 √1、完成路径选择功能是在OSI模型的（）。 A.物理层 B.数据链路层 C.网络层 D.运输层 2、在TCP/IP协议簇的层次中，保证端-端的可靠性是在哪层上完成的（） A.网络接口层 B.互连层 C.传输层 D.应用层 √3、在TCP/IP体系结构中，与OSI参考模型的网络层对应的是（）。 A.网络接口层 B.互联层 C.传输层 D.应用层 4、在OSI七层结构模型中，处于数据链路层与传输层之间的是（）。 A.物理层 B.网络层 C.会话层 D.表示层 √5、计算机网络中可以共享的资源包括（）。 A.硬件、软件、数据 B.主机、外设、软件 C.硬件、程序、数据 D.主机、程序、数据 √6、网络协议组成部分为（）。 A.数据格式、编码、信号电平 B.数据格式、控制信息、速度匹配 C.语法、语义、定时关系 D.编码、控制信息、定时关系 二、填空题 √1、按照覆盖的地理范围，计算机网络可以分为________、________和________。 √2、Internet采用_______协议实现网络互连。 3、ISO/OSI中OSI的含义是________。 √4、计算机网络是利用通信线路将具有独立功能的计算机连接起来,使其能够和 ________ 和________。 5、TCP/IP协议从上向下分为________、________、________和________4层。 6、为了实现对等通信，当数据需要通过网络从一个节点传送到到另一个节点前，必须在数据的头部（和尾部） 加入____________，这种增加数据头部（和尾部）的过程叫做____________或 ____________。 √7、计算机网络层次结构划分应按照________和________的原则。 8、ISO/OSI参考模型将网络分为从低到高的________、________、________、 ________、________、________和 ________七层。 9、建立计算机网络的目的是___________和____________。 三、问答题 1、什么是计算机网络 2、ISO/OSI与TCP/IP有和区别 3、什么是数据的封装、拆包 √4、TCP/IP各层之间有何关系 √5、画出ISO/OSI参考模型和 TCP/IP协议的对应关系，并说明为什么采用层次化的体系结构
计算机网络基础习题 一、单项选择题 1局域网的网络硬件主要包括服务器、工作站、网卡和 ____________ A. 网络拓扑结构 B .计算机 C .网络传输介质 D .网络协议 2. 目前，局域网的传输介质主要是同轴电缆、双绞线和 ____________ A. 电话线 B ?通信卫星 C ?光纤D ?公共数据网 3. __________________________________ 第二代计算机网络是以网为中心的计算机网络 A. 分组交换B .共享交换C .对等服务D .点对点 4?网络节点是计算机与网络的___________ 。 A.接口 B .中心C .转换D .缓冲 5. 上因特网，必须安装的软件是_________________ A. C语言B .数据管理系统 C .文字处理系统 D . TCP/IP协议 6. 下列叙述中正确的是_______________ A. 将数字信号变换为便于在模拟通信线路中传输的信号称为调制 B. 在计算机网络中，一种传输介质不能传送多路信号 C. 在计算机局域网中，只能共享软件资源，不能共享硬件资源 D. 以原封不动的形式将来自终端的信息送入通信线路称为调制解调 7. 为网络提供共享资源的基本设备是_________________ A.服务器 B.工作站C .服务商 D.网卡 A.软件 B .线路C .服务商D 9. 要使用WindowsXP系统电脑上网，首先要对 A. Modem B .工作站C .服务器 10 .若干台有独立功能的计算机，在 _________ .协议 __________ 进行设置 D .网络和拨号连接 的支持下，用双绞线相连的系统属于计算机网络。 A.操作系统 B . TCP/IP协议C .计算机软件D .网络软件 11.计算机网络最突出的优点是_________________ 。 A.共享软、硬件资源 B .处理邮件C .可以互相通信 D .内存容量大 12 .广域网和局域网是按照 __________ 来分的。 &计算机网络系统由硬件、________ 口规程三部分内容组成
第三章 2. 计算机网络采用层次结构的模型有什么好处？答：计算机网络采用层次结构的模型具有以下好处： (1) 各层之间相互独立，高层不需要知道低层是如何实现的，只知道该层通过层间的接口所提供的服务。 (2) 各层都可以采用最合适的技术来实现，只要这层提供的接口保持不变，各层实现技术的改变不影响其他屋。(3) 整个系统被分解为若干个品于处理的部分，这种结构使得复杂系统的实现和维护容易控制。(4) 每层的功能和提供的服务都有精确的说明，这样做有利于实现标准化。 3. ISO在制定OSI参考模型时对层次划分的原则是什么？答:ISO在制定OSI参考模型时对层次划分的 原则是：(1) 网中各结点都具有相同的层次。(2)不同结点的同等层具有相同的功能。(3)同一结点内相 邻层之间通过接口通信。(4)每一层可以使用下层提供的服务，并向其上层提供服务。(5)不同结点的同 等层通过协议来实现对等层之间的通信。 4. 请描述在OSI 参考模型屮数据传输的基本过程。 答: OSI 参考模型中数据传输的基本过程:当源结点的应用进程的数据传送到应用层时，应用层为数据加上本层控制报头后，组织成应用的数据服务单元，然后再传输到表示层；表示层接收到这个数据单元后，加上木层的控制报头构成表示层的数据服务单元，再传送到会话层。依此类推，数据传送到传输层；传输层接收到这个数据单元后，加上木层的控制报头后构成传输层的数据服务单元(报文)；报文传送到 网络层时，由于网络层数据单元的长度有限制，传输层长报文将被分成多个较短的数据字段，加上网络层的控制报头后构成网络层的数据服务申i 元(分组)；网络层的分组传送到数据链路层时，加上数据链路 层的控制信息后构成数据链路层的数据服务单元(顿数据链路层的巾贞传送到物理层后，物理层将以比特 流的方式通过传输介质传输。当比特流到达目的结点时，再从物理层开始逐层上传，每层对各层的控制报头进行处理，将用户数据上交高层，最终将源结点的应用进程的数据发送给目的结点的应用进程。 4. 请描述在OSI 参考模型中数据传输的基本过程。 5. 1.OSI 环境中数据发送过程 1) 应用层 当进程A 的数据传送到应用层时，应用层为数据加上应用层报头，组成应用层的协议数据单元，再传送到表示层。 2) 表示层表示层接收到应用层数据单元后，加上表示层报头组成表示层协议数据单元，再传送到会话层。表示层按照协议要求对数据进行格式变换和加密处理。 3) 会话层会话层接收到表示层数据单元后，加上会话层报头组成会话层协议数据单元，再传送到传输层。会话层报头用来协调通信主机进程之间的通信
计算机网络课后习题答案（第三章） (2009-12-14 18:16:22) 转载▼ 标签： 课程-计算机 教育 第三章数据链路层 3-01 数据链路(即逻辑链路)与链路(即物理链路)有何区别? “电路接通了”与”数据链路接通了”的区别何在? 答：数据链路与链路的区别在于数据链路出链路外，还必须有一些必要的规程来控制数据的传输，因此，数据链路比链路多了实现通信规程所需要的硬件和软件。 “电路接通了”表示链路两端的结点交换机已经开机，物理连接已经能够传送比特流了，但是，数据传输并不可靠，在物理连接基础上，再建立数据链路连接，才是“数据链路接通了”，此后，由于数据链路连接具有检测、确认和重传功能，才使不太可靠的物理链路变成可靠的数据链路，进行可靠的数据传输当数据链路断开连接时，物理电路连接不一定跟着断开连接。 3-02 数据链路层中的链路控制包括哪些功能?试讨论数据链路层做成可靠的链路层有哪些优点和缺点. 答：链路管理 帧定界 流量控制 差错控制 将数据和控制信息区分开 透明传输 寻址 可靠的链路层的优点和缺点取决于所应用的环境：对于干扰严重的信道，可靠的链路层可以将重传范围约束在局部链路，防止全网络的传输效率受损；对于优质信道，采用可靠的链路层会增大资源开销，影响传输效率。 3-03 网络适配器的作用是什么?网络适配器工作在哪一层? 答：适配器（即网卡）来实现数据链路层和物理层这两层的协议的硬件和软件 网络适配器工作在TCP/IP协议中的网络接口层（OSI中的数据链里层和物理层） 3-04 数据链路层的三个基本问题(帧定界、透明传输和差错检测)为什么都必须加以解决？ 答：帧定界是分组交换的必然要求 透明传输避免消息符号与帧定界符号相混淆
第一章概述 1-03 试从多个方面比较电路交换、报文交换和分组交换的主要优缺点。 答：（1）电路交换：端对端通信质量因约定了通信资源获得可靠保障，对连续传送大量数据效率高。 （2）报文交换：无须预约传输带宽，动态逐段利用传输带宽对突发式数据通信效率高，通信迅速。 （3）分组交换：具有报文交换之高效、迅速的要点，且各分组小，路由灵活，网络生存性能好。 1-12 因特网的两大组成部分（边缘部分与核心部分）的特点是什么？它们的工作方式各有什么特点？ 答：边缘部分：由各主机构成，用户直接进行信息处理和信息共享;低速连入核心网。 核心部分：由各路由器连网，负责为边缘部分提供高速远程分组交换。 1-17 收发两端之间的传输距离为1000km，信号在媒体上的传播速率为2×108m/s。试计算以下两种情况的发送时延和传播时延： （1）数据长度为107bit,数据发送速率为100kb/s。 （2）数据长度为103bit,数据发送速率为1Gb/s。 从上面的计算中可以得到什么样的结论？ 解：（1）发送时延：ts=107/105=100s 传播时延tp=106/(2×108)=0.005s （2）发送时延ts =103/109=1μs 传播时延：tp=106/(2×108)=0.005s 结论：若数据长度大而发送速率低，则在总的时延中，发送时延往往大于传播时延。但若数据长度短而发送速率高，则传播时延就可能是总时延中的主要成分。 1-21 协议与服务有何区别？有何关系？ 答：网络协议：为进行网络中的数据交换而建立的规则、标准或约定。由以下三个要素组成：（1）语法：即数据与控制信息的结构或格式。 （2）语义：即需要发出何种控制信息，完成何种动作以及做出何种响应。 （3）同步：即事件实现顺序的详细说明。 协议是控制两个对等实体进行通信的规则的集合。在协议的控制下，两个对等实体间的通信使得本层能够向上一层提供服务，而要实现本层协议，还需要使用下面一层提供服务。 协议和服务的概念的区分： 1、协议的实现保证了能够向上一层提供服务。本层的服务用户只能看见服务而无法看见下面的协议。下面的协议对上面的服务用户是透明的。 2、协议是“水平的”，即协议是控制两个对等实体进行通信的规则。但服务是上层使用所提供的服即服务是由下层通过层间接口向上层提供的。，”垂直的“ 务必须与下层交换一些命令，这些命令在OSI中称为服务原语。 1-26 试解释以下名词：协议栈、实体、对等层、协议数据单元、服务访问点、客户、服务器、
一、选择题 题目1 计算机网络的功能有（）。 选择一项： A. 用户管理 B. 病毒管理 C. 资源共享正确 D. 站点管理 题目分析： 计算机网络的功能有：（1）资源共享；（2）数据通信；（3）集中管理；（4）增加可靠性；（5）提高系统的处理能力和安全功能。其中，资源共享和数据通信是计算机网络最基本的两大功能。 正确答案是：资源共享 题目2 网络资源子网负责（）。 选择一项： A. 信息处理 B. 数据通信 C. 数字认证机制
D. 路由 题目分析： “资源子网”主要负责：（1）全网的信息处理；（2）为网络用户提供网络服务；（3）资源共享功能。 正确答案是：信息处理 题目3 通常按网络覆盖的地理范围分类，可分为局域网、（）和广域网三种。 选择一项： A. 星型网络 B. 有线网 C. 城域网 D. 无线网 反馈 Your answer is incorrect. 题目分析：
计算机网络按网络覆盖的地理范围进行分类可以分为：（1）局域网；（2）城域网；（3）广域网。 正确答案是：城域网 题目4 为了简化计算机网络的分析与设计，有利于网络的硬件和软件配置，按照计算机网络的系统功能，一个计算机网络中实现网络通信功能的设备及其软件的集合称为网络的（）。 选择一项： A. 无线网 B. 通信子网 C. 有线网 D. 资源子网 反馈 Your answer is incorrect. 题目分析： 计算机网络系统是由通信子网和资源子网组成。通信子网：一个计算机网络中实现网络通信功能的设备及其软件的集合。资源子网：网络中实现资源共享功能的设备及其软件的集合。
第七章网络安全 7-01 计算机网络都面临哪几种威胁？主动攻击和被动攻击的区别是什么？对于计算机网 络的安全措施都有哪些？ 答：计算机网络面临以下的四种威胁：截获（），中断()，篡改(),伪造（）。 网络安全的威胁可以分为两大类：即被动攻击和主动攻击。 主动攻击是指攻击者对某个连接中通过的进行各种处理。如有选择地更改、删除、 延迟这些。甚至还可将合成的或伪造的送入到一个连接中去。主动攻击又可进一步 划分为三种，即更改报文流；拒绝报文服务；伪造连接初始化。被动攻击是指观察和分析某一个协议数据单元而不干扰信息流。即使这些数据对 攻击者来说是不易理解的，它也可通过观察的协议控制信息部分，了解正在通信的协议 实体的地址和身份，研究的长度和传输的频度，以便了解所交换的数据的性质。这种被 动攻击又称为通信量分析。 还有一种特殊的主动攻击就是恶意程序的攻击。恶意程序种类繁多，对网络安全威胁 较大的主要有以下几种：计算机病毒；计算机蠕虫；特洛伊木马；
逻辑炸弹。 对付被动攻击可采用各种数据加密动技术，而对付主动攻击，则需加密技术与适当的 鉴别技术结合。 7-02 试解释以下名词：（1）重放攻击；（2）拒绝服务；（3）访问控制；（4）流量分析； （5）恶意程序。 答：（1）重放攻击：所谓重放攻击（）就是攻击者发送一个目的主机已接收 过的包，来达到欺骗系统的目的，主要用于身份认证过程。（2）拒绝服务：( )指攻击者向因特网上的服务器不停地发送大量 分组，使因特网或服务器无法提供正常服务。 （3）访问控制：（）也叫做存取控制或接入控制。必须对接入网络的权限 加以控制，并规定每个用户的接入权限。 （4）流量分析：通过观察的协议控制信息部分，了解正在通信的协议实体的地址和 身份，研究的长度和传输的频度，以便了解所交换的数据的某种性质。这种被动攻击又 称为流量分析（）。 （5）恶意程序：恶意程序（）通常是指带有攻击意图所编写的
实验一网线制作 1、简述自制网线的情况，并分析原因； 2、6类双绞线的制作相对于5类（超5类）线，需要注意的地方有哪些（扩展）； 下面是100M和1000M网线的常见制作方法、千兆网线的施工注意事项。 5类线(100M)的制作： a: 绿白（3）、绿（6）、橙白（1）、蓝（4）、蓝白（5）、橙（2）、棕白（7）、棕（8） b：橙白（1）、橙（2）、绿白（3）、蓝（4）、蓝白（5）、绿（6）、棕白（7）、棕（8） 常见普通线为：b－b 常见对拷线：a－b（1-3、2-6交叉） 6类线的制作（千兆线）： a：橙白（1）、橙（2）、绿白（3）、蓝（4）、蓝白（5）、绿（6）、棕白（7）、棕（8） b: 绿白（3）、绿（6）、橙白（1）、棕白（7）、棕（8）、橙（2）、蓝（4）、蓝白（5） 常见普通线为：b－b 常见对拷线：a－b（1-3、2-6、4-7、5-8交叉）－（与100m的不同）两种网线的线序不同 3、为什么夹线钳剥掉外层护套要让裸漏的网线稍长一点，整好线序后又剪短； 方便整理、排列线序 4、步骤5中保护套为何也要伸入水晶头中； 增强网线的抗拉伸能力，加强网线与水晶头之间的连接 实验二路由器的配置 1、路由器的几种配置方式分别在什么场合使用比较合适？ 1．控制台方式 这种方式一般是对路由器进行初始化配置时采用，它是将PC机的串口直接通过专用的配置连线与路由器控制台端口"Console"相连，在PC计算机上运行终端仿真软件（如Windows 系统下的超有终端），与路由器进行通信，完成路由器的配置。在物理连接上也
可将PC的串口通过专用配置连线与路由器辅助端口AUX直接相连，进行路由器的配置。 2．远程登录（Telnet）方式 这是通过操作系统自带的TELNET程序进行配置的（如Windows\Unix\Linux等系统都自带有这样一个远程访问程序）。如果路由器已有一些基本配置，至少要有一个有效的普通端口，就可通过运行远程登录（Telnet）程序的计算机作为路由器的虚拟终端与路由器建立通信，完成路由器的配置。 3．网管工作站方式 路由器除了可以通过以上两种方式进行配置外，一般还提供一个网管工作站配置方式，它是通过SNMP网管工作站来进行的。这种方式是通过运行路由器厂家提供的网络管理软件来进行路由器的配置，如Cisco的CiscoWorks，也有一些是第三方的网管软件，如HP的OpenView等，这种方式一般是路由器都已经是在网络上的情况下，只不过想对路由器的配置进行修改时采用。 4．TFTP服务器方式 这是通过网络服务器中的TFTP服务器来进行配置的，TFTP （Trivial File Transfer Protocol）是一个TCP/IP简单文件传输协议，可将配置文件从路由器传送到TFTP服务器上，也可将配置文件从TFTP服务器传送到路由器上。TFTP不需要用户名和口令，使用非常简单。 2、你认为本实验中的那几个命令使用频率最大？ Configure、interface、exit、end 实验三交换机的配置 1.交换机的第一层功能是什么？
计算机网络习题及答案精编W O R D版 IBM system office room 【A0816H-A0912AAAHH-GX8Q8-GNTHHJ8】
计算机网络习题及答案 第一章计算机网络的基本概念 一、选择题 √1、完成路径选择功能是在OSI模型的（ ）。 A.物理层 B.数据链路层 C.网络层 D.运输层 2、在TCP/IP协议簇的层次中，保证端-端的可靠性是在哪层上完成的（） A.网络接口层 B.互连层 C.传输层 D.应用层 √3、在TCP/IP体系结构中，与OSI参考模型的网络层对应的是（）。 A.网络接口层 B.互联层 C.传输层 D.应用层 4、在OSI七层结构模型中，处于数据链路层与传输层之间的是（）。 A.物理层 B.网络层 C.会话层 D.表示层 √5、计算机网络中可以共享的资源包括（）。 A.硬件、软件、数据 B.主机、外设、软件 C.硬件、程序、数据 D.主机、程序、数据
√6、网络协议组成部分为（）。 A.数据格式、编码、信号电平 B.数据格式、控制信息、速度匹配 C.语法、语义、定时关系 D.编码、控制信息、定时关系 二、填空题 √1、按照覆盖的地理范围，计算机网络可以分为________、________和________。 √2、Internet采用_______协议实现网络互连。 3、ISO/OSI中OSI的含义是________。 √4、计算机网络是利用通信线路将具有独立功能的计算机连接起来,使其能够和________ 和________。 5、TCP/IP协议从上向下分为________、________、________和________4层。 6、为了实现对等通信，当数据需要通过网络从一个节点传送到到另一个节点前，必须在数据的头部（和尾部） 加入____________，这种增加数据头部（和尾部）的过程叫做____________或 ____________。 √7、计算机网络层次结构划分应按照________和________的原则。 8、ISO/OSI参考模型将网络分为从低到高的________、________、________、 ________、________、________和 ________七层。
第二章习题 2. 3，1 单项选择题 [1 1适合在传输介质上传输的对象是( D )。 A．信息 B．数据(C．信号 D．二进制数 [2] 。 [3]利用一根同轴电缆互连主机构建以太网，则主机间的通信方式为( C )。 A．全双工 B．半双工 C．单工 D．不确定 [4] 一个1Mbps的网卡将1000比特数据全部发送到传输线上需要( D ). A Is B． C. D. [5]E1标准采用的复用方式是( A )。 A．同步时分复用 B．统计时分复用 C．频分复用 D．码分多址 [61若采用同步TDM方式通信，为了区分不同数据源的数据，发送端应该采取的措施是( C )。A．在数据中加上数据源标识 B。在数据中加上时间标识 C．各数据源使用固定时间片 D．各数据源使用随机时间片 [7]若采用同步TDM方式通信，接收端要将信号解复用，接收数据时要按照( B )。 A．时间片上的目的地址 B．数据上的时间标识 C．数据上的数据源标识 D．与源端相同的时间顺序 [8] 若采用统计TDM方式通信，只有当数据源有数据发送时才分配时间片，并在时间片中( )。A．仅附加发送信道序号 B．仅附加接收信道序号 C．附加发送信道序号和接收信道D．无须附加信息 [9]交换机采用的多路复用方式是( )。 A．同步TDM B．统计TDM C．FDM D．WDM [10]现有16路光信号通过过波分复用系统复用到一根光纤上，每条支路的速率为2. 5Gbps，则复用后的速率为( D )。 A. 2. 5Gbps B. lOGbps C. 20Gbps D. 40Gbp [11]传统的模拟电视系统采用的复用方式是( C )。 A．同步TDM B．统计TDM C．FDM D．WDM [12]当对数据率不同的多路信号采用同步TDM方式复用时，通常采用的技术是( )。 A．脉冲填充 B．压缩时隙 C．降低数据率 D．限制数据源 [13] 与同步TDM相比，统计TDM需要解决的特殊殊问题是( A )。 A．性能问题B．线路利用率问题 C．成帧与同步 D．差错控制 [17] 数据通信系统中发送装置的主要功功能是( D )。 A．将信号从信源发送到信宿 B．将信源的数据转发到传输介质上 C．将模拟信号转变成数字信号 D．产生适合在传输系统中传输的信号 [18] 在数据通信系统中，发送装置的作用一般不包括( C )。 A．调制信号 B．适配电压 k C．检错和纠错 D．暂存数据 [19] 以下为数字数据的是( D )。 A．声音 B．电视视频 C．气压值 D。硬盘保存的图像文件 [20] 传输计算机内的文件可用的信号形式有( )。 A．微波信号 B．脉冲信号 C．红外线信号 D．A、B、C都可以 [21] 下面说法正确的是( )。
第一章 一、名词解释 广域网：覆盖范围从几十公里到几千公里，是一个国家、地区或横跨几个洲的网络。城域网：可以满足几十公里范围内的大量企业、机关、公司的多个局域网互联的需要，并能实现大量用户与数据、语音、图像等多种信息传输的网络。 局域网：用于有限地理范围，将各种计算机、外设互联起来的网络。 通信子网：由各种通信控制处理机、通信线路与其他通讯设备组成，负责全网的通信处理业务。 资源子网：由各种主机、外设、软件与信息资源组成，负责全网的数据处理业务，并向网络用户提供各种网络资源与网络服务。 计算机网络：以能够相互共享资源的方式互联起来的自制计算机系统的集合。 分布式系统：存在着一个能为用户自动管理资源的网络操作系统，由它来自动调用完成用户任务所需的资源，整个网络系统对用户来说就像一个大的计算机系统一样。 公用数据网：由邮电部门或通信部门统一组建与管理，向社会用户提供数据通信服务的网络。 广播网络：网络中的计算机或设备使用一个共享的通信介质进行数据传播，网络中的所有结点都能收到任一结点发出的数据信息。 点—点网络：是两台计算机之间通过一条物理线路连接的网络。 四、简答题 1、计算机网络的发展可以划分为几个阶段？每个阶段的特点？ 四个阶段：1）远程联机阶段2)互联网络阶段3）标准化网络阶段4）网络互连与高速网络阶段 远程特点：系统中只有一个计算机处理中心，各终端通过通信线路共享主计算机的硬件和软件资源，主计算机负担过重，终端独占线路，资源利用率低。 互联特点：实现计算机与计算机通信的计算机网络系统，呈现出的是多台计算机处理中心的特点。 标准化网络阶段：网络体系结构与协议标准化的研究广域网、局域网与分组交换技术的研究与应用 网络互联与高速网络阶段：Internet技术的广泛应用网络计算技术的研究与发展宽带城域网与接入网技术的研究与发展网络与信息安全技术的研究与发展。 2. 按照资源共享的观点定义的计算机网络应具备哪几个主要特征？ 建立的主要目的是实现计算机资源的共享；互连的计算机是分布在不同地理位置的多台独立“自治系统”；连网计算机在通信过程中必须遵循相同的网络协议。 3、什么是计算机网络？ 计算机网络是把分布在不同地点，并具有独立功能的多个计算机系统通过通信设备和线路连接起来，在功能完善的网络软件和协议的管理下，以实现网络中资源共享为目标的系统。 4、计算机网络的主要功能是什么？ 1）资源共享：硬件资源、软件资源、数据资源、信道资源；2）网络通信；3）分布式处理；4）集中管理；5）均衡负载 5、计算机网络硬件包括哪些？ 主计算机、网络工作站、网络终端、通信处理机、通信线路、信息变换设备
计算机网络课后习题答 案 文档编制序号：[KKIDT-LLE0828-LLETD298-POI08]
1习题 一、填空题 1．在OSI参考模型中，网络层所提供的服务包括虚电路服务和数据报服务。 2．如果网络系统中的每台计算机既是服务器，又是工作站，则称其为对等网。 3．网络协议主要由语法、语义和规则三个要素组成。 4．OSI参考模型规定网络层的主要功能有：分组传送、流量控制和网络连接建立与管理。 5．物理层为建立、维护和释放数据链路实体之间二进制比特传输的物理连接，提供机械的、电气的、功能的和规程的特性。 6．设置传输层的主要目的是在源主机和目的主机进程之间提供可靠的端到端通信。7．在OSI参考模型中，应用层上支持文件传输的协议是文件传送、存取和管理 FTAM ，支持网络管理的协议是报文处理系统MHS 。 二．选择题 1．按覆盖的地理范围分类，计算机网络可以分成局域网、城域网和广域网。 2．如果某种局域网的拓扑结构是 A ，则局域网中任何一个结点出现故障都不会影响整个网络的工作。 A）总线型结构 B）树型结构 C）环型结构 D）星型结构 3．网状拓扑的计算机网络特点是：系统可靠性高，但是结构复杂，必须采用路由选择算法和流量控制方法。 4．在OSI七层结构模型中，执行路径选择的层是 B 。 A）物理层 B）网络层 C）数据链路层 D）传输层
5．在OSI七层结构模型中，实现帧同步功能的层是 C 。 A）物理层 B）传输层 C）数据链路层 D）网络层 6．在OSI七层协议中，提供一种建立连接并有序传输数据的方法的层是C。 A）传输层 B）表示层 C）会话层 D）应用层 7．在地理上分散布置的多台独立计算机通过通信线路互联构成的系统称为（C）使信息传输与信息功能相结合，使多个用户能够共享软、硬件资源，提高信息的能力。 A）分散系统B）电话网C）计算机网络D）智能计算机 8．若要对数据进行字符转换和数字转换，以及数据压缩，应在OSI（D）层上实现。A）网络层B）传输层C）会话层D）表示层 三、思考题 1．简述计算机网络的定义、分类和主要功能。 计算机网络的定义： 计算机网络，就是利用通信设备和线路将地理位置不同的、功能独立的多个计算机系统互连起来，以功能完善的网络软件（即网络通信协议、信息交换方式、网络操作系统等）实现网络中资源共享和信息传递的系统。 计算机网络可以从不同的角度进行分类： （1）根据网络的交换功能分为电路交换、报文交换、分组交换和混合交换； （2）根据网络的拓扑结构可以分为星型网、树型网、总线网、环型网、网状网等； （3）根据网络的通信性能可以分为资源共享计算机网络、分布式计算机网络和远程通信网络； （4）根据网络的覆盖范围与规模可分为局域网、城域网和广域网；
计算机网络习题及答案 第一章计算机网络的基本概念 一、选择题 √1、完成路径选择功能是在OSI模型的（）。 A.物理层 B.数据链路层 C.网络层 D.运输层 2、在TCP/IP协议簇的层次中，保证端-端的可靠性是在哪层上完成的（） A.网络接口层 B.互连层 C.传输层 D.应用层 √3、在TCP/IP体系结构中，与OSI参考模型的网络层对应的是（）。 A.网络接口层 B.互联层 C.传输层 D.应用层 4、在OSI七层结构模型中，处于数据链路层与传输层之间的是（）。 A.物理层 B.网络层 C.会话层 D.表示层 √5、计算机网络中可以共享的资源包括（）。 A.硬件、软件、数据 B.主机、外设、软件 C.硬件、程序、数据 D.主机、程序、数据 √6、网络协议组成部分为（）。 A.数据格式、编码、信号电平 B.数据格式、控制信息、速度匹配 C.语法、语义、定时关系 D.编码、控制信息、定时关系 二、填空题 √1、按照覆盖的地理范围，计算机网络可以分为________、________和 ________。 √2、Internet采用_______协议实现网络互连。 3、ISO/OSI中OSI的含义是________。
√4、计算机网络是利用通信线路将具有独立功能的计算机连接起来,使其能够和________ 和________。 5、TCP/IP协议从上向下分为________、________、________和________4层。 6、为了实现对等通信，当数据需要通过网络从一个节点传送到到另一个节点前，必须在数据的头部（和尾部） 加入____________，这种增加数据头部（和尾部）的过程叫做____________或____________。 √7、计算机网络层次结构划分应按照________和________的原则。 8、ISO/OSI参考模型将网络分为从低到高的________、________、________、________、________、________和 ________七层。 9、建立计算机网络的目的是___________和____________。 三、问答题 1、什么是计算机网络 2、ISO/OSI与TCP/IP有和区别 3、什么是数据的封装、拆包 √4、TCP/IP各层之间有何关系 √5、画出ISO/OSI参考模型和 TCP/IP协议的对应关系，并说明为什么采用层次化的体系结构
第三章 1.网卡的主要功能有哪些？它实现了网络的哪几层协议？ 网卡工作在OSI模型的数据链路层，是最基本的网络设备，是单个计算机与网络连接的桥梁。它主要实现如下功能： (1) 数据的封装与解封，信号的发送与接收。 (2) 介质访问控制协议的实现。采用不同拓扑结构，对于不同传输介质的网络，介质的 访问方式也会有所不同，需要有相应的介质访问控制协议来规范介质的访问方式，从而使网络用户方便、有效地使用传输介质传递信息。 (3) 串/并行转换。因为网卡通过总线以并行传输方式与计算机联系，而网卡与网络的通 信线路采用串行传输方式联系，所以网卡应具有串/并行转换功能。 (4) 发送时，将计算机产生的数字数据转变为适于通信线路传输的数字信号形式，即编 码；接收时，将到达网络中的数字信号还原为原始形式，即译码。 2.网卡有几种分类方式？ 1．按连接的传输介质分类 2．按照总线类型分类 3.使用粗缆、细缆及双绞线的网卡接口名称分别是什么？ 粗缆网卡使用AUI连接头，用来连接收发器电缆，现在已经看不到这种网卡的使用了。 细缆网卡使用BNC连接头，用来与BNC T型连接头相连，现在也很少使用，在一些布网较早的单位还可以见到。连接同轴电缆的网卡速率一般为10 Mb/s。双绞线网卡是现在最常用的，使用RJ-45插槽，用来连接网线的RJ-45插头。 4.简述安装网卡的主要步骤。 对于台式计算机，若使用USB网卡，则只要将网卡插入计算机的USB接口中就可以了； 若使用ISA或者PCI网卡，则需以下安装步骤： (1) 断开电源，打开机箱。 (2) 在主板上找到相应的网卡插槽，图3.8所示为ISA插槽和PCI插槽。选择 要插入网卡的插槽，将与该插槽对应的机箱金属挡板取下，留下空缺位置 (3)将网卡的金属挡板朝向机箱背板，网卡下方的插条对准插槽，双手均匀用 力将网卡插入插槽内，这时网卡金属挡板正好填补了上一步操作留下的空缺位置 (4) 根据机箱结构，需要时用螺丝固定金属挡板，合上机箱即可。 对于笔记本电脑，网卡的安装较为简单。首先找到笔记本的PCMCIA 插槽，如图3.10所示，然后将PCMCIA网卡有金属触点的一头插入PCMCIA 插槽，这样网卡就安装好了 5.集线器是哪一层的设备，它的主要功能是什么？ 集线器属于物理层设备，它的作用可以简单的理解为将一些机器连接起来组成一个局域网。 6.在集线器或交换机的连接中，级联与堆叠连接方式有什么异同？ 1.级联是通过集线器的某个端口与其他集线器相连，堆叠是通过集线器背板上的专用堆叠 端口连接起来的，该端口只有堆叠式集线器才具备。 2.距离不同堆叠端口之间的连接线也是专用的。堆叠连接线长度很短，一般不超过1 m， 因此与级联相比，堆叠方式受距离限制很大。 3.但堆叠线缆能够在集线器之间建立一条较宽的宽带链路，再加上堆叠单元具有可管理 性，这使得堆叠方式在性能方面远比级联方式好。 7.交换机是哪一层的设备，它的主要功能是什么？ 交换机是二层网络设备（即OSI参考模型中的数据链路层）。
一、填空题（每小题3分，共24分） 1．数据通信系统可以分为源系统，传输系统和目的系统三大部分。 2．网络协议的三个基本要素是指语义，语法和同步。 3．从通信的双方信息交互的方式来看，可以有单工通信，半双工通信和全双工通信三种基本方式。 4．传输媒体可分为导向传输媒体和非导向传输媒体，导向传输媒体有双绞线，同轴电缆和光缆。 5．信道复用技术主要有时分复用、频分复用和统计时分复用等。 6．若IP地址是172．68．35．96，子网掩码为255．255．252．0，则该地址的子网号是8，主机号是864，该子网能连接1022台主机。 7．10BASE-T中10代表10Mb/s的数据率、BASE表示连接线上的信号是基带信号、T表示双绞线。 8．主要的宽带接入技术有FTTx、HFC网、xDSL等。 9．数据链路层的必须解决的三个基本问题是封装成帧、透明传输、错检测。 12．在因特网中，应用层的协议很多，例如，支持万维网应用的HTTP协议，支持电子邮件的STMP协议，支持文件传送的FTP协议。 13．面向连接服务具有三个阶段连接建立、数据传输、连接释放。 14．转发器，网桥和路由器分别在物理层，数据链路层和网络层上使用的中间设备，以实现网络的互连。 17.要发送的数据为11010011，采用CRC的生成多项式是P(X)=X4+X+1，应添加在数据后面的余数是1001。 18．一个3200bit长的TCP报文传到IP层，加上160bit的首部后成为数据报。下面的互连网由两个局域网通过路由器连接起来。但第二个局域网所能传输的最长数据帧中的数据部分只有1200bit。因此，数据报在路由器必须分片。试问第二个局域网向其上层要传送3840（bit）的数据。 19．假设500m长的CSMA/CD网络的数据速率为1Gb/s，且无中继器，信号在电缆中的速度为200000km/s。能够使用此协议的最短帧长是5000（bit）。 20．长度为100字节的应用层数据依次交给运输层、网络层和数据链路层传送，分别需加上20字节的TCP首部、20字节的IP首部、18字节的帧首部和尾部。问：数据的传输效率是63.29%。 21．在传输基带数字信号时，主要有和等编码方法。数据率为10Mb/s的以太网在物理媒体的码元传输速率是20M波特。 11．计算机网络安全的内容主要有保密性、安全协议的设计、访问控制。 10．扩频通信通常有两大类直接序列扩频DSSS、跳频扩频FHSS，CDMA的码片序列属于直接序列扩频DSSS。 15．对计算机网络的威胁可分为被动攻击和主动攻击两大类，主动攻击可进一步划分为更改报文流、拒绝服务、伪造连接初始化等三种。 16．IP数据报首部长度的最大值为60字节，最小值为20字节，IP数据报最大长度为65535字节。 21．在传输基带数字信号时，主要有曼彻斯特编码和差分曼彻斯特编码等编码方法。 22．数字签名必须保证能够实现报文鉴别、报文的完整性、不可否认三个功能。 23．目前因特网提供的音频/视频服务大体可分为流式存储音频/视频、流式实况音频/视频、交互式音频/视频。
计算机网络课后习题答案（第五章） (2009-12-14 18:28:04) 转载▼ 标签： 课程-计算机 教育 第五章传输层 5—01试说明运输层在协议栈中的地位和作用，运输层的通信和网络层的通信有什么重要区别为什么运输层是必不可少的 答：运输层处于面向通信部分的最高层，同时也是用户功能中的最低层，向它上面的应用层提供服务 运输层为应用进程之间提供端到端的逻辑通信，但网络层是为主机之间提供逻辑通信（面向主机，承担路由功能，即主机寻址及有效的分组交换）。 各种应用进程之间通信需要“可靠或尽力而为”的两类服务质量，必须由运输层以复用和分用的形式加载到网络层。 5—02网络层提供数据报或虚电路服务对上面的运输层有何影响 答：网络层提供数据报或虚电路服务不影响上面的运输层的运行机制。 但提供不同的服务质量。 5—03当应用程序使用面向连接的TCP和无连接的IP时，这种传输是面向连接的还是面向无连接的 答：都是。这要在不同层次来看，在运输层是面向连接的，在网络层则是无连接的。 5—04试用画图解释运输层的复用。画图说明许多个运输用户复用到一条运输连接上，而这条运输连接有复用到IP数据报上。 5—05试举例说明有些应用程序愿意采用不可靠的UDP，而不用采用可靠的TCP。 答：VOIP：由于语音信息具有一定的冗余度，人耳对VOIP数据报损失由一定的承受度，但对传输时延的变化较敏感。 有差错的UDP数据报在接收端被直接抛弃，TCP数据报出错则会引起重传，可能带来较大的时延扰动。 因此VOIP宁可采用不可靠的UDP，而不愿意采用可靠的TCP。 5—06接收方收到有差错的UDP用户数据报时应如何处理 答：丢弃 5—07如果应用程序愿意使用UDP来完成可靠的传输，这可能吗请说明理由
计算机网络习题及答案文件编码（008-TTIG-UTITD-GKBTT-PUUTI-WYTUI-8256）
计算机网络习题及答案 第一章计算机网络的基本概念 一、选择题 √1、完成路径选择功能是在OSI模型的（）。 A.物理层 B.数据链路层 C.网络层 D.运输层 2、在TCP/IP协议簇的层次中，保证端-端的可靠性是在哪层上完成的（） A.网络接口层 B.互连层 C.传输层 D.应用层 √3、在TCP/IP体系结构中，与OSI参考模型的网络层对应的是（）。 A.网络接口层 B.互联层 C.传输层 D.应用层 4、在OSI七层结构模型中，处于数据链路层与传输层之间的是（）。 A.物理层 B.网络层 C.会话层 D.表示层 √5、计算机网络中可以共享的资源包括（）。 A.硬件、软件、数据 B.主机、外设、软件 C.硬件、程序、数据 D.主机、程序、数据 √6、网络协议组成部分为（）。 A.数据格式、编码、信号电平 B.数据格式、控制信息、速度匹配 C.语法、语义、定时关系 D.编码、控制信息、定时关系 二、填空题 √1、按照覆盖的地理范围，计算机网络可以分为________、________和 ________。 √2、Internet采用_______协议实现网络互连。 3、ISO/OSI中OSI的含义是________。
√4、计算机网络是利用通信线路将具有独立功能的计算机连接起来,使其能够和________ 和________。 5、TCP/IP协议从上向下分为________、________、________和________4层。 6、为了实现对等通信，当数据需要通过网络从一个节点传送到到另一个节点前，必须在数据的头部（和尾部） 加入____________，这种增加数据头部（和尾部）的过程叫做____________或 ____________。 √7、计算机网络层次结构划分应按照________和________的原则。 8、ISO/OSI参考模型将网络分为从低到高的________、________、________、________、________、________和 ________七层。 9、建立计算机网络的目的是___________和____________。 三、问答题 1、什么是计算机网络 2、ISO/OSI与TCP/IP有和区别 3、什么是数据的封装、拆包 √4、TCP/IP各层之间有何关系 √5、画出ISO/OSI参考模型和 TCP/IP协议的对应关系，并说明为什么采用层次化的体系结构 答案 一、选择题 123456 C C B B A C 二、填空题 1、局域网城域网广域网
一、填空题 1.网络协议主要由三个要素组成：（）、（）和同步。 2.假设两个主机A,B通过一个路由器进行互联，提供主机A和主机B的应用进程之间 通信的层是（），通过主机与主机之间通信的层是（）。 3.通信的目的是传送消息，如语音、文字和图像等。（）是运送消息的实体。 4.在计算机网络中的信道使用多种复用技术，（）是指所有用户在不同的时间 占用相同的频带。（）是指所有的用户在同样的时间占用不同的频带宽度。 5.数据链路层使用的信道主要有两种类型：（）信道和（）信道， 前者最常使用的协议是PPP, 后者最常使用的协议有CSMA/CD。 6.以太网使用CSMA/CD协议可以使很多计算机以多点接入的方式连接在一根总线上， 协议的实质是（）和（）。 7.当网桥刚接入到以太网时，它的转发表是空的，这时若网桥收到一帧数据，网桥就 是按照（）算法处理收到的帧，然后把帧转发出去。 二、选择题 1.IP数据报穿越Internet过程中可能被分片。在IP数据报分片以后，下列哪些设备 负责IP数据报的重组（）。 A.源主机 B.目的主机 C.分片途径的路由器 D.分片途径的路由器和目的主机 2.下列哪个地址属于C类地址（） A. B.10.10.1.2 C. D. 3.关于RIP协议和OSPF协议，下列说法正确的是（）。 A.都是基于链路状态的外部网关协议。 B.RIP是基于链路状态的内部网关协议，OSPF是基于距离向量的内部网关协议。 C.都是基于距离向量的内部网关协议。 D.RIP是基于距离向量的内部网关协议，OSPF是基于链路状态的内部网关协议。 4.下列哪一个选项不属于路由选择协议的功能（） A.获取网络拓扑结构的信息。 B.选择到达每个目的网络的最优路径 C.构建路由表 D.发现下一跳的物理地址。 5.每一条TCP连接唯一的被通信两端的两个端点所确定，TCP连接的端点是指 （）。 地址地址 C.端口号 D.套接字 6.一条TCP连接的建立过程包括（）次握手。