Computer Network Homework2’s hard question

Computer Network Homework2’s hard question

2. What is the signal which is used to modulate the original signal?

A. analog signal

B. digital signal

C. carrier signal

D. base signal

(2 points)

Answer:

C

Explanation:

original signal指最原始的想要被传给别人的信号

5.In which of the following fibers，light rays follow sinusoidal paths along the fibers?

A.Single-mode fiber

B.step-index multimode fiber

C.graded-index multimode fiber

Answer:

C

7.Which of the following UTP(unshielded twisted pair) is most suitable for 1000Mbps network?

A.cat3

B.cat 4

C.cat 5

D.cat6

Answer:

D

8. Which of the following is the function of data link layer?

A.transfering original bit stream

B.transfering data between processes

C.routing

D.transfer data in a physical network

Answer:

D

11. If the selective repeat protocol (sliding window) is used, how much is the timeout set to at least?

A. somewhat more than 4 times RTT

B. somewhat more than 2 times RTT

C. somewhat more than RTT

D. somewhat less than RTT

Answer:

B

Explanation:

收到错序帧返回NAK，发送要重传的帧返回确认帧，超时时间要大于这两个来回的时间以防止超时重传。

13. For the PPP protocol, what protocol is applied to get IP address?

A. LCP

B. IPCP

C. IPXCP

D. CHAP

(2 points)

Answer:

B

Suppose the sequence numbers can be reused in a Seletive Repeat protocol, the sending window size is SWS and the receiving window size is RWS, how many sequence numbers are required at least? Please explain it.

Answer:

SWS+RWS  （未批改）

Explanation:

因为如果发送方发送了新的（未收到确认）SWS帧，接收方收到后依次上传给上层协议并发送确认帧，此时接收窗口必须移到新的RWS个序号，否则，发送方重新发送的帧会落在接收窗口内被错误接收。这样的话重发的SWS个帧的序号与接收窗口的序号必须不同，因为至少需要SWS+RWS个序号。

4. If a stop-and-wait protocol is used on a 10Mbps point-to-point link，the maximum length of a frame is 4000 bytes, the link length is 2000km (propagation speed is 200000km/s)，how much is the maximum utilization of bandwidth(maximum throughput/bandwidth)?

Answer:
13.8%

Explanation:
throughput=包长/(发送时间+RTT)
= 4000bytes/(4000bytes/10Mbps
+ 2*2000km/200000km/s)
=4000*8bits/(3.2ms+20ms)
=32/23.2Mbps
=1.38Mbps
(因为ACK很短，这里忽略了ACK的发送时间)
util=1.38/10=13.8%

5. If the sliding window protocol is applied to the above question and the sending window size is equal to 5, how much is the maximum utilization of bandwidth?

Answer:
69%

Explanation:
throughput=包长/(发送时间+RTT)
= 5*4000bytes/(5*4000bytes/10Mbps
+ 2*2000km/200000km/s)
=5*4000*8bits/(5*3.2ms+20ms)
=5*32/36 Mbps
=4.44Mbps
(因为ACK很短，这里忽略了ACK的发送时间)
util=4.44/10=44.4%
=======
上面有错。因为流水方式，当发送窗口满了时收到每帧的确认都会立即发出一个新的帧，所以，每个回合应该只考虑一帧的发送时间。5*4000*8bits/(3.2ms+20ms)=6.9
util=6.9/10=69%

8. In PPP protocol, which protocol will be used to determine whether the authentication step is needed?

Answer:
LCP

1. For a stop-and-wait protocol, how many sequence numbers are required at least? Please explain it.

Answer:
2个或者1比特。（未批改）

Explanation:
正常情况是不需要序号的。出错的情况有三种：（1）数据帧丢失（2）确认帧丢失（3）确认帧在超时时间之后返回。如果没有序号，（2）（3）会出现重复接收，（3）还会出现错误确认。用2个序号就可以防止这些错误出现。

5. With the Selective Repeat (SR) protocol, the receiver must send an ACK (it can delay some time or combine serveral ACKs) when receiveing an Data Frame any time, true or false?  Please explain it.

Answer:
TRUE （未批改）

Explanation:
一个帧落在接收窗口之内当然要发送确认帧，以防止前面的确认丢失。即使一个帧落在接收窗口之外也必须发送确认帧，因为会出现RWS的确认帧全部丢失的情况，以后重发的帧都会落在接收窗口之外。