期中测试

1、填空题：

‎[10 Marks] Consider sending a large file of 10 Gbits from Host A to Host B. There are 4 links (and 3 switches) between A and B, and the links are uncongested (that is, no queueing delays). Segments are transmitted in the style of store and forward. Host A segments the file into segments of S bits each and adds 320 bits of header to each segment, forming packets of L =320 + S bits. Each link has a transmission rate of 100 Mbps. Find the value of S that minimizes the delay of moving the file from Host A to Host B. Disregard propagation delay. Show the detailed calculation steps.

‌

‎AB

‌答案: 【 Time at which the 1st packet is received at the destination = (4×（320+S）)/(100×〖10〗^6 )

After this, one packet is received at destination every (320+S)/(100×〖10〗^6 )

Thus delay in sending the whole file (4×（320+S）)/(100×〖10〗^6 ) +(⌈(10×〖10〗^9)/s⌉ -1)×(320+S)/(100×〖10〗^6 )
= (3×（320+S）)/(100×〖10〗^6 ) + (100×（320+S）)/s = 32000/s + 3s×〖10〗^(-8) + 9.6×〖10〗^(-6)+100

为了求极值，对上面求导，并令其等于0。得到：

3×〖10〗^(-8)- 32000/s^2 = 0
s^2=32000/(3×〖10〗^(-8) )=(32×〖10〗^11)/3

S=1.0328×〖10〗^6

S的最优值是1.0328 M bit，可以让延时最小。】

2、填空题：
​‌[10 Marks] Consider that only a single TCP (Reno) connection uses one 100 Mbps link which does not buffer any data. Suppose that this link is the only congested link between the sending and receiving hosts. Assume that the TCP sender has a huge file to send to the receiver, and the receiver’s receive buffer is much larger than the congestion window. We also make the following assumptions: each TCP segment size is 1,000 bytes; the two-way (round trip) propagation delay of this connection is 200 msec; and this TCP connection is always in congestion avoidance phase, that is, ignore slow start.‌a. What is the maximum window size (in segments) that this TCP connection can achieve?‌b. What is the average window size (in segments) and average throughput (in bps) of this TCP connection?‌c. How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?‌
答案: 【 a) Let W denote the max window size measured in segments.
Then, W*MSS/RTT = 100Mbps,
as packets will be dropped if the maximum sending rate exceeds link capacity.
Thus, we have W*1000*8/0.2=100*10^6,
then W is 2500 segments.

b) As congestion window size varies from W/2 to W, then the average window size is 0.75W=1875 segments.
Average throughput is 1875*1000*8/0.2=75 Mbps.

c) as the number of RTTs (that this TCP connections needs in order to increase its window size from W/2 to W) is given by W/2. Recall the window size increases by one in each RTT.
(2500/2) *0.2= 250 seconds,】

随堂测验01

1、单选题：

下图描述的网络协议要素是

‍

‍选项：
A: 语法
B: 语义
C: 时序
D: 格式
答案: 【 时序】

2、单选题：

‏在下图所示的采用“存储–转发”方式的数据报网络中，所有链路的数据传输速率为100 Mbps，分组大小为1 000 B，其中分组头大小为20 B。若主机H1向主机H2发送一个大小为980 000 B的文件，则在不考虑分组拆装时间和传播延迟的情况下，从H1发送开始到H2接收完为止，需要的时间至少是

​

‏

​选项：
A: 80 ms
B: 80.08 ms
C: 80.16 ms
D: 80.24 ms
答案: 【 80.24 ms】

3、判断题：
‍电路交换更适合实时数据流传输。‎
选项：
A: 正确
B: 错误
答案: 【 正确】

4、填空题：
​实现网络互连的传输媒介（如光纤、双绞线、微波等）称为‌
答案: 【 链路】

5、填空题：
‌约定计算机网络通信实体间如何通信的是‍
答案: 【 协议】

第一单元测验

1、单选题：
​下列选项中，不属于协议要素的是‏
选项：
A: 层次
B: 语法
C: 语义
D: 时序
答案: 【 层次】

2、单选题：
‌在分组交换网络中，主要取决于网络拥塞程度的时间延迟是‍
选项：
A: 节点处理延迟
B: 排队延迟
C: 传输延迟
D: 传播延迟
答案: 【 排队延迟】

3、单选题：
​在一个CDMA网络中，某站点正接收另一码序列为（-1,1,1,-1,-1,-1,1,-1）的站点发送的数据，若该站点收到（-111-1-1-11-1 1-1-1111-11 1-1-1111-11 -111-1-1-11-1），则该站点收到的数据是‏
选项：
A: 0001
B: 0110
C: 1001
D: 1000
答案: 【 1001】

4、单选题：

‎在下图所示的采用“存储–转发”方式的分组交换网络中，所有链路的数据传输速率为100 Mbps，分组大小为1 000 B，其中分组头大小为20 B。若主机