Homework 3
3220104929 241019 作业 3.pdf
1
A bit string, 0111101111101111110, needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing?
011110 111110 0111110 10
2
What is the remainder obtained by dividing x^7 + x^5 + 1 by the generator polynomial x^3 + 1? (give your answer as bit string)
10100001 000/1001 = 10110111 ... 111
3
A channel has a bit rate of 4 kbps and a propagation delay of 20 msec. For what range of frame sizes does stop-and-wait give an efficiency of at least 50 percent?
(2*20msec) / (F/4kbps + 2*20msec) \(\le\) 0.5
\(\Rightarrow\) F \(\ge\) 160 bit
故 frame 大小应大于等于 160bit
4
Consider an error-free 64-kbps satellite channel used to send 512-byte data frames in one direction, with very short acknowledgements coming back the other way. What is the maximum throughput for window sizes of 1, 7, 15? The earth-satellite propagation time is 270 msec. (give your answer as an integer)
a
for window size = 1
512byte/64kbps = 64msec
64 + 270*2 = 604msec
512byte / 604msec = 6781bps
b
for window size = 7 ，注意，第一个 ACK 回来时就完成了一个 cycle，后面的 ACK 与发射正好卡点了
64 * 7 = 448 msec < 604 msec
448 + 270*2 = 988 msec
512 byte * 7 / 604 msec = 47470 bps
c
for window size = 15
64 * 15 = 960 msec > 604 msec
这意味着可以不间断发射，则吞吐量为 64kbps
5
A 100-km-long cable runs at the T1 data rate. The propagation speed in the cable is 2/3 the speed of light in vacuum. How many bits fit in the cable?
100km / (2/3 * 3* 10^8m/s) = 0.5 msec
T1 data rate = 1.544Mbps
0.5 msec * 1.544Mbps = 772 bit
6
A CRC generator polynomial is G(X)= X^16+X^15+X^2+1. How many bits will the checksum be?
16 bit
7
Assume the sequence number has 3 bits. What is the maximum number of outstanding sending frames for a go back N protocol?
2^3-1 = 7
8
Assume the sequence number has 5 bits. What is the maximum number of outstanding sending frames for a selective repeat protocol?
2^(3-1) = 4
9
For a station X, which is not the rule of MACA?
A. If X received RTS of station A, X must remain silent for a short time.
B. If X received RTS, but didn’t receive CTS, then X may not transmit its data.
C. If X has not received RTS, but received CTS, then X may not transmit its data.
D. If X has received both RTS and CTS, then X may not transmit its data.
B
10
After the sender first sends frames from 0 to 6 and at the end of timeout receives the acknowledgements for frame 1, 3, and 5, the next frame it will re-transmit is which frame? (assume the protocol is go-back-N)
frame 6