Q. A bit string, 0111101111101111110, needs to be transmitted at the data link layer. What is the string actually transmitted after bit stuffing?

五个1插个0

011110111110011111010


Q. 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 / 1001 =  101110 … 111

111是余数


Q.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?
A. bits

发送一帧的时间等于信道传播延迟的2倍,信道利用率为50.

现在发送速率为4kbp/s,发一位需要0.25微秒

(20*10-3*2)/(0.25*10-6) = 160000 bits

 


Q.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.A. for window size=1: bps
A.A. for window size=7: bps
A.for window size=15: bps

注意512-byte = 512 * 8 bits

Tframe = 512 * 8bits/64kbps = 64ms

Transmission cycle: T = 270ms * 2 + 64ms = 604ms

Transmission window: 604ms/64ms = 9.4, 大于9的时候channel就满了

W = 1 ; throughout = 512 * 8b / 604ms = 6781bps = 6.78kbps

W = 2 ; throughout = 7 * 512 * 8b / 604ms

W = 15 > 9 ; throughout = 65kbps


Q.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?
A. bits

电缆中传播速度是每秒200000km,每毫秒200km,100km将会在0.5ms内填满。

T1速率125微秒传送193位帧,0.5ms可以传送4个T1帧,即193*4 = 772bits


Assume the sequence number has 4 bits. What is the maximum number of outstanding sending frames for a go back N protocol?

Box 1: Enter your answer as a number. Examples: 3, -4, 5.5
Enter DNE for Does Not Exist, oo for Infinity

序列号有4位,回退N协议,最多没有完成发送帧有多少?

四位可以表示0~15共16个数字 最多没有完成15个。


Assume the sequence number has 5 bits. What is the maximum number of outstanding sending frames for a selective repeat protocol?

Box 1: Enter your answer as a number. Examples: 3, -4, 5.5
Enter DNE for Does Not Exist, oo for Infinity

序列号有4位,选择重传协议,最多没有完成发送帧有多少?

五位可以表示0~31共32个数字 最多没有完成(max+1)/2 = 33/2 = 32/2 = 16


Which is not the CSMA/CA rule of 802.11?
  • If station X received RTS of station A, X must remain silent for a short time
  • If station X received RTS, but did not receive CTS, then X may not transmit its data.
  • If station X has not received RTS, but received CTS, then X may not transmit its data
  • If station X has received both RTS and CTS, then X may not transmit its data

Box 1: Select the best answer

B


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 frame ________. (assume the protocol is go-back-n)
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