Solution of Home Work 7 1) There are two computers, called S (Source) and D (Destination). They are connected by a 20 km long link (full duplex, used only between S and D). The link has a data rate of 100 Mbit/sec and the
speed of the signal on the link is (2/3)*c .
Practically speaking, we have only 1500 Byte dataframes from S to D, and only 50 Byte ACKframes from D to S. This includes headers,
trailers, ``wrappers'', etc. D acknowledges every dataframe it
receives as soon as the whole frame is in.
If S and D use a ``Sliding Window'' protocol, what window size (in frames) would you recommend? WHY?
(The ``why'' is much more important than the actual answer, so try to be simultanously concise and very clear!)
Answer:
Serialization delay of a data frame is:
1500 * 8
------------------ = .12msec
100, 000, 000
Propagation delay is: 20
------------------ = .1msec
(2/3)*3,00,000
Serialization delay of an ACK frame is: 400
--------------- = .004msec
100,000,000
At time t, when S starts sending a data frame, the tail of the frame leaves S at (t + .12) msec. The same frame arrives at D after the time (t + .12 + .1) msec.
Assume an ACK is generated in D at the same time it receives a data frame. Then the ACK will leave D at the time, (t + .12 + .1 + .004) msec.
So, the ACK for the data frame will arrive at S after (t + .12 + .1 + .004 + .1) msec, which is (t + 0.324) msec.
So, S needs to have data frames to send at t, (t + .12) and (t + .324). So, the window size must be 3 to prevent the bandwidth from remaining unused.
2) In a bit-oriented protocol, we use the flag 01111110 , to indicate beginning and end of a frame.
(I usually would write this as 0111 1110 , for readability). A. Bitstuff the following pattern: 01001111111001111100110011110 ^ ^ | | The answer is: 0100111110110011111000110011110 B. ``Unstuff'' the following pattern: 0110111110111110110011111001010 - - - Answer is: 0110111111111111001111101010 C. What is your reaction if asked to unstuff the following pattern? 01001111110010