Question

You are designing a reliable byte stream transport layer protocol (not TCP) to operate over a 1x107 bps network and it is using a sliding window for flow control. The time for keeping the transmission pipe full is taken to be the RTT of the network which is 50ms. Each number in the advertised window or sequence number represents 2 bytes of data.

a) What is the minimum number of bits necessary for the Advertised Window field of your protocol header? Provide this minimum number as a whole number (round up).

b) How would you choose the Sequence number size in accordance with the Advertised Window size determined in A) c) Given this sequence number size, how long will it take for it to wrap around?

Answer 1

Solution:

Given,

=>Bandwidth(B) = 1*107 bps

=>Round trip time (RTT) = 50 ms

=> Each number in the advertized window or sequence number represents 2 bytes of data.

(a)

Explanation:

=> We know that window size(W) = Bandwidth*RTT

=>W = 107 bps *50 ms

=> W = (107/8) Bps *50*10-3 sec                   (1 byte = 8 bits and 1 second = 1000 msec)

=> W = 62500 bytes

Minimum number of bits required = seal(log262500) = 16 bits

So the minimum number of bits required for advertised window = 16 bits

(b)

Explanation:

=> As we have found advertised window is of 16 bits and the sequence number field is of 2 bytes.

=>Sequence number field = 2 bytes = 16 bits            (1 bytes = 8 bits)

(c)

Explanation:

We know that,

=> Sequence number bits = log2(Wrap around time*bandwidth)

Here wrap around time (WAT) is in seconds and bandwidth is in bytes.

=> 16 = log2(WAT*107/8)

=>WAT = 216 *8/107 seconds

=>WAT = 219/107 seconds

=> WAT = 0.0524288 seconds

I have explained each and every part with the help of formulas as well as statements attached to it.