a sliding window protocol of 4mbps point to point link has propagation delay of 0.5sec. assume that each frame carries 2kb of data. what is the minimum no. of bits used for sequence number field?
Answers
Answer:
8 bit
Explanation:
propagation delay(Tp)=0.5 sec
bandwidth(B)= 4 Mbps
length of data =2 KB
Transmission delay(Tt)=length of data / bandwidth = 2 KB/ 4 Mbps= 2000 sec
we need no. of bits used in sequence number field, so that we have to calculate how many sequence no. are there in sender window.
sequence number =1 +2*(Tp/Tt)
= 1 + 2 *(0.5/2000)
= 251
bit required = ㏒₂(251)
= 7.97
so,min. no of bits required for sequence no. field is 8.
Answer:
8 bits
Explanation:
If length of data = 2KB (given)
Transmission delay(Tt)=length of data / bandwidth = 2 KB/ 4 Mbps = 4.096 ms (1KB = 2^10 * 8 bits and 1mbps =10^6 bps).
We need no. of bits used in sequence number field, so that we have to calculate how many sequence no. are there in sender window.
Total sender window size =1 +2*(Tp/Tt)
= 1 + 2 *(0.5/4.096)
= 251
Suppose if we are using go back n protocol, sequence no.>= sender window size + receiver window size to avoid any data duplicacy.
So, min sequence no. = 251 +1 = 252 (bits = 8) given below.
Suppose if we are using selective repeat, min sequence no. = 251 + 251 =502 (bits = 9)
For minimum bits required = ㏒₂(252)
= 7.97
so, minimum no. of bits required for sequence no. field is 8.