self doubt

Question

to calculate min no. of bits required for sequence number field if given for

1.sliding window 2.go back-N 3. selective reject what we do exactly?????

we know Availble sequence no.>= sender window size + receiver window size or we simply take seq no = sender window size

Comments

We know for all cases ;

Availble sequence no.(n-bit) >= sender window size(Ws) + receiver window size(Wr)

Total Possible Sequnce number is : 2 power n .

Cases :

a) Stop and Wait : Ws = 1 ; Wr =1 ( for 1 packet sending at sender side )

b) GO BANCK - N : Ws ; Wr =1  For Ws packet at Sender : Ws + Wr = (2 power n)  => Ws = (2 power n)  - 1 and Wr=1

c) SR Protocol : In SR both Sender and reciver side window size is same .

Ws = Wr = (2 power k) (suppose k bit for Sender Side) . For Ws at Sender  and Wr at reciever : Ws + Wr = (2 power n)  => (2 power k) + (2 power k) = (2 power n) => (2 power k) = (2 power (n-1)) => k = (n-1) .

So , Ws = Wr = (2 power (n-1))

Answer 1

seq no >=sender window size+recievers window size....to avoid duplicate packets we must have this much of seq nos for successful transmission of our data packets....
now when we see GBN....here sender window size is N and reciver's always 1...so we need atleast N+1 sequence nos ..else there is a possiblity of redundancy in packets..
in SR....if i assume senders window size to be N..reciever's wil also be N...means we need atleast 2N seq nos here...

Answer 2

The maximum amount of data that can be sent to the network is BandWidth Delay Product .

So,maximum window size that can be sent is BDP which is BW * RTT.

If N is the number of bits in sequence number field.

SWP=> 2^N = max .window size

GBN= >max.window size + 1 = 2^N

SR= > 2*max.window size = 2^N

Comments

why u r taking max window size

we can take sender window size + receiver window size <=available sequence number(2^n)

where n is the no. of bits for sequence no. field

for each protocol we will use this formula ryt???