self dobutCN

Question

Can u plz explain For sliding window protocol,what is expression for following:

i)Window size

ii)Sequence number

iii)max efficiency

 

 

Somewhere i saw Window size= 1+ 2a [a=tp/tt]     seq no =1+2a and max eff =window size/(1+2a)   how it can be possibe ?eff will be 1.

Comments

@Magma @Gupta731

---

Sender window size will be $1+2a$ for maximum efficiency. Otherwise you will be specified with $w_s$ as in Go back N, here N is the sender window size with which you will calculate the efficiency η$=\frac{N}{1+2a}$.

Minimum number of Sequence Number required is calculated by $1+2a$ and the number of bits required to represent them is calculated by taking log of it as you know.

In sliding window protocol the window size is not actually $1+2a$, it is even limited by the number of sequence number available.

---

@Gupta731 maximum efficiency become 1 if Sender window size 1+ 2a ?? Minimum no of sequence for maximum efficiency become ln(1+2a) else ln(N+1) for go back N and ln(N+N) for selective repeat???

 

---

That is $log_2$ not $ln$ i.e $log_e$.

For Go back N  $log_2 (1+2a)$

For Selective repeat $log_22(1+2a)$

Answer 1

Let me clarify you

There are three types of sliding window protocols

1.stop and wait

2.go back n

3.selective repeat

In stop and wait at sender window size is 1 and receiver window size is 1

efficiency formula = TT/TT+2PT if you divide with TT num and denominator it would be 1/1+2a (a=TP/TT)

Throughput = efficiency * bandwidth

cycle time i.e total time taken to send a packet = TT+TP(data)+TP(ACK)+TT(ACK)[It is negligible]

hence cycle time = TT+2TP

only 1 packet is transmitted onto the link and then sender waits for acknowledgement from the receiver. The problem in this setup is that efficiency is very less as we are not filling the channel with more packets after 1st packet has been put onto the link.

Let's see Go Back N: In This Sender window size is 'N'

It Sends N Packets in a cycle time hence

Efficiency = N*TT/TT+2TP i.e (N/1+2a)         N is Window size

In Go Back N sender window size is N and receiver window size = 1

hence total seq num required = N+1

bits in seq num field = ceil log base2 (N+1)

where as is selectives repeat window size at sender = N , Receiver window size =N

Hence efficiency = N/1+2a

seq num required is N+N=2N  (N is window size)

Bits in seq num field = ceil log base (2N)