Efficiency of Stop and Wait Protocol
Last Updated :
10 May, 2023
Stop and Wait is a flow control protocol. In which the sender sends one packet and waits for the receiver to acknowledge and then it will send the next packet. In case if the acknowledgement is not received, the sender will retransmit the packet. This is the simplest one and easy to implement. but the main disadvantage is the efficiency is very low. Total time taken to send one packet,
= Tt(data) + Tp(data) + Tq + Tpro + Tt(ack) + Tp(ack)
Since,
Tp(ack) = Tp(data)
And,
Tt(ack) << Tt(data).
So we can neglect Tt(ack)
Tq = 0 and Tpro = 0
Hence,
Total time = Tt(data) + 2 * Tp
Where,
Tt(data) : Transmission delay for Data packet
Tp(data) : propagation delay for Data packet
Tq: Queuing delay
Tpro: Processing delay
Tt(ack): Transmission delay for acknowledgment
Tp(ack) : Propagation delay for acknowledgment
We know that the Efficiency (η),
= Useful time / Total cycle time.
= Tt / (Tt + 2*Tp)
= 1 / (1+2*(Tp/Tt))
= 1 / (1+2*a)
where,
a = Tp / Tt
if Tt(ack) :Transmission delay for acknowledgment , Tq: Queuing delay and Tpro: Processing delay is mention
We know that the Efficiency (η),
=Useful time / Total cycle time
=Tt(data) /Tt(data) + Tp(data) + Tq + Tpro + Tt(ack) + Tp(ack)
Above formula is applicable for any condition, if any of the things are not given we assume it to be 0.
Other name of Efficiency are Line Utilization or Link Utilization or Sender Utilization.
Throughput: Number of bits send per second, which is also known as Effective Bandwidth or Bandwidth utilization.
Throughput=Length of frame/Total Cycle time
=L/(Tt(data) + Tp(data) + Tq + Tpro + Tt(ack) + Tp(ack)) —1
=(L/BW)*BW)/(Tt(data) + Tp(data) + Tq + Tpro + Tt(ack) + Tp(ack))
=Tt(data)*BW/(Tt(data) + Tp(data) + Tq + Tpro + Tt(ack) + Tp(ack))
= η*BW –2
we know that Tt(data)=L/BW
where Tt(data)=Transmission Delay
BW=Band Width
Above formula is applicable for any condition, if any of the things are not given we assume it to be 0.
Other name of Throuhput are Effective Bandwidth or Bandwidth Utilization or Maximum data rate possible.
Throughput,
= L/(Tt + 2*Tp)
= ((L/BW)*BW)/(Tt + 2*Tp)
= Tt/(Tt + 2*Tp) * BW
= 1/(1 + 2a) * BW
Hence, Throughput
= η * BW
where,
BW : BandWidth
L : Size of Data packet
Factors affecting Efficiency:
n = 1/(1 + 2*(Tp/Tt)
= 1/(1 + 2*(d/v)*(BW/L))
where,
d = distance between source and receiver
v = velocity
Lets see an example. Example: Given,
Tt = 1ms
Tp = 2ms
Bandwidth = 6 Mbps
Efficiency(η)
= 1/(1 + 2*a)
= 1/(1 + 2*(2/1))
= 1/5
= 20 %
Throughput
= η * BW
= (1/5) * 6
= 1.2 Mbps
Note: As we can observe from the above given formula of Efficiency that:
- On increasing the distance between source and receiver the Efficiency will decrease. Hence, Stop and Wait is only suitable for small area network like LAN. It is not suitable for MAN or WAN, as the efficiency will be very low.
- If we increase the size of the Data packet, the efficiency is going to increase. Hence, it is suitable not for small packets. Big data packets can be send by Stop and Wait efficiently.
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