(a) Stop and wait flow control

(b) Sliding window flow control with window sizes of 4 and 7 

Link specifications:

Frame length = 1000 bits/frame

Velocity of propagation = 2 x 10^8 m/sec

Link distance = 20 km

Data rate = 20 Mbps

Electronics > Sem 6 > Computer Communication Networks

Protocol performance:

• The throughput efficiency is the measure of the performance of an ARQ protocol. For any channel a certain bandwidth and bit error rate are specified.
• For such a channel there will be an optimum operating condition that will support for the maximum "Net Data Throughput" (NDT).
• NDT indicates the number of usable characters detected at the receiver.
• It indicates the number of correct bits detected in a specified period of time.
• This is done by distinguishing between the total number of bits received ( including the check bits) other number of correct bits. Throughput efficiency is defined as:

$\eta = t_f/ (t_f + 2t_p)$

where

$t_f$ = transmission time required to transmit a frame

$t_p$ = propagation time required to reach destination for a transmitted bit

N = frame size in bits

R = data rate

Suppose A is a sender and B is a receiver then the assumptions are as follows

Assumptions: Receiver sends an immediate acknowledgement on the reception of a data frame. Size of acknowledgement frame is very small. Flow is unidirectional. Sender receives acknowledgement after tf+tp+tp time. It can send data immediately after receiving the acknowledgement.

If tf and tp constant, tp/tf is constant. let $A = t_p/t_f$ Therefore, $\eta = 1/(1+2A)$

Propagation time is equal to distance d of the link divided by velocity of propagation v.

$t_p = d/v$

Transmission time is equal to the length of the frame in bits, divided by the rate R.

$t_f = L/R$

$A = (d/v)÷(L/R) = Rd/Lv$

Propagation delay time = $V/d = 0.1 msec$

1. $\eta = L / (L+BT) $

   $\eta = 1000/ (1000+20*10^6*0.1*10^{-3})$

   $\eta = 0.333$ or 33.33%

2. For window size of 7

   $\eta = L / (L+BT) $

   $\eta = (1000*7)/ (1000*7+20*10^6*0.1*10^{-3})$

   $\eta = 0.77$ or 77.77 %

3. For window size of 4

   $\eta = L / (L+BT) $

   $\eta = (1000*4)/ (1000*4+20*10^6*0.1*10^{-3})$

   $\eta = 0.66$ or 66.66%