Question 7941 – Computer-Networks

The value of parameters for the Stop-and-Wait ARQ protocol are as given below:

          Bit rate of the transmission channel = 1 Mbps.
          Propagation delay from sender to receiver = 0.75 ms.
          Time to process a frame = 0.25 ms.
          Number of bytes in the information frame = 1980.
          Number of bytes in the acknowledge frame = 20.
          Number of overhead bytes in the information frame = 20.

Assume that there are no transmission errors. Then, the transmission efficiency (expressed in percentage) of the Stop-and-Wait ARQ protocol for the above parameters is _________ (correct to 2 decimal places).

Correct Answer: A

Given Data:
B = 1Mbps, L = 1980Bytes, Overhead = 20Bytes
T_Proc = 0.25ms, L_Ack = 20Bytes
T_p=0.75ms
Total Data size(L) = (L + overhead) = 1980+20 = 2000Bytes
Efficiency of Stop & Wait ARQ?
T_t = L/B = 2000Bytes/1Mbps = (2000×8bits)/(10⁶ b/s) = 16msec
T_Ack = L_Ack/B = (20×8bits)/(10⁶ bits/sec) = 0.16msec
∴ In Stop and Wait ARQ, efficiency
ƞ = T_t/(T_t+T_Ack+2T_p+T_Proc) = 16ms/(16+0.16+2×0.75+0.25ms) = 16ms/17.91ms = 0.8933