Question

Stop-and-Wait Flow Control & Asynchronous Transmission: A node wants to communicate with another node that is 3000 meters away. The transmitted signal propagates at a speed of 3x108 m/s. The node sends frames of length 1500 bytes at a data rate of 1 Gbps. The node employs Stop-and-Wait flow control.

a) What is the utilization of the system?

b) What is the maximum distance a receiver node can be at such that the utilization is at least 0.75?

c) Asynchronous Transmission: A receiver’s clock drifts at a rate of 0.5 seconds every day. The receiver can synchronize itself with the transmitter at the beginning of each frame but not during the frame. It samples the middle of a bit to decode whether it’s a 0 or 1. Will the system described above work well with such a receiver? What is the maximum frame size that will work fine?

Answer 1

(a)

distance between two nodes = 3000 meters

speed = 3*10⁸ m/s

length of frame = 1500 bytes = (1500*8) bits = 12000 bits

bandwidth = 1Gbps = 10⁹ bps

We have to find utilization of the system. Utilization can be given by the below formula :

utilization = 1/(1+2*)

is the ratio of propagation delay(T_p) to the transmission time(T_t).

Transmission time(T_t) = length of frame / bandwidth = 12000 / 10⁹ = 12 * 10^-6 seconds

Propagation delat(T_p) = distance/speed = 3000/ 3*10⁸ = 10^-5 seconds

= T_p /T_t = 10^-5 / 12 * 10^-6 = 5/6

utilization = 1/ (1+2*(5/6)) = 0.375

Answer a: utilization = 0.375

(b)

utilization = 0.75

we have to find the distance. Let distance be d meters.

speed = 3*10⁸ m/s

length of frame = 1500 bytes = (1500*8) bits = 12000 bits

bandwidth = 1Gbps = 10⁹ bps

Transmission time(T_t) = length of frame / bandwidth = 12000 / 10⁹ = 12 * 10^-6 seconds

Propagation delat(T_p) = distance/speed = (d / 3*10⁸ ) seconds

using the formula :

utilization = T_p / T_p + 2*T_t

0.75 = (d / 3*10⁸) / ((d / 3*10⁸) + 12 * 10^-6)

On solving we get, d = 10800 meters

Answer b: maximum distance for a utilization of 0.75 is 10800 meters.

(c)

since the clock is drifting at the receiver side the system will not work well. As the receiver is not able to merge the packets and forward them. The maximum frame size is the one whose transmission time is less than the drift time.

if T_t = 0.5 seconds

bandwidth = 1Gbps = 10⁹ bps

Transmission time(T_t) = length of frame / bandwidth

length of frame = T_t* bandwidth = 0.5 * 10⁹ = 5*10⁸ bits

converting length into bytes =  (5*10⁸) / 8 = 6.25 * 10⁷ bytes

Answer c: the maximum frame size that will work fine is 6.25 * 10⁷ bytes