Consider a full-duplex transmission link which is 200 km long and has a transmis

Question

Consider a full-duplex transmission link which is 200 km long and has a transmission speed of 10 Mbps in both directions. The propagation delay is 5 msec. per 1000 km. The Stop-and-Wait protocol is being used to transmit data packets which are 2,000 bits long in one direction. You may assume that ACKs returned in the opposite direction are small in size so their transmission time can be ignored. You may also ignore processing times and queuing delays for both data packets and ACKs. (e) What is the round-trip Bandwidth-Delay product? Now compute the utilization from the Bandwidth- Delay product and the packet length (f) If everything else is the same as above, what would be the length of the link if the utilization was only 0.01? (g) If everything else is the same as above, how long should the data packets be for the utilization to be 0.5?

Explanation / Answer

a)

Transmission speed, R = 10Mbps

Link distance, d = 200km

Propagation delay, P1000 = 5ms/1000km Therefore, P200 = (5ms/5)/200km = 1ms/200km

Therefore, 1 bit takes 2ms to make a round trip.

Bandwidth-Delay product = 2ms x 10Mbps = 10000.

Therefore, the system can send 10000 bits during the time it takes for the data to go from the sender to the receiver and back again.

But, the system sends only 2000 bits in one direction i.e. 4000 bits in both directions.

Therefore, Utilization from Bandwidth-Delay product is 4000/10000 = 0.4 (or 40 %).

Also, U = 1/ (1+2a) where, a = (Rd)/ (VL), where

V – Velocity of Propagation = 200km/1ms = 200M m/s, L – Frame length

U = 0.4 i.e. 0.4 = 1/ (1+2a) i.e. (1+2a) = 2.5 i.e. 2a = 1.5 i.e. a = 0.75

(Rd)/ (VL) = 0.75 i.e. (10M x 200k) /(200M x L) = 0.75

i.e. L = 13333 bits (Packet Length)

b)

U = 0.01, L = 13333, V = 200M m/s, R = 10Mbps, d=?

U = 1/ (1+2a) where, a = (Rd)/ (VL)

(Rd)/ (VL) = 49.5 i.e. (10M x d) = 49.5 x (200M x 13333)

i.e. d = 13200 km (Length of link)

c)

P1000km = (5ms)/1000km i.e. P13200km = (5*13.2ms)/13200km = 66ms/13200km

Therefore, 1 bit takes 132ms to make a round trip.

The Bandwidth-delay product = 10Mbps x 132ms = 1320000.

Therefore, the system can send 1320000 bits during the time it takes for the data to go from the sender to the receiver and back again.

But, the system sends only 2000 bits in one direction i.e. 4000 bits in both directions.

Utilization from Bandwidth-Delay product is 0.5.

Therefore, to achieve such utilization, the packet should consist of 0.5 x 1320000 = 660000 bits.

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