You are given a 1000 mile link with a 1 Gb/s data rate between a sender and rece

Question

You are given a 1000 mile link with a 1 Gb/s data rate between a sender and receiver. All data is sent in 5000 byte packets. For the Stop-and-Wait (SAW) protocol determine the link utilization assuming that 1) the sender always has data to send, 2) the overhead from processing and Ack transmission time is neglible, 3) the overhead from packet headers is negligible, and the link is error-free (no packets are lost). a) Determine the link utilization for the above described scenario. b) Change the above original scenario to be a 1000 foot link, determine the link utilization. c) Change the above original scenario (the 1000 mile link case) to use a Sliding Window protocol with windows size 10 packets, determine the link Utilization. d) For the above original scenario, what is the window size needed for a Sliding Window protocol to achieve 100% link utilization?

Explanation / Answer

Link utilization in slinding window protocol=(WSTt) / (Tt+2Tp)

where WS = Window size

Tt (transition time) = L / B (length of packetss/ bandwidth)

Tp (Prpagation time) = D / V (distance / velocity)

a)

L = 5000 bytes = 5000 * 8 bits = 4 * 104 bits

B = 1 Gb / s = 10 9  bps

D = 1000 miles = 1600 kms

Tt = (4 * 104 ) / 10 9 = 0.4 us = 0.04 ms

Tp = 5 ms based on propagation of about 1 foot per nanosecond and about 5000 feet in a mile.

WS for SAW = 1

LU = (WSTt) / (Tt+2Tp) = (Tt) / (Tt+2Tp) = 0.04 / (0.04 + 10) = 0.04/ 10.04 = 0.0039

c)

L = 10 packets = 10 * 5000 * 8 bits = 4 * 10 6 bits

B = 1Gb /s

Tt = 4 * 106 / 10 9 = 4 ms

Tp = 5ms

LU = (WSTt) / (Tt+2Tp)

LU = (Tt) / (Tt+2Tp) = 4 / (4 * 2(5)) = 4/14 = 0.285

d)

For SW we know that

LU = (WSTt) / (Tt+2Tp)

when LU = 1

Ws = (Tt + 2 * Tp) / Tt = (10.04) / 0.04 = 251

window size = 251

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