From Kurose, 5th edition. Ch. 3, P43: Consider that only a single TCP (Reno) con

Question

From Kurose, 5th edition.
Ch. 3, P43: Consider that only a single TCP (Reno) connection uses one 10Mbps link which does not buffer any data. Suppose that this link is the only congested link between the sending and receiving hosts. Assume that the TCP sender has a huge file to send to the receiver, and the receiver’s receive buffer is much larger than congestion window. We also make the following assumptions: each TCP segment size is 1,500 bytes; the two-way propagation delay of this connection is 100 msec; and this TCP connection is always in congestion avoidance phase, that is, ignore slow start.

(a) What is the maximum window size (in segments) that this TCP connection can achieve?
My answer:
To find the maximum size of the window, N, we must assume we want 100% usage:
• d-trans = L/R = 12000 bits / 10^10 = 0.00000012 secs = 0.00012 ms = .12 microseconds.
• U = (L/R)/(L/R + RTT) = 0.00012ms/30.00012ms = 0.0000012
• N*U = 1.0 ? N = 1/0.0000012 = 833333 packets.

(b) What is the average window size (in segments) and average throughput (in bps) of this TCP connection?
(c) How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?


What I'm thinking:
I have to use the window formulas somehow, but I don't know the applications of them. The equation that I see is: N*(L/R)/(L/R + RTT) where N is the window amount, L is packet size, R is transmission rate, and RTT is the two-way transmission delay.

Explanation / Answer

Given Link capacity C = 10 Mbps = 10 x 10^6 bps RTT = 100 msec = 0.1 sec MSS = 1500 bytes = 1500 x 8 bits Maximum window size, say Wmax = (RTT x C) / MSS = (0.1 x 10 x 10^6) / 1500 x 8 = 83.3333 = 84 segments Average window size, say W = 0.75 x Wmax = 0.75 x 84 = 63 segments Average throughput = (W x MSS) / RTT = (63 x 1500 x 8) / 0.1 = 7.56 bps Time taken = Wmax / 2 x RTT = (84 / 2) x 0.1 = 4.2 sec Hope this will help you

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