EBK COMPUTER NETWORKING
7th Edition
ISBN: 8220102955479
Author: Ross
Publisher: PEARSON
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Expert Solution & Answer
Chapter 3, Problem P46P
a.
Explanation of Solution
Given:
Link capacity= 10 Mbps
TCP segment size= 1500 bytes
Finding maximum window size:
Let the variable “X” represent the maximum of window’s size in segments.
Consider the expression,
Substitute the given values as follows:
b.
Explanation of Solution
Given:
Link capacity= 10 Mbps
TCP segment size= 1500 bytes
Let the variable “X” represent the maximum of window’s size in segments.
Finding average throughput:
For example the congestion window size “cwnd” varies from “X/2” to “X”, then the average window size is as follows:
c.
Explanation of Solution
Given:
Link capacity= 10 Mbps
TCP segment size= 1500 bytes
Let the variable “X” represent the maximum of window’s size in segments.
Time taken for recovering from packet loss:
When there is a loss while sending packets, “X” becomes “X/2”, that is
Expert Solution & Answer
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Students have asked these similar questions
TCP congestion control example. Consider the figure below, where a TCP sender sends 8
TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8. Suppose the initial value of the sequence number is 0
and every segment sent to the receiver each contains 100 bytes. The delay between the
sender and receiver is 5 time units, and so the first segment arrives at the receiver at t = 6.
The ACKs sent by the receiver at t = 6, 7, 8, 10, 11, 12 are shown. The TCP segments (if any)
sent by the sender at t = 11, 13, 15, 16, 17, 18 are not shown. The segment sent at t=4 is lost,
as is the ACK segment sent at t=7.
TCP
sender
t=1 T
t=2
t=3
t=4+
t=5-
t=6+
t=11
t=12
t=13
t=14
t=15
t=16
t=17
t=18
I
data segment
data segment
data segment
data segment
data segment
data segment
data segment
data segment
ACK
ACK
ACK
ACK
ACK
ACK
Ty
A A
V V
htt
TCP
receiver
t=6
t=7
t=8
t=9
t=10
t=11
t=12
t=13
What does the sender do at t=17? You can assume for this question that no timeouts have
occurred.
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 the 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 150 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?
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?
TCP congestion control example. Consider the figure below, where a TCP sender sends 8 TCP segments at t = 1, 2, 3, 4, 5, 6, 7, 8. Suppose the initial value of the sequence
number is 0 and every segment sent to the receiver each contains 100 bytes. The delay between the sender and receiver is 5 time units, and so the first segment arrives at the
receiver at t = 6. The ACKS sent by the receiver at t = 6, 7, 8, 10, 11, 12 are shown. The TCP segments (if any) sent by the sender att = 11, 13, 15, 16, 17, 18 are not shown. The
segment sent at t=4 is lost, as is the ACK segment sent at t=7.
t=1 T
data segment
t=2+
data segment
data segment--
t=3
TCP
sender
TCP
receiver
t=4+
t=5+
data segment
- data segment
t=6+
t36
data segment
t=7
data segment
t=8
data segment
t=9
ACK
+ t=10
k --
ACK
t=11
t=11
t=12
t=12
t=13
t=13
t=14
ACK
-ACK
ACK
t=15
t=16
t=17
ACK
t=18
What does the sender do at t=17? You can assume for this question that no timeouts have occurred.
Chapter 3 Solutions
EBK COMPUTER NETWORKING
Ch. 3 - Prob. R1RQCh. 3 - Prob. R2RQCh. 3 - Consider a TCP connection between Host A and Host...Ch. 3 - Prob. R4RQCh. 3 - Prob. R5RQCh. 3 - Prob. R6RQCh. 3 - Suppose a process in Host C has a UDP socket with...Ch. 3 - Prob. R8RQCh. 3 - Prob. R9RQCh. 3 - In our rdt protocols, why did we need to introduce...
Ch. 3 - Prob. R11RQCh. 3 - Prob. R12RQCh. 3 - Prob. R13RQCh. 3 - Prob. R14RQCh. 3 - Suppose Host A sends two TCP segments back to back...Ch. 3 - Prob. R16RQCh. 3 - Prob. R17RQCh. 3 - Prob. R18RQCh. 3 - Prob. R19RQCh. 3 - Prob. P1PCh. 3 - Prob. P2PCh. 3 - UDP and TCP use 1s complement for their checksums....Ch. 3 - Prob. P4PCh. 3 - Prob. P5PCh. 3 - Prob. P6PCh. 3 - Prob. P7PCh. 3 - Prob. P8PCh. 3 - Prob. P9PCh. 3 - Prob. P10PCh. 3 - Prob. P11PCh. 3 - Prob. P12PCh. 3 - Prob. P13PCh. 3 - Prob. P14PCh. 3 - Prob. P15PCh. 3 - Prob. P16PCh. 3 - Prob. P17PCh. 3 - Prob. P21PCh. 3 - Prob. P22PCh. 3 - Prob. P25PCh. 3 - Prob. P26PCh. 3 - Prob. P27PCh. 3 - Host A and B are directly connected with a 100...Ch. 3 - Prob. P29PCh. 3 - Prob. P30PCh. 3 - Prob. P31PCh. 3 - Prob. P33PCh. 3 - Prob. P34PCh. 3 - Prob. P35PCh. 3 - Prob. P37PCh. 3 - Prob. P38PCh. 3 - Prob. P39PCh. 3 - Prob. P41PCh. 3 - Prob. P42PCh. 3 - Prob. P43PCh. 3 - Prob. P44PCh. 3 - Prob. P45PCh. 3 - Prob. P46PCh. 3 - Prob. P47PCh. 3 - Prob. P48PCh. 3 - Prob. P49PCh. 3 - Prob. P51PCh. 3 - Prob. P53PCh. 3 - Prob. P55P
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