Respuesta :
- The maximum window size (in segments) that this TCP connection can achieve is 125
- The average window size (in segments) and average throughput (in bps) of this TCP connection are 93.75 and [tex]7.5 * 10^6 bp[/tex]
- To take for this TCP connection to reach its maximum window again after recovering from a packet loss is about 9.375 seconds.
Explanation:
Transmission Control Protocol (TCP) is responsible to break up the message (in this case is the data from application layer) into TCP Segments and then reassembling them at the receiving side
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?
[tex]\frac{(Window Size) (Maximum Segment Size)}{(Round Trip Time)} = \frac{(W) (MSS)}{(RTT)} = rate[/tex]
[tex]\frac{(W) (1500 * 8)}{150 *10^-{3}} = 10 * 10^6[/tex]
[tex]W=125[/tex]
b. What is the average window size (in segments) and average throughput (in bps) of this TCP connection?
Average window size [tex]= 0.75W = (0.75) (125) = 93.75[/tex]
Average throughput [tex]= \frac{(Wavg ) (MSS)}{(RTT) } = \frac{(93.75) (1500 * 8)}{150 *10^-3} \\=7.5 * 10^6 bp[/tex]
c. How long would it take for this TCP connection to reach its maximum window again after recovering from a packet loss?
The number of RTTs needed in order to increase the window size from [tex] W/2[/tex] to [tex]W = \frac{W}{2}* RTTs[/tex]
[tex]W= (\frac{125}{2} ) (150 * 10^{-3}) =9.375 sec[/tex]
Learn more about TCP connection https://brainly.com/question/14281718
#LearnWithBrainly
A) The maximum window size (in segments) that this TCP connection can achieve is; 125
B) The average window size (in segments) and average throughput (in bps) of this TCP connection are; 93.75 and 7.5 * 10⁻⁶ bps
C) The time it will take for this TCP connection to reach its maximum window again after recovering from a packet loss is; 9.375 sec
What is TCP Connection Protocol?
A) The maximum window size (in segments) that the TCP connection can achieve is given by the formula;
Window size = (Rate * Round Trip Time)/Maximum Segment size
Plugging in the relevant values from the question gives;
Window size = ((10 * 10⁻⁶ * 150 * 10⁻³)/(1500 * 8)
Window size = 125
B) The average window size (in segments) is given by;
Average = 0.75 * Window size
Average = 0.75 * 125
Average Window size = 93.75
Formula for the average throughput (in bps) is;
Average Throughput = (Average Window Size * Maximum Segment size)/Round trip time
Average Throughput = (93.75 * 1500 * 8)/(150 * 10⁻³)
Average Throughput = 7.5 * 10⁻⁶ bps
C) The time needed to increase the window size to its maximum is;
W_max = (Window size/2) * Round trip time
W_max = (125/2) * (150 * 10⁻³)
W_max = 9.375 sec
Read more about TCP Connections at; https://brainly.com/question/8156649
