Computer Networking Management
Task -1
a) For each segment, making any necessary assumptions, calculate the following and plot the data on a graph:
• An appropriate timeout interval for use when transmitting the next TCP segment.
Estimated RTT =
We can calculate Estimated RTT by using the above calculation Formula
If α = 0.1 in this problem
We can derive exponentially weighted moving average (EWMA) of RTT
For First 10 segments:
1) (0.9)9 x 127 + (0.1) x (0.9)) 8 x 134 + (0.1) x (0.9)) 7 x 121+ (0.1) x (0.9)) 6x 139+ (0.1) x (0.9)) 5 x 139+ (0.1) x (0.9)) 4 x 131 + (0.1) x (0.9)) 3 x 124 + (0.1) x (0.9)) 2x 122 + (0.1) x (0.9)) 1x 128
+ (0.1) x130
For Second 10 segments:
(0.9)9 x 139 + (0.1) x (0.9)) 8 x 134 + (0.1) x (0.9)) 7 x 123+ (0.1) x (0.9)) 6x 291+ (0.1) x (0.9)) 5 x 137+ (0.1) x (0.9)) 4 x 141 + (0.1) x (0.9)) 3 x 134 + (0.1) x (0.9)) 2x 126 + (0.1) x (0.9)) 1x 139
+ (0.1) 267
Figure: Segment vs. EWMA
For Third 10 segments:
(0.9)9 x 127 + (0.1) x (0.9)) 8 x 134 + (0.1) x (0.9)) 7 x 121+ (0.1) x (0.9)) 6x 139+ (0.1) x (0.9)) 5 x 139+ (0.1) x (0.9)) 4 x 131 + (0.1) x (0.9)) 3 x 124 + (0.1) x (0.9)) 2x 122 + (0.1) x (0.9)) 1x 128
+ (0.1) x130
For reference, alpha = .1
• The variability of RTT (sometimes referred to as Dev RTT).
Estimated RTT =54.96
Sample RTT =127
Difference = Sample RTT – Estimated RTT
=>127-54.96
=> 54.96
As per Jacobson/Karels Algorithm,
Deviation = Deviation + d ( |Difference| – Deviation)), where d is a fraction between 0 and 1
If, Deviation = 5, d= 0.01
5 + 0.01 ( 54.96 – 5))
5.4996
Deviation =5.4996
An appropriate timeout interval for use when transmitting the next TCP segment.
First need to Measure Sample RTT for each segment/ACK