pratice questions
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batz107
Apr 13, 2010, 4:19:08 PM4/13/10
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השאלה שלי לקוחה מהשאלות חזרה שפרסמה חיה באתר.
בשאלה 2 סעיף א', יש הכפלה ב
log50
בשאלה 2 סעיף א', יש הכפלה ב
log50
מישהו יכול להסביר את החישוב ?
תודה
yaron winter
Apr 14, 2010, 7:43:03 AM4/14/10
to 89...@googlegroups.com
It's an approximation for HTTP that was developed in the class (see the HTTP lecture presentation).
In summary:
It was proven the if
File Size < 2*R*RTT + MSS
Then the total transferring time of the file by the HTTP can be well approximated by
T ~ (File Size)/R + RTT*log2(File Size/MSS)
Now, in our example the condition (size < 2*R*RTT + MSS) is not satisfied, but the above is still a good approximation (which is what we were asked for).
Yaron
In summary:
It was proven the if
File Size < 2*R*RTT + MSS
Then the total transferring time of the file by the HTTP can be well approximated by
T ~ (File Size)/R + RTT*log2(File Size/MSS)
Now, in our example the condition (size < 2*R*RTT + MSS) is not satisfied, but the above is still a good approximation (which is what we were asked for).
Yaron
2010/4/13 batz107 <bat...@gmail.com>
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batz107
Apr 14, 2010, 8:18:25 AM4/14/10
to 89350
שלום ירון,
לפי מה שאני רואה בהרצאה 9
לפי מה שאני רואה בהרצאה 9
T<= RTT + RTT' * log2(Filse Size/MSS)
או
T~ RTT * log2 (fileSize/MSS)
אך לפי התשובה שפורסמה, האיברים אינם מסתדרים לי בנוסחא
RTT - זה לשם הקמת קשר
אך שידור האובקייטים וקצב ההתפשטות אינו מסתדר בנוסחא.
האם אתה יכול להבהיר זאת ?
תודה
yaron winter
Apr 14, 2010, 8:46:22 AM4/14/10
to 89...@googlegroups.com
Hi,
It is a matter of crude approximation and some algebra:
the time, in the exercise, is approximated as follows:
connection establishment time + transmission time + propagation time * 2 * log2(50)
RTT = propgation time * 2
and
50 = File Size / MSS,
thus we have the component
RTT*log2(size/MSS)
as desired
Also, in the lecture they used the term RTT', which is RTT + MSS/R
Thus we obtain:
RTT ' * log2(size/MSS) ~ RTT*log2(size/MSS) + transmission time
The first component (RTT) corresponds to the connection establishment time, thus we got a completely equivalent expression!
Since according to this question's circumstances the RTT is negligible with respect to transmission time, they use the formula although its condition (i.e. file size < 2*R*RTT + < MSS) is not satisfied.
I hope it's clearer now,
Yaron
It is a matter of crude approximation and some algebra:
the time, in the exercise, is approximated as follows:
connection establishment time + transmission time + propagation time * 2 * log2(50)
RTT = propgation time * 2
and
50 = File Size / MSS,
thus we have the component
RTT*log2(size/MSS)
as desired
Also, in the lecture they used the term RTT', which is RTT + MSS/R
Thus we obtain:
RTT ' * log2(size/MSS) ~ RTT*log2(size/MSS) + transmission time
The first component (RTT) corresponds to the connection establishment time, thus we got a completely equivalent expression!
Since according to this question's circumstances the RTT is negligible with respect to transmission time, they use the formula although its condition (i.e. file size < 2*R*RTT + < MSS) is not satisfied.
I hope it's clearer now,
Yaron
2010/4/14 batz107 <bat...@gmail.com>
תודה
batz107
Apr 14, 2010, 8:54:23 AM4/14/10
to 89350
תודה רבה, ההסבר היה ברור מאוד!
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