Consider a scenario that we derieved five SampleRTT values : 80 ms, 120 ms, 140
ID: 3924990 • Letter: C
Question
Consider a scenario that we derieved five SampleRTT values : 80 ms, 120 ms, 140 ms, 90 ms, and 110 ms. Compute the EstimatedRTT after each of these SampleRTT values is obtained, using a value of = 0.125 and assuming that the value of EstimatedRTT was 100 ms just before the first of these five samples were obtained. Compute also the DevRTT after each sample is obtained, assuming a value of = 0.25 and assuming the value of DevRTT was 5 ms just before the first of these five samples was obtained. Last, compute the TCP TimeoutInterval after each of these samples is obtained.
Explanation / Answer
we know that, EstimatedR TT = xSampleRTT + (1 x ) EstimatedR TT
value of EstimatedRTT was 100 ms and value of = 0.125,
After obtaining first sampleRTT :
EstimatedRTT = (0.125 * 80 )+ (0.875 * 100) = 97.5ms
we know that, DevRTT = y |SampleRTT EstimatedRTT| + (1 y ) DevRTT
value of = 0.25 and assuming the value of DevRTT was 5 ms just before
the first of these five samples was obtained:
Here y=
DevRTT = 0.25 |80 97.5| + (1-0.25)* 5 = 8.125ms
we know that, TimeoutInterval = EstimatedRTT + 4 * DevRTT
TimeoutInterval = 97.5 + 4 * 8.125 = 130ms
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After obtaining second sampleRTT = 120ms:
EstimatedRTT = 0.125 * 120 + 0.875 * 97.5 = 100.31ms
DevRTT = 0.25 |120 100.31|+ 0.75 * 8.125 = 11.02ms
TimeoutInterval = 100.31+ 4 * 11.02 = 144.39ms
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After obtaining Third sampleRTT = 140ms:
EstimatedRTT = 0.125 * 140 + 0.875 * 100.31 = 105.27ms
DevRTT = 0.25 |140 105.27| + 0.75 * 11.02= 16.95ms
TimeoutInterval = 105.27 + 4 *16.95= 173.07ms
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After obtaining fourth sampleRTT = 90ms:
EstimatedRTT = 0.125 * 90 + 0.875 * 105.27 =103.36 ms
.
DevRTT = 0.25 |90 103.36| + 0.75 *16.95 = 26.32ms
TimeoutInterval = 103.36 + 4 *26.32 = 208.64ms
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After obtaining fifth sampleRTT = 110ms:
EstimatedRTT = 0.125 * 110 + 0.875 * 103.36= 104.19ms
DevRTT = 0.25 |115 104.19| + 0.75 *26.32 = 22.44ms
TimeoutInterval = 104.19 + 4 *22.44 = 193.95ms
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