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kap26 [50]
3 years ago
9

Suppose that TCP's current estimated values for the round trip time (estimatedRTT) and deviation in the RTT (DevRTT) are 400 mse

c and 25 msec, respectively (see Section 3.5.3 for a discussion of these variables). Suppose that the next three measured values of the RTT are 210, 400, and 310 respectively.
Compute TCP's new value of estimatedRTT, DevRTT, and the TCP timeout value after each of these three measured RTT values is obtained. Use the values of α = 0.125 and β = 0.25. a. Measured value RTT=350 msec Measured value RTT=325 msec c. Measured value RTT=250 msec estimatedRTT = ? DevRTT = ? TimeoutInterval = ?
Computers and Technology
1 answer:
Masja [62]3 years ago
5 0

Answer:

617.5 msecs, 580.89 msecs, 582.36 msecs

Explanation:

Lets look at the following formulas:

EstimatedRTT = α * SampleRTT+(1- α) * EstimatedRTT

DevRTT = β * | SampleRTT- EstimatedRTT|+(1- β)* DevRTT

TimeoutInterval = EstimatedRTT +4* DevRTT

Given values are:

EstimatedRTT = 400 msec

DevRTT = 25 msec

α = 0.125

β = 0.25

After the first RTT estimate: (210)

EstimatedRTT = 0.125 * 210 + (1 - 0.125) * 400 = 376.25 msecs

DevRTT = 0.25 * |210 - 376.25| + (1 - 0.25) * 25 = 60.3125 msecs

TimeoutInterval = 376.25 + 4 * 60.3125 = 617.5 msecs

After the second RTT estimate: (400)

EstimatedRTT = 0.125 * 400 + (1 - 0.125) * 376.25 = 379.218 msecs

DevRTT = 0.25 * |400 - 379.218| + (1 - 0.25) * 60.3125 = 50.42 msecs

TimeoutInterval = 379.218 + 4 * 50.42 = 580.89 msecs

After the third RTT estimate: (310)

EstimatedRTT = 0.125 * 310 + (1 - 0.125) * 379.218 = 370.56 msecs

DevRTT = 0.25 * |310 - 370.56| + (1 - 0.25) * 50.42 = 52.95 msecs

TimeoutInterval = 370.56 + 4 * 52.95 = 582.36 msecs

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-------------------------------------------------------------------------

Process exited after 0.01643 seconds with return value 0

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