retagged by
3,289 views
1 votes
1 votes

Consider Jacobson's algorithm for setting the time out timer. Let's assume that Initial Round Trip time is 10 ms with initial deviation of 5 ms. What will be the value of  Time Out Timer after receiving 20 ms, 30 ms, 40 ms round trip time respectively. (Take α= 0.5)

Please solve and what does initial deviation means here of 5ms(how to take it into account)?

retagged by

1 Answer

Best answer
1 votes
1 votes

IRTT = 10ms (Given)

ID = 5ms (Given)

$\alpha$ = 0.5

Given these information, we calculate the Timeout (The time within which the acknowledgement should reach te sender to avoid retransmission).

According to Jacobson's Algorithm, TO = 4 * ID + IRTT

So, TO = 4 * 5 + 10 = 30ms

Now, ARTT = 20ms (The acknowledgement reached after 20ms which is less than te TO time, so we can proceed without Karn's modification)

Actual Deviation, AD = | IRTT - ARTT | = 10ms

New RTT =  $\alpha$ IRTT + (1 - $\alpha$)ARTT = 0.5 * 10 + (1 - 0.5) * 20 = 15ms

New Deviation, ND = $\alpha$ ID + (1 - $\alpha$)AD = 0.5 * 5 + (1 - 0.5 ) * 10 = 7.5ms

Thus, we calculate the RTT and Deviation for the next transmission. We proceed as above with the subsequent transmissions.

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

IRTT = 15ms

ID = 7.5ms

TO = 4 * 7.5 + 15 = 45ms

ARTT = 30ms (The acknowledgement reached after 30ms which is less than TO time)

Actual Deviation, AD = | IRTT - ARTT | = 15ms

New RTT = $\alpha$ IRTT + (1 - $\alpha$ )ARTT = 0.5 * 15 + (1 - 0.5) * 30 = 22.5ms

New Deviation, ND = $\alpha$ ID + (1 - $\alpha$ )AD = 0.5 * 7.5 + (1 - 0.5) * 15 = 11.25ms

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

IRTT = 22.5ms

ID = 11.25ms

So, TO = 4 * 11.25 + 22.5 = 67.5ms

Now, ARTT = 40ms (The acknowledgement reached after 40ms which is less than TO time)

Actual Deviation, AD = | IRTT - ARTT | = 17.5ms

New RTT = $\alpha$ IRTT + (1 - $\alpha$)ARTT = 0.5 * 22.5 + (1 - 0.5) * 40 = 31.25ms

New Deviation, ND = $\alpha$ ID + (1 - $\alpha$ )AD = 0.5 * 11.25 + (1 - 0.5) * 17.5 = 14.375ms

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

IRTT = 31.25ms

ID = 14.375

TO = 4 * 14.375 + 31.25 = 88.75ms

selected by

No related questions found