Prerequisite – Computer Network | TCP Timers
Calculating Time out timer (TOT) at transport layer is tricky as propagation delay is not constant i.e. path may change continuously and traffic is dynamic. So, static TOT cannot be used at TCP. And unnecessarily retransmitting the same data packet multiple times may cause congestion.
Solution for this is we need dynamic TOT which can adjust to changes in round trip time (RTT).
Algorithm for Dynamic TOT calculation:
- Basic algorithm
- Jacobson’s algorithm
- Karn’s modification
1. Basic algorithm –
We assume initial round trip time i.e PRTT. On sending out each packet TOT = 2 * PRTT. The next round trip time is calculated using PRTTn+1 = α PRTTn + (1 - α)ARTTn where PRTT = predicted round trip time ARTT = actual round trip time α = smoothing factor such that 0<= α <=1
Example – Let PRTT1 = 10ms and α = 0.5
TOT = 2 * PRTT1 = 20ms Let ARTT1 = 15ms Then, PRTT2 = (0.5 * 10) + (0.5 * 15) = 12.5ms TOT = 2 * 12.5 = 25ms Let ARTT2 = 20ms PRTT3 = (0.5 * 12.5) + (0.5 * 20) = 16.25ms TOT = 2 * 16.25 = 32.5ms And so on TOT is calculated.
- better than static TOT.
- TOT is flexible to dynamic round trip time.
- It takes into consideration all packets to derive new PRTT.
- TOT = 2 * PRTT is used to allow a grace time for returning acknowledgement.
- It is wasteful.
2. Jacobson’s algorithm –
Calculates TOT value more intuitively than basic algorithm.
We assume initial round trip time i.e. PRTT. PRTTn+1 = α PRTTn + (1 - α)ARTTn PDn+1 = α PDn + (1 - α)ADn where ADn = |PRTTn - ARTTn| AD = Actual deviation PD = predicted deviation On sending out each packet, TOT = (4 * PD) + PRTT.
Iteration 1 Given α = 0.5, PRTT1 = 10ms, PD1 = 5ms and ARTT1 = 20ms TOT = (4 * 5) + 10 = 30ms AD1 = |10 - 20| = 10ms Iteration 2 PRTT2 = α PRTT1 + (1 - α)ARTT1 = (0.5 * 10) + (0.5 * 20) = 15ms PD2 = α PD1 + (1 - α)AD1 = (0.5 * 5) + (0.5 * 10) = 7.5ms TOT = (4 * 7.5) + 15 = 45ms Given ARTT2 = 30ms AD2 = |15 - 30| = 15ms Iteration 3 PRTT3 = α PRTT2 + (1 - α)ARTT2 = (0.5 * 15) + (0.5 * 30) = 22.5ms PD3 = α PD2 + (1 - α)AD2 = (0.5 * 7.5) + (0.5 * 15) = 11.25ms TOT = (4 * 11.25) + 22.5 = 67.5ms Given ARTT3 = 10ms AD2 = |22.5 - 10| = 12.5ms And so on TOT is calculated.
Problem with Basic and Jacobson’s Algorithm
In both, PRTTn+1 = α PRTTn + (1 – α)ARTTn
i.e both depend on previous segment ARTT. But if initial time out timer times out then what next TOT will be chosen since the acknowledgement is delayed i.e its coming after time out so ARTT is not available.
3. Karn’s Modification –
Whenever the timer times out do not apply either of Basic or Jacobson algorithm as ARTT is not available instead double the time out timer(TOT) whenever the timer times out and a retransmission is made.
- One Time Password (OTP) algorithm in Cryptography
- Longest Remaining Time First (LRTF) CPU Scheduling Algorithm
- EIGRP Cost Calculation
- Factorial calculation using fork() in C for Linux
- Timer in C++ using system calls
- Dynamic Frequency Scaling and Dynamic Voltage Scaling
- Dynamic NAT (on ASA)
- Dynamic Trunking Protocol (DTP)
- How do Dynamic arrays work?
- GATE CS 2016 Sec 5 – Dynamic Programming
- Difference between Static and Dynamic Web Pages
- Difference between Static and Dynamic Routing
- Difference between Static and Dynamic IP address
- Dynamic Host Configuration Protocol (DHCP)
If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to email@example.com. See your article appearing on the GeeksforGeeks main page and help other Geeks.
Please Improve this article if you find anything incorrect by clicking on the "Improve Article" button below.