Understanding TCP/IP Model Internals and Interfaces

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1508 Understanding TCP/IP Model Internals and Interfaces Analyze TCP New Reno 4. Select Scenarios -> Switch to Scenario to switch to scenario New_Reno. 5. TCP New Reno supports fast retransmission and fast recovery. However, contrary to TCP Reno, it halves the congestion window only once during recovery process. a. Fast Recovery has been already enabled on both the workstation and the server by setting TCP Parameters->Fast Recovery to New Reno: 3. Repeat steps 2 through 3 from page 16 to run the simulation, and load results into templates. You should see the following results: Results Analysis Congestion Window Examine TCP Connection.Congestion Window Size (bytes) graph to answer the following. To get an exact value, point the mouse to the desired point and read the tooltip. 1. What was the congestion window size before the packet loss was detected? ___________ 2. What did the congestion window size value change to after detecting packet loss? _________ CONFIDENTIAL – RESTRICTED ACCESS: This information may not be disclosed, copied, or transmitted in any format without the prior written consent of OPNET Technologies, Inc. © 2010 OPNET Technologies, Inc. 26
1508 Understanding TCP/IP Model Internals and Interfaces 3. How did the server recover from packet loss, by using fast retransmit or timeout retransmission? ______________ 4. How many times was a slow start performed during the session lifetime? ___________ Retransmission Pattern Examine sequence number of forwarded and dropped segments graph to answer the following. 5. How many segments were retransmitted? ___________ 6. Were any segments retransmitted unnecessarily? ___________ 7. When is the last segment sent? ___________ Detailed Analysis The size of the congestion window just before fast retransmission was 12,328 bytes. After fast retransmission, the congestion window is reset to 12,864 / 2 + 3 * MSS = 6,432 + 3 * 536 = 8, 040 Note: In this calculation, when the congestion window is halved, it is rounded down to an integral multiple of the MSS. In this example, 12,864 is 24 * 536, the result of the division is also an integral multiple of the MSS. If the window had been 23 * 536 = 12,328, then the result would be 11.5 * 536 = 6,164, which must be rounded down to 11 * 536 = 5,895. Note: 12,328/2 is rounded down to the increments of MSS, rather than to the actual value (6,164). The process then continues receiving more duplicate ACKs and increases its congestion window by one MSS for each received ACK. Unlike in Reno, the ACK for the retransmitted segment 23 does not take New Reno out of the recovery process. New Reno will keep increasing its congestion window for each received ACK until the ACK for the last segment sent before retransmission is received (which is segment 46). When the client receives the retransmitted segment 28, then it will send an acknowledgement for all the outstanding segments. When the sender gets this acknowledgement, fast recovery ends, and the sender enters slow start. CONFIDENTIAL – RESTRICTED ACCESS: This information may not be disclosed, copied, or transmitted in any format without the prior written consent of OPNET Technologies, Inc. © 2010 OPNET Technologies, Inc. 27
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Understanding TCP/IP Model Internals and Interfaces

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