PerformanceIssues

PerformanceIssues - Performance Issues in HighSpeed...

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Performance Issues in High- Speed Networks Dr. Sanjay P. Ahuja, Ph.D. Professor School of Computing, UNF
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2 Performance Problems in Computer Networks 1. Performance can degrade when there is structural imbalance. E.g. Gigabit line attached to low-end machines where CPU can’t process packets fast enough which leads to packets being lost and more retransmissions. 2. Overloads can be synchronously triggered. E.g. If a TCP (or UDP) segment contains a bad port # and this segment is broadcast to 10,000 machines, then receivers will send back a flood of error notifications. 3. Poor performance can occur due to lack of system tuning. E.g. a machine has plenty of CPU and memory but not enough memory has been allocated for buffer space, then segments can be lost. Or if he CPU scheduling algorithm does not give high enough priority to the process which is processing incoming segments, some of the segments can be lost. 4. Gigabit networks bring with them new performance problems. These are caused by running older protocols over gigabit lines. E.g. transmitting data from San Diego to Boston at 1 Gbps on a fiber line. Receiver’s buffer capacity = 64 KB (so window size = 64 KB). One-way prop. delay = 20 msec (speed of light). In just 0.5 msec (64 KB / 1 Gbps), all segments are on the fiber. After 20 msec, the first segment hits Boston and is ACKed. Finally, 40 msec after starting, the first ACK gets back to sender and 2 nd burst can be sent. So line utilization = 0.5 msec / 40 msec = 1.25% !
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3 Performance Problems in Computer Networks The state of transmitting one megabit from San Diego to Boston. (a) At t = 0. (b) After 500 μ sec. (c) After 20 msec. (d) After 40 msec.
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4 Bandwidth-delay product Bandwidth-delay product = bandwidth (bps) * RT delay (in seconds) This is the capacity of the pipe from sender to receiver and back (in bits) and is the # of bits that can fit in the pipe. For the example discussed in the previous slide, the Bandwidth-delay product = 1 * 10 9 * 40 msec = 1 * 10 9 * 40 * 10 -3 = 40 * 10 6 bits = 40 million bits = 5 MB Sender would’ve to transmit a burst of 40 million bits to be able to fill the pipe with bits both ways until the first ACK comes back. So for TCP with a window size of 64 KB = 64 * 1024 * 8 = 0.5 * 10 6 bits. This gives a 1.25% utilization of the line, i.e. 0.5 * 10 6 / 40 * 10 6 = 0.0125 or 1.25%. Conclusion: The receiver’s window size needs to be preferably larger than the Bandwidth-delay product. So for a 1-Gbps transcontinental line, a window size = 40 Mbits = 5MB is needed. Note: If the line utilization is so poor for sending a megabit of data (0.5 * 10
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This note was uploaded on 07/29/2011 for the course CNT 6707 taught by Professor Ahuja during the Spring '11 term at UNF.

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PerformanceIssues - Performance Issues in HighSpeed...

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