L4: Basic Networking Calculations

EECS 3213, F14 1

L4: Basic Networking Calculations

Sebastian Magierowski York University

EECS 3213 Fall 2014

L4: Basic Calculations

EECS 3213, F14 2

•  Calculating basic physical properties of networks –  What do bits look like? –  What’s R –  What’s bandwidth –  What’s the delay –  The value of partitioning data –  Queuing –  Throughput –  Errors

Outline


EECS 3213, F14 3

•  Consider what bits look like –  What’s the R (data rate, data bandwidth)

4.1 Data Rate


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EECS 3213, F14 4

•  The extent of signal energy as a function of frequency

4.2 Signal Bandwidth


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EECS 3213, F14 5

•  propagation time •  transmission time

4.3 Delay


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EECS 3213, F14 6

•  Finding the spectrum of a periodic signal

4.4 Delay of Packets Through Network


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EECS 3213, F14 7

•  λ: packets arriving per second •  µ: packets processed per second •  ρ (G): offered load = λ/µ •  Rate at which a queue is emptied? •  Avg. time to empty a queue (or to get out of node)? •  Avg. wait time to get to front of queue? •  Avg. number of packets in queue?

4.5 Queuing Delay


EECS 3213, F14 8

Queuing Delay Details


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EECS 3213, F14 9

4.6 Link Sharing

•  Slow in parallel •  One fast


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EECS 3213, F14 10

•  N = λ•T •  N: avg. number of packets in the system (i.e. network) •  T: avg. time spent in the system •  λ: avg. packet arrival per second in the system

4.7 Little’s Formula


EECS 3213, F14 11

•  Rate of data transfer but… –  …accounts for overhead of communication and network details

•  3 MB file takes 2 minutes to download –  T is?

4.8 Throughput


12 2. PRINCIPLES

rate of 768Kbps. The difference comes from the fact that the download corresponds to a sequenceof packets and there are gaps between the packets.

Figure 2.2 shows two typical situations where the throughput is less than the link rate. Theleft part of the figure shows a source S that sends packets to a destination D. A packet is said to beoutstanding if the sender has transmitted it but has not received its acknowledgment. Assume thatthe sender can have up to three outstanding packets. The figure shows that the sender sends threepackets, then waits for the acknowledgment of the first one before it can send the fourth packet,and so on. In the figure, the transmitter has to wait because the time to get an acknowledgment– RT T , for round-trip time – is longer than the time the transmitter takes to send three packets.Because of that, waiting time during which the transmitter does not send packets, the throughput ofthe connection is less that the rate R of the link. Note that, in this example, increasing the allowednumber of outstanding packets increases the throughput until it becomes equal to the link rate. Themaximum allowed number of outstanding bytes is called the window size. Thus, the throughput islimited by the window size divided by the round-trip time.

The right part of the figure shows devices A, C, D attached to a router B. Device A sendspackets at rate R. Half the packets go to C and the other half go to D. The throughput of theconnection from A to C is R/2 where R is the rate of the links. Thus, the two connections (from A

to C and from A to D) share the rate R of the link from A to B. This link is the bottleneck of thesystem: it is the rate of that link that limits the throughput of the connections. Increasing the rateof that link would enable to increase the throughput of the connections.

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Figure 2.2: The throughput can be limited by the window size (left) and by a bottleneck link (right).

2.2.4 DELAYDelay refers to the elapsed time for a packet to traverse between two points of interest. If the twopoints of interest are the two end points (e.g., the pair of communicating hosts), we refer to thedelay between these two points as the end-to-end delay. End-to-end delay typically comprises thetransmission and propagation times over the intervening links, and queuing and processing times

12 2. PRINCIPLES

rate of 768Kbps. The difference comes from the fact that the download corresponds to a sequenceof packets and there are gaps between the packets.

Figure 2.2 shows two typical situations where the throughput is less than the link rate. Theleft part of the figure shows a source S that sends packets to a destination D. A packet is said to beoutstanding if the sender has transmitted it but has not received its acknowledgment. Assume thatthe sender can have up to three outstanding packets. The figure shows that the sender sends threepackets, then waits for the acknowledgment of the first one before it can send the fourth packet,and so on. In the figure, the transmitter has to wait because the time to get an acknowledgment– RT T , for round-trip time – is longer than the time the transmitter takes to send three packets.Because of that, waiting time during which the transmitter does not send packets, the throughput ofthe connection is less that the rate R of the link. Note that, in this example, increasing the allowednumber of outstanding packets increases the throughput until it becomes equal to the link rate. Themaximum allowed number of outstanding bytes is called the window size. Thus, the throughput islimited by the window size divided by the round-trip time.

The right part of the figure shows devices A, C, D attached to a router B. Device A sendspackets at rate R. Half the packets go to C and the other half go to D. The throughput of theconnection from A to C is R/2 where R is the rate of the links. Thus, the two connections (from A

to C and from A to D) share the rate R of the link from A to B. This link is the bottleneck of thesystem: it is the rate of that link that limits the throughput of the connections. Increasing the rateof that link would enable to increase the throughput of the connections.

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Figure 2.2: The throughput can be limited by the window size (left) and by a bottleneck link (right).

2.2.4 DELAYDelay refers to the elapsed time for a packet to traverse between two points of interest. If the twopoints of interest are the two end points (e.g., the pair of communicating hosts), we refer to thedelay between these two points as the end-to-end delay. End-to-end delay typically comprises thetransmission and propagation times over the intervening links, and queuing and processing times

EECS 3213, F14 12

•  Message size, error, rates and hops

4.9 Errors


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L4: Basic Networking Calculations