4. Evaluation Networks

8/9/2019 4. Evaluation Networks

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BasicEvaluation of

Networks

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PeriodicTransmissions

Delay ChartsPropagation Delay

Transmission Delay

Processing Delay

Typical Charts

Pipelining

Basic Evaluation of Networks

Jose L. Muñoz, Juanjo Alins, Oscar Esparza, Jorge MataTelematics Engineering

Universitat Politècnica de Catalunya (UPC)

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Outline

1 Units and Parameters

2 Link MetricsMetrics

Periodic Transmissions

3 Delay ChartsPropagation DelayTransmission Delay

Processing DelayTypical ChartsPipelining

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Transmission Delay

Processing Delay

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Pipelining

Transmission and storage units

• Transmission rate units :• Power of 10 (in bits).• Kilo=103 Mega=10 6 Giga=10 9 Tera=10 12 , etc.•

Example 1 Mbps.• Storage units :

• Power of 2 (in bytes).• Kilo=210 (1024 ), Mega=2 20 (1024 · 1024 ),

Giga=2 30 ,Tera=2 40 ,etc.• Example: 1 MB=2

3· 2

20bits.

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Evaluation Parameters I• L[bits ]. Message length (bits or bytes).• T tot [s]. Total time of a message transmission (seconds).• P [bits ]. Packet (or frame) size (bits or bytes).• H [bits ]. Packet (or frame) header size (overhead) (bits

or bytes).• B [bps ]. Link or network transmission rate (bps).• T x [s]. Packet (or frame) transmission time (seconds).• T h [s]. Packet (or frame) header transmission time

(seconds).

• T d [s]. Packet (or frame) data transmission time(seconds).

• d [m]. Link distance (meters).• D [s]. Propagation delay (seconds).• V [m/ s]. Propagation speed (m/s).

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Evaluation Parameters II

• T p [s]. Packet (or frame) processing time (seconds).• T c [s]. Cadence Time. In a periodic transmission, time

between packets or frames (seconds).• RTT [s]. Round Trip Time , time since the sender sends

data and receives data from the other end (seconds).• ST [s]. Slot Time .• λp [pps ]. Cadence in packets (frames) per second (pps).• λb [bps ]. Cadence in bits per second (bps).• λ [bps ]. Goodput or user information bit rate in bits per

second (bps).• S [s]. Virtual or real circuit establishing and releasing

time (seconds).

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Transmission Delay

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Typical ChartsPipelining

Evaluation Parameters III

• N [adimensional ]. Number of links between origin anddestination.

• U [ %]. Link or network utilization( %).• ηlink [ %]. Link efciency ( %).• ηproto [ %]. Protocol efciency ( %).

• cs . Acronym: “ circuit switching ”.• ps . Acronym: “ packet switching ”.• dt . Acronym: “ datagram ”.• vc . Acronym: “ virtual circuit ”.

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Transmission Delay

Processing Delay


Outline


2 Link MetricsMetricsPeriodic Transmissions



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H Data H Data H Data

Tobs

Link utilization (0 - 1)

U = T use T obs

Goodput (bps)

λ = Data T obs

Link efciency (0 - 1)

ηlink = T data T obs

Protocol efciency (0 - 1)

ηprot = Data Data + Overhead Note. At this moment, “protocol” is the data link layer protocol (layer 2), the

“packet” or PDU is a frame and data is data encapsualted by some user or

upperlayer in the data eld of the frame.9/42


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Transmission Delay

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Periodic Transmissions

H Data H Data

Th

Td

Tx

Tc

• Time between frames or cadence time: Tc• Cadence rate ( λ c ): λc = 1/Tc [frames / sec]• Header Time: T h • Data Time: T d

U = T h + T d T c = T x T c ηlink =

T d T c λ =

Data T c ηprot =

T d T x

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Transmission Delay

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Periodic TX Example I• A is sending a bit ow of 64 kbps to C. For this purpose,

B is used as intermediate node.

A

B

C

Circuit Network

B AB

=64Kbps

Packet NetworkBBC=384Kbps

• From A to B, we use acircuit network thatprovides us with a circuitof 64 kbps.

• From B to C, we use apacket network of asingle link of 384 kbps.

• The packet network only uses the data link layer.•

B encapsulates the data of the ow into frames (withoutlosing any bit).• Frames are generated by adding a 5 Bytes header to

the user information (data of the received ow).• A frame is generated every 500 µs.

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Periodic TX Example II

a) Calculate U , ηlink , ηprot , λ [bps ] of the BC link. A

B

C

Circuit Network

B AB

=64Kbps

Packet NetworkB

BC=384Kbps

H Data H Data

Th

Td

Tx

Tc=500μs

Data=64 kbps*500μs=32 bitsT

d=32 bits/384 kbps= 83.33μs

Tx=72 bits/384 kbps= 187.5μs

U = 187 . 5500 = 0.375 (37 .5 %)ηlink = 83 . 33500 = 0.166 (16 .6 %)

ηprot = 3272 = 0.444 (44 .4 %)λ = 32 b500 µ s = 64 kbps

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Periodic TX Example III

b) Recalculate the parameters if we want to encapsulatethe ow using an utilization U=100 % in the packetnetwork (BC link).

U =

1 ⇒ T c =

T x

T x = 40bits + Data 384 Kbps

Data = T c · 64 Kbps

Data = 8bits T c = T x = 125 µs

U = 125125 = 1(100 %)ηprot = 848 = 0.166 (16 .6 %)

ηlink = 20 . 83125 = 0.166 (16 .6 %)λ = 8bits 125 µ s = 64Kbps

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Transmission Delay

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Transmission Delay

Processing Delay


Delay charts

• Delay-charts are a basicevaluation tool fordeterministic datatransmissions.

• Deterministic means thateverything is known.

• These charts represent:• The time on the vertical

axis.• The nodes and links in

the horizontal axis.

n

m e

n

F r a m e 1

F r a m e 1

F r a m e 2

F r a m e 2

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B i


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Transmission Delay

Processing Delay


Propagation delay

• The propagation Delay(D) is the amount oftime that it takes for thehead of the signal totravel from the sender tothe receiver through acertain medium.

DF r a m e

• D is the sum of two magnitudes D = D 1 + D 2 .• D 1 is calculated as a ratio between the link length d

(meters or kilometers) and the propagation speed of themedium V : D 1 = d V .• D 2 : It’s the delay introduced by physical repeaters.

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Basic


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Transmission Delay

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Transmission Delay

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Transmission time

• The transmission time(T x ) is the amount oftime required to push allthe bits of informationthat we want to transmitinto the transmissionmedium.

Tx

F r a m e

• T x is calculated as the ratio between the length of theinformation L (in bits) to transmit and the transmissionrate of the link B (in bits per second or bps): T x = LB .

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Transmission Delay

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Transmission Delay

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Processing delay I

• The processing delay ( T p ) is the amount of time that a

node needs to process the overhead .

link-1

link-4

link-2

link-3

…

Demuxlink-1

If-0

…Demuxlink-n

If-1

OS

Muxlink-1

Muxlink-n

Typical model

• Typically, processing the overhead is partly done by thenetwork card (network interface) and partly by the OS.

• The network interface processes the layer 2 overhead(layer 2 CPU), while the Operating System (centralCPU) processes the upper layers.

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Transmission Delay

Processing Delay


Processing delay II• Store & Forward (S&F) :

•

To process the frame, thenode must receive theentire frame.

• A node cannot processmore than one frame at atime.

• While the node isprocessing a frame, it maybe receiving or transmittingother frames.

Tp

F r a m e

F r a m e

• Another mechanism is cut & throught (C&T), in whichthe header is processed as soon as it is received andthe frame might be forwarded without waiting to itscomplete reception.

• Unless otherwise stated, we will always consider S&F .

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Transmission Delay

Processing Delay


Store & Forwardof Messages

B

M

B B

• The message must be completely received beforebeing forwarded.

• Simple example: indirect communication through 3links:

•

With T p ≈ 0 and negligible propagation delay ( D ≈ 0).• Link B = 1.5 Mbps• Message M length L = 7500000 bits• It takes L/ B = 5s to transmit the message on each link.• Total time to transmit the message = 15 s

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Transmission Delay

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Store & Forwardof Frames

B B B

. . .P1P5000

• Following the previous example:• Message length M size 7500000 bits.• Frame P has a size of 1500 bits.• B = 1.5 Mbps• Avoiding header’s size, we have to send 7.5 M / 1.5 K =

5000 frames.• Each frame is transmitted in 1ms.• With S&F, T p ≈ 0 and D ≈ 0• The total time to transmit the message = 5.002s

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Basicl f


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Units andParameters

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Outline





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BasicE l i f


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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Circuit switching

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Units andParameters

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Datagram packet switching

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Units andParameters

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Virtual circuit packet switching

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Units andParameters

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Outline





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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining

• Packet networks can be understood as a pipeline .• In general, a pipeline is:

• A set of data processing elements connected in series.• The output of one element is the input of the next one.• The elements of a pipeline execute in parallel.• Some amount of buffer storage is often inserted

between elements.• Each element in the pipeline requires a certain amount

of time to process the input and obtain the output.• In the case of networks, inputs and outputs are PDUs.• The processing delay is due to processing the overhead

of the PDU, or to the transmission of the PDU.

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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining Toy Example I

A

Tires andchassis

1 hour 3 hours 2 hours

B

Engine C

Body

• Some questions may arise:• How much it takes to get the rst car from the factory?• How many cars per hour can we produce?•

What happens with the queues?• Who has spare time?• How can we optimize this pipeline?• How can we analyze it? We will use a delay chart .

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BasicEvaluation of

Pi li i T E l III


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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining Toy Example III

5

1 1h

2

3

4

A

5

6

7

8

0h

1h

8h

C

1

2

2

3

1

4

90h

1h

3

4

5

9

9

B out

3h

2h

...

...

...

...

i

i23

j45678

k5678

6

7

8

...

7h

j

k

• The rst car:1h+3h+2h=6h

• Cars per hour (cph):0.33cph

• What happens with thequeues? Up to 4 units

in the queue of node Cand up to 3 units in thequeue of node B.

• Who has spare time? Aand C. Now C also hashis spare time alltogether.

• How can we optimizethis pipeline? Let’s trywith other elements inthe pipeline.

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BasicEvaluation of

Pi li i T E l IV


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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining Toy Example IV• More elements or more powerful elements?

1 hour 3 hours

1 hour 1 hour 1 hour

1 hour

(a) Equivalent toserial CPUs

(b) Equivalentto a faster CPU

1Ghz 1Ghz 1Ghz

1Ghz 3Ghz

1Ghz

1 hour

or

• Several serial elements in the pipeline (a) of a powerequivalent to a single element (b) (parallel operations)provide the same cadence time.

• However, (a) introduces more initial delay than (b).

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BasicEvaluation of

Pi li i ith F I


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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining with Frames I

1 0.122ms

2

3

4

src

5

6

7

8

0

11.22ms

i234567

9

10

11

12

13

...

i

891011

int dstn

1.22ms

n

• Link 1 has a bit rateB 1 = 100 Mbps .

• Link 2 has a bit rateB 2 = 10 Mbps ..

• We transmit frames of size

1526 Bytes a .• Processing delay is not

considered: T p ≈ 0.• Propagation delay is not

considered: D ≈

0.• T 100 Mbps x = 0.122 ms .• T 10 Mbps x = 1.22 ms .a This is the size of a typical Ethernet fra

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BasicEvaluation of

N k Pi li i g ith F II


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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Pipelining with Frames II• Consider that the two links

have a bit rate B 1 = 100 Mbps .• Propagation delay: D ≈ 0.• If T x = X , we consider a

processing delay: T p = 2 · X .• This processing delay affects

the network cadence? YESbecause the processing delayis per frame (packet).

• When processing doesn’t limit

cadence, we can achieve a100 % link utilization .

• Devices whose processingdoesn’t limit cadence are saidto run at “wire speed” .

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BasicEvaluation of

N t k Pipelining with Frames III


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Networks

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Transmission Delay

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Typical Charts

Pipelining

Pipelining with Frames III

src0

int dst

link1

Tx=X

D=2X

Tc=Tx=X

......

1234 1

23

X X

link2 • The two links have a bitrate B 1 = 100 Mbps .

• Processing delay: T p ≈ 0.• T x = X and the

propagation delay

D = 2 · X .• The propagation delay

affects the networkcadence? NO because

propagation introduces adelay in the reception ofthe whole signal, but it isnot a per frame delay.

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BasicEvaluation of

Networks In Summary


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Networks

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

In Summary

• The path that frames follow through a network can be

viewed as a pipeline.• The elements of the pipeline include:

• Overhead processing . Processors do S&F with theframe (packet) processing its overhead.

• Frame transmission . Links with different transmission

rates produce different transmission times for the sameframe (packet).• The maximum frame cadence λmax c is determined by

the slowest operation performed by any element:• The largest transmission time over the slowest link.•

The largest processing time in the slowest CPU.• T max c is max (T i x , T i p ), where i is a crossed link.• Then, λmax p = 1/ T max c .• Propagation delay doesn’t affect cadence.

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BasicEvaluation of

Networks Streaming Example I


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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Streaming Example I

950B50B

Frame for video streamencapsulation:

• Video streaming can be

analyzed as a pipeline.• A user wants to see a lm

of 540 MB and 90 min thatis stored on a server.

• To send the movie to the client, the server generatesframes of size 1000 Bytes with 50 Bytes of overhead.

• The server needs 6 ms to encapsulate the video dataon each frame.

• The network is formed by a link with B1=800 kbps thatconnects the server and an intermediate node and byanother link with B2=1 Mbps that connects theintermediate node with the client.

• The intermediate node requires 10 ms of processing.

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BasicEvaluation of

Networks Streaming Example II


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Transmission Delay

Processing Delay

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Pipelining

Streaming Example II

• In the client host, there is a video player that is able to

start playing the movie before it has been completelyreceived. This is called video streaming .

• Let’s consider that the player can set the number ofminutes to wait before starting to play the movie.

a) Calculate the lm data rate (in Kbps) that the networkcan provide.

b) Calculate the time that destination node has to wait tostart the movie in order to play it without cuts, (nomissing information at any time). This is called playouttime .

Note. Consider that frames are transmitted as quick aspossible, and links are dedicated exclusively to our videostreaming transmission.

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BasicEvaluation of

Networks Streaming Example III


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etwo s

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Transmission Delay

Processing Delay

Typical Charts

Pipelining

Streaming Example III

a) Film data rate:•

To play the movie, a data rate of540 · 8 · 220 / (90 · 60 ) = 838 .86 kbps are needed.• Let’s compute the goodput provided by the network.• The server takes 6 ms to generate a frame.• Link1 needs (1000 · 8)/ 800 kbps = 10 ms to transmit a

frame.• The intermediate node takes 10 ms to process a frame.• Therefore, the cadence provided by the network is

1 frame/10 ms. In kbps: 950 · 8/ 0.01 = 760 kbps .b) Playout time:

•

The missing data rate should be covered byaccumulating bytes before playing the lm.• T accum · 760 kbps = ( 838 .86 kbps − 760 kbps ) · (90 · 60 sec ).• T accum = 9.3387 min.

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4. Evaluation Networks

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