CS485 Sylabus

Performance Analysis of the unslotted 0-persistent CSMA protocol

Note on this webpage
- This webpage contains the analysis only for the unslotted non-persistent CSMA
- The analysis for slotted non-persistent CSMA is in the subsequent webpage.
- The analysis for 1-persistent CSMA protocols will come later....

Definitions and notations

Definitions:

Arrival rate λ = number of new messages that arrives per time unit
Offered load (rate) G = number of (old and new) messages that contend for the channel per time unit
Throughput S = number of messages that are successfully transmitted per time unit
A message is successfully transmitted if its transmission does not collide with any other message transmission.
Maximum propagation delay τ = the maximum time needed for a transmission from any source to reach all nodes in the CSMA system
The max. prop. delay τ depends on the length of the cables

Simplifying assumptions

Simplifying assumptions:

All packets have the same length T
(So the transmission time for any packet is the same duration)
Assume that the propagation delay between any two nodes is equal to τ seconds
Like this:

Define:

τ a = --- T

which is the normalized propagation delay

The packet arrival process is a Poisson process with rate λ
The offered load (new arrivals and retransmission attempts combined) is also Poisson distributed with rate G
(We will see that the offered load G is dependent on the arrival rate λ)

Overview of the system operation

Abstractly speaking, the non-persistent CSMA protocol will cause the nodes to transmit in the following manner:

Notes:

The network is initially idle

Then one or more nodes will transmit

If one node transmits, then the busy period will contains a successful transmission
If more than one node transmit, then the busy period will contains a collision (unsuccessful transmission)

Note:

Because CSMA does not have collision detection, when a node starts transmitting, it will transmit for T sec (until the packet is transmitted)

After the transmission(s) finishes, there will be an idle period when the nodes will sense the channel before they transmit.
This idle period has the same characteristic as the first idle period

Throughput analysis - part 1

The following diagram gives the factors that determine the throughput of the non-persistent CSMA protocol:

Observations:

There is a repeating pattern:
The length of a busy period is variable
(depending on the last starting transmission in the busy period)
The length of an idle period is variable (random)
(depending on the time of the next transmission that starts the subsequent busy period)

Abstraction: represent the repeating pattern by one average pattern

Definitions and notations:

B = the average duration of a busy period in the non-persistent CSMA cycle
A busy period can be useful (i.e., successful transmission) or not useful (i.e., collision)
I = the average duration of a idle period in the non-persistent CSMA cycle

Throughput of the unslotted non-persistent CSMA protocol:

U S_{non-persistent CSMA} = ------ ...... (1) B + I

All we need to do now is to find U, B, and I :-)

The average duration of an idle period (I)

Characteristics of an idle period:

Notes:

The length (duration) of an idle period is a random variable
An idle period ends with the first transmission
Within the idle period there are 0 message arrivals !

The cumulative density function of an idle period I is therefore:

Ҏ[ I ≤ x ] = 1 - Ҏ[ I > x ] = 1 - Ҏ[ no packet arrives in x sec ]

(Gt)^k Ҏ[ k arrivals in t sec ] = ---- e^-Gt k!

(Gx)⁰ Ҏ[ I ≤ x ] = 1 - ---- e^-Gx 0! <=> Ҏ[ I ≤ x ] = 1 - e^-Gx ...... (2)

The probability density function of an idle period I is then:

d Ҏ[ I ≤ x ] f_I(x) = --------------- d x d (1 - e^-Gx) <==> f_I(x) = -------------- d x <==> f_I(x) = G e^-Gx ...... (3)

Now we can calculate the average duration I:

I = ₀∫^∞ x f_I(x) dx (= the definition of expected value) <=> I = ₀∫^∞ x G e^-Gx dx 1 | x = ∞ <=> I = -x e^-Gx - --- e^-Gx | G | x = 0 1 <=> I = ( 0 - 0 ) - ( 0 - --- ) G 1 <=> I = --- ..... (4) G

One down, two more to go....

Average length of a useful period U

Recall that the length of each transmission is equal to T

2 kinds of busy periods:

Useful

A busy period that contains exactly one transmission is useful (no collision)
The length (duration) of a busy period that contains exactly one transmission is equal to T (= length of one transmission)
The contribution of such busy perido to the total useful period is T

Not useful

A busy period that contains more than one transmission is not useful (collision)
The length (duration) of a busy period that contains multiple transmissions is random (and difficult to compute)
However, the contribution of such busy perido to the total useful period is 0 (zero) !!!

Average length of the useful period:

The average length of the useful periods is the following weight average:

U = T × Ҏ[ transmission successful ] + 0 × (1 - Ҏ[ transmission successful ])

A busy period is successful, i.e., contains exactly one transmission if an only if:

Notes:

Some node starts transmitting
Within τ sec of the start of this transmission, no other nodes will start a transmission
(Then after τ sec, when a node senses the transmission medium it will detect an on-going transmission and will refrain from starting its transmission)

Thus, Ҏ[ transmission successful ] is equal to:

Notes to the figure above:

Ҏ[ transmission successful ] = Ҏ[ 0 arrivals within "vulnerable period" ] = Ҏ[ 0 arrivals within τ sec ]

(Gt)^k Ҏ[ k arrivals in t sec ] = ---- e^-Gt k!

(Gτ)⁰ Ҏ[ transmission successful ] = ----- e^-Gτ 0! <=> Ҏ[ transmission successful ] = e^-Gτ ..... (5)

Therefore:

U = T × Ҏ[ transmission successful ] + 0 × (1 - Ҏ[ transmission successful ]) = T × e^-Gτ + 0 × (1 - e^-Gτ)

U = T × e^-Gτ ...... (6)

Two down, one more to go....

Average length of a busy period B - part 1

Anatomy of a (simple) busy period:

A busy period begins with some node starting a transmission:
The transmission reaches the end of the cable at t = τ:
The node finishes transmitting after t = T sec:
Notice that the transmission medium is not yet idle...
I.e., the busy period has not yet ended !!!
The transmission clears the cable at t = T + τ:

The length of a busy period is a random variable due to possible nearly-simultaneous starting transmissions:

Notes:

If there is one transmission in the busy period, the length of the busy period will be equal to (T + τ) sec
But when there are multiple (nearly simultaneous) transmissions, the length of the busy period will be equal to (T + y + τ) sec, where:

Average length of a busy period B

From the discussion above: B = T + τ + y ...... (7) where: T = length of a packet transmission (constant) τ = (max) end-to-end delay (constant) y = time lag of the last transmission (random)
From Theory of probability: E[B] = E[ T + τ + y ] <=> E[B] = T + τ + E[y] <=> B = T + τ + y ...... (8)

In order to compute the expected value E[y] (y), we need to find the pdf (probab. distribution function) of y first.

The pdf of the random variable y

The random variable y is the lag time of the last transmission in the busy period.
Since the propagation delay is at most τ, the lag time of any transmission is at most τ
(Because after τ sec, a node that wants to transmit will detect that the channel is busy and will refrain from transmitting)

Situation:

Therefore:

Ҏ[ y ≤ t ] = Ҏ[ 0 arrivals in (τ - t) sec ]

(Gt)^k Ҏ[ k arrivals in t sec ] = ---- e^-Gt k!

Therefore: (G(τ-t))⁰ Ҏ[ y ≤ t ] = ----------- e^-G(τ-t) 0! <=> Ҏ[ y ≤ t ] = e^-G(τ-t) (with t ∈ [0, τ)) ...... (9)

We can now obtain the probability density function (pdf) f_y: (see: click here )

d Ҏ[ y ≤ t ] f_y(t) = -------------- d t d e^-G(τ-t) <=> f_y(t) = ----------- d t <=> f_y(t) = e^-G(τ-t) × -G (-1) <=> f_y(t) = G e^-G(τ-t) (with t ∈ [0, τ)) ...... (10) ***

Problem with the pdf that we found....

Every pdf f(t) must satisfy the following property:
(because the cumulative outcome of every possible value is equal to certainty (1))

This property does not hold for f_y(t) in the equation (10):

_-∞∫^∞ f_y(t) dt = _-∞∫^∞ G e^-G(τ-t) dt = ₀∫^τ G e^-G(τ-t) dt -+ τ = e^-G(τ-t) | -+ 0 = e^-G(τ-τ) - e^-G(τ-0) = 1 - e^-Gτ ≠ 1 !!!

Reason of this behavior:

y is a hybrid random variable:

f_y(t) is continuous on (0,τ), but...

f_y(t) is discontinuous in t = 0...

In fact, in t = 0, the random variable y behaves like a discrete random variable:

Ҏ[ y = 0 ] = Ҏ[ no "last arriving" transmission ] = Ҏ[ one transmission in busy period ] = Ҏ[ successful transmission ] = e^-Gτ (see Eq. (5) - click here)

Note:

The correct pdf of the random variable y:

f_y(t) = e^-Gτ × &delta(t) + G e^-G(τ-t) (with t ∈ [0, τ)) ...... (11)

The function &delta(t) is defined as:

&delta(t) = 1 if t = 0 &delta(t) = 0 otherwise

The expected value of the random variable y

We can now compute the expected value E[y]:

E[y] = _-∞∫^∞ t × f_y(t) dt = ₀∫^τ t × f_y(t) dt

Split off the discontinuous point t = 0

=> E[y] = 0 × Ҏ[ y = 0 ] + ₀₊∫^τ t × f_y(t) dt = 0 × e^-Gτ + ₀₊∫^τ t × G e^-G(τ-t) dt = G e^-G(τ)₀₊∫^τ t × e^-G(-t) dt = G e^-Gτ₀₊∫^τ t × e^Gt dt

Maple: > integrate( t * exp(G*t), t=0..tau ); 1 - exp(G tau) + exp(G tau) G tau --------------------------------- 2 G

1 - e^Gτ + Gτe^Gτ <=> E[y] = G e^-Gτ ----------------- G² 1 - e^Gτ + Gτe^Gτ = e^-Gτ ----------------- G e^-Gτ - 1 + Gτ = --------------- G 1 - e^-Gτ = τ - --------- ........ (12) G

Average length of a busy period B - part 2

Recall that the average length of a busy period B was: (see: click here)

B = T + τ + y ...... (8)

With Equation (12), we can now complete the derivation:

1 - e^-Gτ B = T + τ + ( τ - --------- ) G 1 - e^-Gτ = T + 2τ - --------- ...... (13) G

Throughput analysis of unslotted non-persistent CSMA - part 2

Recall that: throughput of the unslotted non-persistent CSMA protocol: (see: click here)

U S_{non-persistent CSMA} = ------ ...... (1) B + I

We have found that:

1 I = --- ...... (4) G U = T × e^-Gτ ...... (6) 1 - e^-Gτ B = T + 2τ - --------- ...... (13) G

Therefore:

U S = ------- B + I T × e^-Gτ <=> S = ----------------------------- (× G/G) T + 2τ - (1 - e^-Gτ)/G + 1/G G T × e^-Gτ <=> S = ----------------------------- (-1 + 1 = 0) G(T + 2τ) - (1 - e^-Gτ) + 1 G T × e^-Gτ <=> S = ------------------ (sub: a = τ/T, or τ = aT) G(T + 2τ) + e^-Gτ G T × e^-GaT <=> S = ------------------- (move GT out) G(T + 2aT) + e^-GaT GT × e^-aGT <=> S = ------------------- GT(1 + 2a) + e^-aGT