RELATION BETWEEN PURE AND SLOTTED ALOHA IN NETWORKING - PART 2

PREVIOUS POST : WHAT IS ALOHA ? IN NETWORKING PART 1

Now in this post we understand about PURE ALOHA and SLOTTED ALOHA NUMERICALLY.

Some terminology is very important when we want to calculate following things regarding ALOHA.

1. PERCENTAGE OF SUCCESSFUL TRANSMISSION.

2. PERCENTAGE OF COLLISION IN CHANNEL.

3. CHANNEL UTILISATION.

4. EFFICIENCY OF ALOHA.

Before knowing about all above things we must know about following terms :

1. FRAME TIME:

Time required to transmit standard time frame.

2. CHANNEL LOAD (G) OR OFFERED LOAD :

Mean / average number of frames including new frame and re-transmitted frame per frame time.

3. THROUGHPUT (S) : (including only newly generated frame)

    Define 1: Mean number of newly generated frames per frame time.

    Define 2: mean number of successful transmission per unit time.

    S < = 1 [ This is because at a time only one frame is generated ]

There are basically three type of frame time.

A. SUCCESSFUL FRAME TIME :

    When one frame is transferred at a time.

B. COLLISION FRAME TIME :

    When at a time more than one frame is transmitted.

C. EMPTY FRAME :

    hen no any frame is transmitted.

RELATION BETWEEN S AND G:

    G : (NEW + TRANSMITTED FRAME)

    G >  1

    When S < = 1

    Therefore    S < = G

NOTE : Only after successful transmission new frame is generated.

YOU MUST REMEMBER ALL ABOVE TERMS

NOW WE WILL FIND THE EFFICIENCY OF ALOHA BY CALCULATING  PROBABILITY OF SUCCESSFUL TRANSMISSION (P) AND THROUGHPUT (S).

PURE ALOHA:

In pure ALOHA any time we can transmit the message.

Let the probability of success is P0

    S = G . P0

    Where ;    G = Channel load

                     S = Throughput

No one can transmit between two frame otherwise collision will occur.

therefore Vulnerable period= 2 frames.

Now as we know ; According to Poisson's formula (Markov chain model)

    P_n(t) =e^-λt (λt)ⁿ / !n

P_n(t) : It is the probability that in time duration.

t       : There will be exactly (n) activity

λ      : Mean /Average no of activity per unit of time.

By this formula we can find out how many activity will perform in the next T time , If we know the average number of activity per unit of time.

IN THE CASE OF  PURE ALOHA :

    P₀(2 frame time) = e^-2G (2G)⁰ / !0

Probability of success is :

    P₀ = e^-2G 

Then the throughput is :    

    S = G . P₀

     S = G .  e^-2G

Then differentiation S with respect to G and equating to zero like:    

      ds/dg=0

      dS/dG=0

     dS/dG ⇒ G ( -2 ) e^-2G₊ e^-2G _{= 0}

             _{⇒ -2 G} e^-2G ₊ e^-2G _{= 0}

             ⇒ ( -2 G + 1 ) = 0

           ⇒ G = 1/2   → The value of G is on an average

For  G = 1/2  , S = maximum

Then the value S will be :

    S_max = (1/2) e^{-2x 1/2}

               ^{= 1/ 2e} 

               ^{= 0.184}

    S_max = 0.184

So here only 18.4 %  is successful transmission and the other time is waste due to collision or ideal.

So here we can say that it is a poor utilisation of channel.

GRAPHICAL REPRESENTATION :

When sufficient Input is not given to channel then under loaded condition will occur.

        U = S / ( S + C + E )

        where : S is Success

                    C is Collision

                    E is Empty

In the case of overload collision will increase.

Here in 0.5 G the efficiency of ALOHA is 0.184.

SLOTTED ALOHA:

In slotted ALOHA we can transmit in time slot.

Let probability of success is P_0.

    S = G . P₀

Where :

               G = Channel load

               S = Throughput

Here if transmission start in 3.15 sec of frame time than it will complete in 3.20 sec of frame time and if any transmission want to start at 3.16 or 3.17 or 3.18 sec of frame time than it will only start after completion of previous transmission. i.e after 3.20 sec of frame time.

    Vulnerable period = One frame time.

Now as we know ; According to Poisson's formula (Markov chain model)

    P_n(t) =e^-λt (λt)ⁿ / !n 

    P_n(t) : It is the probability that in time duration.

     t       : There will be exactly (n) activity

      λ     : Mean /Average no of activity per unit of time.

Here : t =1 , λ = G , n =0

Then put the value of t , λ , n in above formula we get

        P₀ = e^-G  / 1

     P₀ = e^{-G        (} i.e:  Probability of success ⁾

      S = G. P₀      ( i.e Throughput )

Differentiate with respect to  G and equating to zero.

        _{dS/dG ⇒ - G . e}^{- G}_{= 0}

       dS/dG ⇒ - G . e^{- G}  +  e^{- G} _{= 0}

Now we get  G = 1   on an average

For G = 1  ,   S will be maximum

Than the value of  Smax = G . e^{- G}

                                                   = 1. e- ¹

                                                    =1 / e

                                Smax  = 0.368

So here 36.8% is successful transmission which is more then PURE ALOHA

GRAPHICAL REPRESENTATION :

If  G < 1  Under loaded

If  G > 1  Over loaded

Now in this post it is very clear about PURE ALOHA AND SLOTTED ALOHA with derivation. By the help of derivation and formula we can find the efficiency of ALOHA.

We will meet in next post with Some numerical examples of ALOHA.