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Consider a simplified time slotted MAC protocol, where each host always has data to send and transmits with probability $p$ = $0.2$ in every slot. There is no backoff and one frame can be transmitted in one slot. If more than one host transmits in the same slot, then the transmissions are unsuccessful due to collision. What is the maximum number of hosts which this protocol can support if each host has to be provided a minimum throughput of $0.16$ frames per time slot?

1. $1$
2. $2$
3. $3$
4. $4$
edited | 4.2k views
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in place of minimum if they put maximum throughput is 0.16frames/slot?

then how to solve that ques?
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Could anybody explain this problem clearly as I am not getting the answer ?

Let there be $N$ such hosts.
Then when one host is transmitting then others must be silent for successful transmission.
So the throughput per host

$0.16 = 0.2 \times 0.8^{N-1}$

$\implies 0.8 = 0.8 ^{N-1}$

on comparing the exponents, since base are identical

$N-1=1, N=2.$

Correct Answer: $B$
edited
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Please clear above doubt.+1 to that
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did u get the clarification?
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Nope
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@bikram sir, please clarify the above doubt.
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Can you compile the doubt properly ?

+2

Sure sir. The doubt id as follows:-

1)If there are n host then total throughput will be = 0.16n .
probability at which a frame can transmit successfully = np(1−p)n−1 .

why these two values have to be equal ?

2) Is the method given in the last answer by Rahul Sharma correct?

+14

1.   Let ,

Probability of transmitting for a single station is p. And also assume that there is n number of stations.

The probability of one station can transmit in a given slot with probability of transmission p from n stations is

nc1 * p *(1-p)(n-1)    ....(i)

now, nc1 = n

so , (i) becomes n*p* (1-p)(n-1)

Now in problem

each host has to be provided a minimum throughput of 0.16 frames per time slot

so for n host ,  throughput becomes 0.16n.

and each host always has data to send and transmits with probability p = 0.2 in every slot

so, for n host . total data send = n p (1-p)n-1

put p = 0.2 in above equation,

we get , n * 0.2 *(0.8)(n-1)

as both are deals with total amount of data send in a given time period ..

0.16 n = n * 0.2 *(0.8)(n-1)

--------------

2.  Yes, the method applied by @rahul is correct.

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There is a small misunderstanding in point no. 1, that is the probability that a station can transmit its frame

is p(1-p)^(n-1)

and not np(1-p)^(n-1)        ------(i).

(i) is the resultant probability, we can say that it is entire throughput.

Also the resultant throughput should be 0.16*n

so they are equated.

+3
@Arjun Sir ......can u explain  how  throughput on one side is equated with probability on other side.....because on left hand side is 0.16 whose unit is frames / sec but on right hand side we have probability with no unit....

Please sir explain it..and also what does this p=0.2 means in layman's term
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If there are N hosts then why didn't take nC1*(.2)(.8)^n-1
We know maximum through put of slotted aloha =.368.

Assume frame size =x.Means sender will transfer x byte frame in one time slot.And slotted aloha will be able to successfully process only .368*x.

Now lets say n stations are there in the network. Now n stations each will transmit .16*x frames.Means in one slot data sent =.16*x*n

Now we know .16*x*n <= .368*x

Max. value which satisfies above is 2
Another way to look @ This problem =>

For single station we get 0.2 Throughput.

For 2 stations =>

We get collusion 0.04 time (Both Transmit)

We get idle channel 0.64 time (Both Do NOt Transmit)

0.32 Time When one of them Transmit.

0.32/2 => 0.16 Throughput.
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just second last line -

when one of them transmit =  0.2 +  0.2 - 0.04 = 0.36 rt?

+2
No, you are incorrect. You can not do like that !

It is

1 - (0.04+0.64) => 1 - (both transmit + none transmit !)
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Oh​h , I calculated either of both transmit..
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Ok, if p = 0.5 & n = 1 then what will be the throughput of the system ?
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how can u just divide the probability of one of them transmitting (0.32) by 2 directly and say that for 1 station the Throughput is 0.16 ?
+1 vote  Option B