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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. $4$

in place of minimum if they put maximum throughput is 0.16frames/slot?

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

very imp line

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

then n host = 0.16n

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: $B$

@gaurav yadav thanks bhai i was stuck for so long ,i dont know why people not voting for this comment
Nice explanation

Thanks for explaining, very good explanation.

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.

Ok, if p = 0.5 & n = 1 then what will be the throughput of the system ?
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 ?
from where this .32 came????
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

Option B