FPS=1000, ie: (number of frames)/time=1000, right?

Which is Number of frames per T?

Which is Number of frames per T?

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Purple
asked
in Computer Networks
Jan 26, 2016

20,533 views
2 votes

Answer :

As we know efficiency in slotted aloha = G * e^-G

where G=no of stations who can transmit in Transmission Time

Here transmission time = L/BW = 20 bit / 20 Kbps = 1ms

As BW is 20 Kbps so bits that can be transferred in 1 ms = 20 bits

This means only 1 station can transmit data in time equal to one transmission time

i.e G=1

As throughput is asked in percentage which indicates they are asking indirectly for efficiency

So throughput in percentage = 1 * e ^ -1 =0.3678 i.e 36.78 %

Extending the question :

Throughput = Efficiency * Bandwidth

= 0.3678 * 20 Kbps

= 7.35 Kbps

As we know efficiency in slotted aloha = G * e^-G

where G=no of stations who can transmit in Transmission Time

Here transmission time = L/BW = 20 bit / 20 Kbps = 1ms

As BW is 20 Kbps so bits that can be transferred in 1 ms = 20 bits

This means only 1 station can transmit data in time equal to one transmission time

i.e G=1

As throughput is asked in percentage which indicates they are asking indirectly for efficiency

So throughput in percentage = 1 * e ^ -1 =0.3678 i.e 36.78 %

Extending the question :

Throughput = Efficiency * Bandwidth

= 0.3678 * 20 Kbps

= 7.35 Kbps

1 vote

$\large S_{slotted} = G \times e^{-G}$

$\large G = N \times fps \times T_{fr}$

$\text{frame size} = 20 \ \text{bit} $

Here number of stations are not given and frames generated by whole system is given which is $1000 fps$

so, $G = 1000 \ \times \frac{20}{ 20 \times 10^3} $

$G = 1$

$\large S_{slotted} = 1 \times e^{-1} = 0.368 = 36.8 \text{%}$

$\large G = N \times fps \times T_{fr}$

$\text{frame size} = 20 \ \text{bit} $

Here number of stations are not given and frames generated by whole system is given which is $1000 fps$

so, $G = 1000 \ \times \frac{20}{ 20 \times 10^3} $

$G = 1$

$\large S_{slotted} = 1 \times e^{-1} = 0.368 = 36.8 \text{%}$