5,673 views
2 2 votes

Plz explain it .

its made easy question, they have provided solution for A  and marked C as correct

1 Answer

Best answer
1 1 vote

$Total channel load = \frac{Total number of request / second}{Total number of slots / second}$

As average station makes 36 request / hour. So each station makes a request every 3600 sec / 36 request = 100 sec.

Total load is 10000 requests per 100 sec, i.e., 10000 / 100 = 100 request per second.

Number of slots in 1 second = 1 sec / 100 microsec = 10000 slots per second.

$Total channel load = \frac{100}{10000} = \frac{1}{100}$

• selected by
Position:
Show:

Related questions

1 1 vote
1 1 answer
255
255 views
ASH1198 asked Mar 11
255 views
Consider a slotted ALOHA with 2Mbps ethernet, what is the throughput (in Kbps) for slotted ALOHA at G=3?A. 218.5B. 242.8C. 320.7D. 298.6
3 3 votes
1 1 answer
348
348 views
ASH1198 asked Feb 4
348 views
In a shared 75 kbps channel, each station transmits a 20-bit frame every 2000 ms. If the networkuses Slotted Aloha, what is the maximum number of stations that can be sup...
5 5 votes
2 2 answers
414
414 views
GO Classes asked Oct 22, 2025
414 views
A network designer is configuring a time-slotted MAC protocol for a dedicated system with exactly 8 hosts. Each host will use the same transmission probability, $p$, in e...
2 2 votes
1 1 answer
347
347 views
GO Classes asked Oct 22, 2025
347 views
In a time-slotted MAC protocol system, there are 5 hosts competing for the channel. Each host always has data to send and attempts to transmit in any given time slot with...