Computer Science - Networking - Discussion
Discussion Forum : Networking - Section 2 (Q.No. 16)
16.
A 8-Mbps token ring has a token holding timer value of 10 msec. What is the longest frame (assume header bits are negligible) that can be sent on this ring?
Answer: Option
Explanation:
At 8 Mbps, a station can transmit 80,000 bits or 10,000 bytes in 10 msec. This
is an upper bound on frame length. From this amount, some overhead must be
subtracted, giving a slightly lower limit for the data portion. Therefore, 10,000 B frame is the correct answer.
Discussion:
4 comments Page 1 of 1.
Gr.mathew said:
5 years ago
8mbps=80,000 bits frame.
1byte= 8bits.
We take 80,000bits=? byte
So, Answer is 10,000B.
1byte= 8bits.
We take 80,000bits=? byte
So, Answer is 10,000B.
(1)
Abhishek pandey said:
1 decade ago
8 - mbps and 10msec this is eq to 10/1000sec.
Maxmimum = 8 mbps*.001 = 80,000b frame or 10000 B frame.
Maxmimum = 8 mbps*.001 = 80,000b frame or 10000 B frame.
Sundar said:
1 decade ago
80,000 Bits frame
or
10,000 B Frame
Therefore, Option D is the correct answer.
or
10,000 B Frame
Therefore, Option D is the correct answer.
SARIKA said:
1 decade ago
How can we solve this?
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