Question Bank - Computer Networks

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Suppose we want to download-text documents at the rate of 100 pages per second. Assume that a page consists of an average of 24 lines with 80 characters in each line. What is the required bit rate of the channel?

A.
192 kbps
B.
512 kbps
C.
1.248 Mbps
D.
1.536 Mbps

Solution:

The correct answer is "option 4".CONCEPT: Bit rate means no. of bits per second.Data: Download rate: 100 pages/secondNo. of lines per page: 24 lines/pageNo. of characters per line: 24 character/lineFormula:Bit rate of channel = No. of pages/second × No. of lines/page × No. of characters/line × No. of bitsCALCULATION: Bit rate of channel = 100 pages × 24 lines × 80 characters × 8 bits = 1536000 bits/second or 1.536MbpsHence, the bit rate is 1.536Mbps.Mistake Points Never consider Mbps & MBPS as the same. Mbps - Mega Bits Per SecondMBPS - MegaBytes Per Second1 Byte - 8 bits

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