Required Bit Rate for a Document Download: 25-Comp-B5 Computer Communications, December 2014, Q2

Question 2: Required Bit Rate for a Document Download (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A document of 100 pages; each page has 24 lines of text; each line is 80 characters long; each character is assumed encoded with 8 bits (plain ASCII text over a byte-aligned channel — the standard assumption for this class of problem when a character width is not stated).

Given data
Quantity	Value
Lines per page	24
Characters per line	80
Bits per character (assumed)	8
Pages to download	100

Find. The required bit rate (bits per second) for the channel to deliver the entire 100-page document.

CheckThe question gives no target download time. A bit rate cannot be computed from a data volume alone (rate = volume ÷ time), so this is a genuine omission in the exam's data. We report the document's exact size (fully determined by the given data) and the general rate formula, then illustrate a definite numeric answer using an explicitly assumed download time of 1 second; rescale $R = \text{TotalBits}/T$ for any other assumed $T$ the grader intends.

Approach. Compute the total number of bits in the document from the page geometry and the assumed character width, then divide by the (assumed) time budget to obtain the required bit rate.

Characters per page. Each page has 24 lines of 80 characters each: $$\text{chars/page} = 24 \times 80 = 1920\ \text{characters}.$$
Bits per page. At 8 bits per character: $$\text{bits/page} = 1920 \times 8 = 15{,}360\ \text{bits}.$$
Total document size. Over 100 pages: $$\text{Total bits} = 15{,}360 \times 100 = \boxed{1{,}536{,}000\ \text{bits} = 1.536\ \text{Mbit}}.$$ This figure is fixed by the given data alone and does not depend on any timing assumption.
Required bit rate. A bit rate is a volume divided by a time budget, $R = \text{Total bits}/T$. Adopting the illustrative $T = 1\ \text{s}$ stated in the check callout above: $$R = \frac{1{,}536{,}000\ \text{bits}}{1\ \text{s}} = \boxed{1.536\ \text{Mbps}}.$$ For any other assumed download time, rescale directly — e.g. a 10 s budget requires only $153.6\ \text{kbps}$, and a 1-minute budget requires only $25.6\ \text{kbps}$.

Final results — Question 2
Quantity	Result
Characters per page	1920
Bits per page (8 bits/char)	15,360 bits
Total document size (100 pages)	1,536,000 bits = 1.536 Mbit
Required bit rate (assumed $T=1\,\text{s}$)	1.536 Mbps — general form $R = 1.536\times10^{6}/T$ bps

Topic: Data volume vs. bit rate — the missing-time-budget trap.

Key relations: $\text{Bits} = \text{pages} \times \text{lines/page} \times \text{chars/line} \times \text{bits/char}$; bit rate $R = \text{Bits}/T$.

Why this works: a "rate" is inherently a quantity per unit time; without a stated (or otherwise recoverable) time budget, one and the same document can be sent at any bit rate at all — the calculation only becomes well-posed once a transfer-time constraint is supplied, so the number of bits to be transmitted (the size) is the only quantity the given data alone determines.

Common pitfall: reporting the total bit count as though it were itself "the bit rate" (bits vs. bits/second are different physical quantities), and silently picking a character width without stating the assumption — 7-bit ASCII without a parity/framing bit gives 1,344,000 bits instead, a difference of about 12.5%.

Source: Kurose & Ross, Computer Networking: A Top-Down Approach, 7th ed., Ch.1 §1.4 (Delay, Loss, and Throughput in Packet-Switched Networks).

Question 2 of 7: Required Bit Rate for a Document Download

Question 2: Required Bit Rate for a Document Download (20 marks)