If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is ______
10 kbps
The question asks us to determine the data rate if a file containing 50,000 characters is sent in 40 seconds. The data rate is essentially how much data is transferred per unit of time. It is commonly measured in bits per second (bps) or kilobits per second (kbps).
To calculate the data rate in bits per second, we first need to convert the total number of characters into bits. A standard assumption for character encoding (like ASCII) is that one character is represented by 8 bits (which is equal to 1 byte, but the options are in bits, so we'll work with bits directly).
Total number of characters = 50,000
Number of bits per character = 8 bits
Total data in bits = Number of characters $\times$ Bits per character
Total data in bits = $50,000 \times 8 \text{ bits}$
Total data in bits = $400,000 \text{ bits}$
Now that we have the total data in bits and the time taken to send it, we can calculate the data rate in bits per second (bps).
Time taken = 40 seconds
Data Rate (bps) = $\frac{\text{Total data in bits}}{\text{Time taken in seconds}}$
Data Rate (bps) = $\frac{400,000 \text{ bits}}{40 \text{ seconds}}$
Data Rate (bps) = $10,000 \text{ bits/second}$
The options provided are in kilobits per second (kbps). The prefix 'kilo' in computing networking usually refers to 1000 (though in some contexts like memory, it can mean 1024). For network speeds like kbps, Mbps, Gbps, the base is typically 1000.
1 kbps = 1000 bps
To convert bits per second (bps) to kilobits per second (kbps), we divide the value in bps by 1000.
Data Rate (kbps) = $\frac{\text{Data Rate in bps}}{1000}$
Data Rate (kbps) = $\frac{10,000 \text{ bps}}{1000 \text{ bps/kbps}}$
Data Rate (kbps) = $10 \text{ kbps}$
Thus, the data rate is 10 kbps.
| Metric | Value | Units |
|---|---|---|
| File Size (Characters) | 50,000 | characters |
| Time Taken | 40 | seconds |
| Bits per Character (Standard) | 8 | bits/character |
| Total Data | 400,000 | bits |
| Data Rate | 10,000 | bps |
| Data Rate | 10 | kbps |
| Term | Definition | Common Units |
|---|---|---|
| Data Rate | The amount of data transferred per unit of time. | bps, kbps, Mbps, Gbps |
| Bit | The smallest unit of data, represented as 0 or 1. | N/A |
| Character | A symbol (like a letter, number, or punctuation mark) often represented by a fixed number of bits (e.g., 8 bits). | N/A |
| bps | Bits per second (1 bps = 1 bit per second). | N/A |
| kbps | Kilobits per second (1 kbps = 1000 bits per second). | N/A |
| Mbps | Megabits per second (1 Mbps = 1,000,000 bits per second). | N/A |
It's important to distinguish between units used for data storage (like bytes, kilobytes, megabytes, etc.) and units used for data rate (like bits per second, kilobits per second, megabits per second). Storage units usually use prefixes where 'kilo' means $2^{10} = 1024$, 'mega' means $2^{20} = 1024^2$, and so on (referred to as kibibytes, mebibytes, etc., though often incorrectly labeled KB, MB). However, for network transmission speeds, the decimal prefixes are standard: 1 kilobit = 1000 bits, 1 megabit = 1,000,000 bits.
One byte is equal to 8 bits. So, 50,000 characters (assuming 1 character = 1 byte) is 50,000 bytes. In bits, this is $50,000 \text{ bytes} \times 8 \text{ bits/byte} = 400,000 \text{ bits}$. This confirms our initial conversion step.
Which of the following protocols is responsible for converting higher level protocol addresses to physical network addresses?
Which of the following devices takes data sent from one network device and forwards it to the destination node based on MAC address?
Which of the following statements is/are true with regard to various layers in the Internet stack?
P: At the data link layer, a packet of transmitted information is called a frame
Q: At the network layer, a packet of transmitted information is called a segment
Given below are two statements:
Statement I: The laws of nature put two fundamental limits on data rate of a channel. The H.Nyquist limit restricts the number of independent samples per second to twice the band-width in a Noiseless channel.
Statement II: Shannon's major result about noised channel is that maximum data rate of a channel whose band width is H Hz, and whose signal-to-noise ratio is S/N is given by:
Maximum number of bits/sec \(= H \;log_2 \left(1+\frac{S}{N}\right)\) is given by:
In the light of the above statements. choose the correct answer from the options given below
Which of the following statements are true?
(a) Three broad categories of Networks are:
(i) Circuit Switched Networks
(ii) Packet Switched Networks
(iii) Message Switched Networks
(b) Circuit Switched Network resources need not be reserved during the set up phase.
(c) In packet switching there is no resource allocation for packets.