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Question

If a file consisting of 50,000 characters takes 40 seconds to send, then the data rate is ______

The correct answer is

10 kbps

Understanding Data Rate Calculation

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).

Converting Characters to Bits

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 = $400,000 \text{ bits}$

Calculating Data Rate in bps

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) = $10,000 \text{ bits/second}$

Converting bps to kbps

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.

Step-by-Step Calculation Summary

  1. Identify the total data size in characters: 50,000 characters.
  2. Identify the time taken: 40 seconds.
  3. Convert characters to bits (assuming 8 bits/character): $50,000 \text{ characters} \times 8 \text{ bits/character} = 400,000 \text{ bits}$.
  4. Calculate data rate in bits per second (bps): $\frac{400,000 \text{ bits}}{40 \text{ seconds}} = 10,000 \text{ bps}$.
  5. Convert bps to kilobits per second (kbps) by dividing by 1000: $\frac{10,000 \text{ bps}}{1000} = 10 \text{ kbps}$.
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

Revision Table: Data Rate Concepts

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

Additional Information: Data Measurement

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.

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Important Questions from Data Link Layer

  1. Which one of the following functionality is not performed by data link layer?

  2. It is necessary to design a link-layer protocol between two hosts that are directly connected over a lossless link of length 3000 kilometers. Assume that the link bandwidth is $10^8$ bits per second and that the propagation delay in the link is 5 nanoseconds per meter. Every transmitted data byte is assigned a unique sequence number.

    Let $N$ be the minimum number of bits needed for the sequence number field in the protocol header such that
    i. the sequence numbers do not wrap around before 60 seconds, and
    ii. the maximum utilization of the link is achieved.

    The value of $N$ is ______. (answer in integer)
  3. In a LAN environment, switches improve network efficiency primarily by:

  4. Which of the following protocols is responsible for converting higher level protocol addresses to physical network addresses?

  5. Which of the following are controlled-access protocols
    A. Reservation
    B. Polling
    C. TDMA
    D. Token Passing
    E. CSMA/CA
    Choose the correct answer from the options given below:
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