The transmission time required for sending a 512-byte packet using a 56…
2023
The transmission time required for sending a 512-byte packet using a 56 kilobits (1 kilo = 1000) per second modem is about
Answer: C. 73 milliseconds — CONCEPTTransmission delay is the time needed to place all packet bits onto a link. It equals the packet length in bits divided by the link bit rate. Convert…
- A.
7 milliseconds
- B.
43 milliseconds
- C.
73 milliseconds
- D.
80 milliseconds
Attempted by 2 students.
Show answer & explanation
Correct answer: C
CONCEPT
Transmission delay is the time needed to place all packet bits onto a link. It equals the packet length in bits divided by the link bit rate.
Convert bytes to bits by multiplying by 8, keep the rate in bits per second, and convert seconds to milliseconds by multiplying by 1000.
APPLICATION
The packet length is 512 bytes × 8 bits per byte = 4096 bits.
The modem rate is 56 kilobits per second = 56,000 bits per second because 1 kilo = 1000.
Transmission time = 4096 ÷ 56,000 seconds = 0.073142857 seconds.
In milliseconds, 0.073142857 × 1000 = 73.142857 milliseconds, which is about 73 milliseconds.
CROSS-CHECK
At 56,000 bits per second, 73 milliseconds carries 56,000 × 0.073 = 4088 bits, which is approximately the 4096-bit packet after rounding.
RESULT
Therefore, the required transmission time is about 73 milliseconds.