Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps. a. Calcula

Question

Suppose a file of 10,000 bytes is to be sent over a line at 2400 bps. a. Calculate the overhead in bits and time in using asynchronous communication. Assume one start bit and a stop element of length one bit, and 8 bits to send the byte itself for each character. The 8-bit character consists of all data bits, with no parity bit. b. Calculate the overhead in bits and time using synchronous communication. Assume that the data are sent in frames. Each frame consists of 1000 characters = 8000 bits and an overhead of 48 control bits per frame. c. What would the answers to parts (a) and (b) be for a file of 100,000 characters? d. What would the answers to parts (a) and (b) be for the original file of 10,000 characters except at a data rate of 9600 bps?

Explanation / Answer

File size: 10,000 Bytes = 10,000*8 bits = 80,000 bits

Link Bandwidth:2400 bps = 2400 bits per second

a) We are using asynchronous transmission.

To send a single byte( i.e, 8 bits)to recipient, we had to send 2 bits extra along with that byte.

means 2 bits are redundant which is a overhead in our transmission.

Calculating No. of overhead bits :

per single byte, there are 2 overhead bits,

for 10,000 bytes, overhead bits are 10,000*2 bits( i.e, 20,000 bits)

Calculating overhead Time:

Actually we need to send 80,000 bits which can be send in 80,000 / 2400 seconds(i.e, 33.333 seconds)

But we are sending 80,000 bits along with overhead bits which are 20,000 in count.

Total sending bits = 80,000+20,000 = 1,00,000 bits

These 1,00,000 bits will take 1,00,000 / 2400 seconds(i.e, 41.666 seconds)

So, overhead time in seconds is 41.666 - 33.333 = 8.333 seconds

b) Here we are using synchronous transmission.

The frame size = 8000 bits

control bits per frame = 48 bits

Note: control bits are overhead bits

Calculating No. of overhead bits :

Per 8000 data bits we are having 48 overhead bits.

so, for 80,000 data bits, we will have 10*48 overhead bits(i.e, 480 bits)

Calculating overhead Time:

We able to send 80,000 bits in 33.333 seconds

Total bits we are actually sending are 80,000+480 bits.

Total bits = 80,000+480 = 80,480 bits

to send those many bits through this link, it takes 80,480 / 2400 seconds(i.e, 33.533 seconds)

so, overhead time in seconds = 33.533 - 33.333 = 0.2 seconds

c) File size: 1,00,000 char = 8,00,000 bits

Using Asynchronous Transmission:

Calculating No. of overhead bits :

2 bits per 8 data bits

so, 1,00,000*2 bits are overhead bits

Overhead bits = 2,00,000 bits

Calculating overhead Time:

We can send 8,00,000 bits in 8,00,000/2400 seconds(i.e, 333.333 seconds)

But we are sending 8,00,000+2,00,000=10,00,000 bits

so, it takes 10,00,000/2400 seconds(i.e, 416.666 seconds)

Overhead time = 83.333 seconds

Using Synchronous Transmission:

48 bits per 8,000 data bits

so, 100*48 bits are overhead bits

Overhead bits = 4,800 bits

Calculating overhead Time:

We can send 8,00,000 bits in 8,00,000/2400 seconds(i.e, 333.333 seconds)

But we are sending 8,00,000+4,800=8,04,800 bits

so, it takes 8,04,800/2400 seconds(i.e, 335.333 seconds)

Overhead time = 2 seconds

d)

File size: 10,000 Bytes = 10,000*8 bits = 80,000 bits

Link Bandwidth:9600 bps = 9600 bits per second

Using Asynchronous Transmission:

Calculating No. of overhead bits :

2 bits per 8 data bits

so, 10,000*2 bits are overhead bits

Overhead bits = 20,000 bits

Calculating overhead Time:

We can send 80,000 bits in 80,000/9600 seconds(i.e, 8.333 seconds)

But we are sending 80,000+20,000=1,00,000 bits

so, it takes 1,00,000/9600 seconds(i.e, 10.416 seconds)

Overhead time = 2.083 seconds

Using Synchronous Transmission:

48 bits per 8,000 data bits

so, 10*48 bits are overhead bits

Overhead bits = 480 bits

Calculating overhead Time:

We can send 80,000 bits in 80,000/9600 seconds(i.e, 8.333 seconds)

But we are sending 80,000+480=80,480 bits

so, it takes 80,480/9600 seconds(i.e, 8.383 seconds)

Overhead time = 0.05 seconds

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