All categories

3 years ago

Which is a connectionless protocol in the transport layer? What are the small chunks of data called?

Computers and Technology

2 answers:

xeze [42]3 years ago

6 0

Answer:

User Datagram Protocol (UDP) , transport-layer segment.

Explanation:

The User Datagram Protocol is popularly known as UDP. It is defined as the communication protocol which is used across the internet for any time sensitive transmission like the DNS lookup or the video playback.

The UDP provides a unreliable and connectionless service to a invoking application.

The transport layers on sending side converts the application $\text{layer }$ messages which it $\text{receives}$ from the $\text{sending application process}$ into a transport layer segment called as the transport layer segments. This is achieved by breaking down the application messages into a smaller chunks and then adding the transport layer header into each chunk so as to create a transport layer segment.

slava [35]3 years ago

3 0

UDP and the small chunks of data are called packets

You might be interested in

1. How many bits would you need to address a 2M × 32 memory if:

Dominik [7]

Answer:

a) 23 b) 21
a) 43 b) 42
a) 0 b) 0

Explanation:

1) How many bits is needed to address a 2M * 32 memory

2M = 2^1*2^20, while item =32 bit long word

hence ; L = 2^21 ; w = 32

a) when the memory is byte addressable

w = 8; L = ( 2M * 32 ) / 8 = 2M * 4

hence number of bits = log2(2M * 4)= log2 ( 2 * 2^20 * 2^2 ) = 23 bits

b) when the memory is word addressable

W = 32 ; L = ( 2M * 32 )/ 32 = 2M

hence the number of bits = log2 ( 2M ) = Log2 (2 * 2^20 ) = 21 bits

2) How many bits are required to address a 4M × 16 main memory

4M = 4^1*4^20 while item = 16 bit long word

hence L ( length ) = 4^21 ; w = 16

a) when the memory is byte addressable

w = 8 ; L = ( 4M * 16 ) / 8 = 4M * 2

hence number of bits = log 2 ( 4M * 2 ) = log 2 ( 4^1*4^20*2^1 ) ≈ 43 bits

b) when the memory is word addressable

w = 16 ; L = ( 4M * 16 ) / 16 = 4M

hence number of bits = log 2 ( 4M ) = log2 ( 4^1*4^20 ) ≈ 42 bits

3) How many bits are required to address a 1M * 8 main memory

1M = 1^1 * 1^20 , item = 8

L = 1^21 ; w = 8

a) when the memory is byte addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

hence number of bits = log 2 ( 1M ) = log2 ( 1^1 * 1^20 ) = 0 bit

b) when memory is word addressable

w = 8 ; L = ( 1 M * 8 ) / 8 = 1M

number of bits = 0

5 0

3 years ago

Write the following numbers in binary. 1. 2. 3. 4. 5. 6. 7. 8. 9. 10. 11. 12. 13. 14. 15. 16. 17. 18. 19. 20. 21. 22. 23. 24. 25

AfilCa [17]

0 1 10 11
Explanaition:

4 0

3 years ago

Email Communication has it own set of etiquette guidelines for users to follow if they want to be effective communicators. Which

Olin [163]

You answer is most likely D.

5 0

3 years ago

The common channel signaling (CCS) system provides a separate network dedicated to control and signaling over the PSTN. This ena

dezoksy [38]

The correct answer is A.True

8 0

3 years ago

Create a program in Matlab that prints the numbers from 1-100.

Burka [1]

Answer:

numbers = 1:1:100;

for num=numbers

remainder3 = rem(num,3);

remainder5 = rem(num,5);

if remainder3==0

disp("Yee")

else

if remainder3 == 0 && remainder5 == 0

disp ("Yee-Haw")

else

if remainder5==0

disp("Haw")

else

disp("Not a multiple of 5 or 4")

end

Explanation:

Initialize the numbers variable from 1 to 100.
Loop through the all the numbers and find their remainders.
Check if a number is multiple of 5, 3 or both and display the message accordingly.

4 0

3 years ago