0 — Test 388

Correct Answer: B. B. Local forwarding
Explanation: This is the function of local forwarding.

Correct Answer: C. C. Post office protocol
Explanation: ‘POP’ stands for Post Office Protocol.

Correct Answer: B. B. 4
Explanation: The first 4-bits show the version i.e. 0100 -> version 4.

Correct Answer: B. B. 8
Explanation: The right 4-bits multiplied by 4 is the header length (0010=2). 2 x 4 = 8.

Correct Answer: C. C. 32
Explanation: HLEN = 1000 = 8. 8 x 4 = 32 bytes.

Correct Answer: D. D. 20
Explanation: Total number of bytes in the header = 5 x 4 = 20 bytes. Total length = 0028_16 = 40 bytes. Data carried by the packet = (40 – 20) = 20 bytes.

Correct Answer: D. D. 1
Explanation: In an IPv4 header, the TTL is the 9th byte (or the 17th and 18th hexadecimal digits) which is one when counted from right to left. (or the 17th and 18th hexadecimal digits).

Correct Answer: B. B. 128.11.3.31
Explanation: Converting the binary value to their respective decimal values yields 128.11.3.31.

Correct Answer: A. A. 129.11.11.239
Explanation: Converting the binary value to their respective decimal values yields 129.11.11.239. Step-by-Step ConversionFor each octet, we multiply each bit by $2^n$ where $n$ is its position (from 7 down to 0).First Octet: 10000001$1 times 2^7 = 128$$0 times 2^6 = 0$$0 times 2^5 = 0$$0 times 2^4 = 0$$0 times 2^3 = 0$$0 times 2^2 = 0$$0 times 2^1 = 0$$1 times 2^0 = 1$Total: $128 + 1 = mathbf{129}$Second Octet: 00001011$0 times 128 = 0$$0 times 64 = 0$$0 times 32 = 0$$0 times 16 = 0$$1 times 8 = 8$$0 times 4 = 0$$1 times 2 = 2$$1 times 1 = 1$Total: $8 + 2 + 1 = mathbf{11}$Third Octet: 00001011This is identical to the second octet.Total: 11Fourth Octet: 11101111$1 times 128 = 128$$1 times 64 = 64$$1 times 32 = 32$$0 times 16 = 0$$1 times 8 = 8$$1 times 4 = 4$$1 times 2 = 2$$1 times 1 = 1$Total: $128 + 64 + 32 + 8 + 4 + 2 + 1 = mathbf{239}$Final ResultBy combining the four decimal values with periods, we get:Dotted-Decimal Notation: 129.11.11.239

Correct Answer: A. A. 01101111 00111000 00101101 01001110
Explanation: Replacing each decimal number with its binary equivalent we get 01101111 00111000 00101101 01001110.