TCP/IP Questions and Answers

Answer: B

To turn a binary number into decimal, you just have to add the values of each bit that is a 1. The values of 10011101 are 128, 16, 8, 4, and 1. 128 + 16 + 8 + 4 + 1 = 157.

Hexadecimal is a base-16 number system. The values of hexadecimal are 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F-16 characters total, from which to create all the numbers you'll ever need. So, if 1001 in binary is 9, then the hexadecimal equivalent is 9. Since we then have 1101, which is 13 in binary, the hexadecimal answer is D and the complete hexadecimal answer is 0x9D.

Answer: B

Correct answer: TCP

Telnet needs the characters to arrive in order and without loss, so it runs over TCP, on port 23. IP is a layer 3 protocol, TCP/IP is the name of the suite rather than a protocol, and UDP would deliver a terminal session with characters missing or out of sequence.

Answer: B

Correct answer: 2 and 3

Statements 2 and 3 are correct: ICMP carries information about network problems, and its messages are encapsulated within IP datagrams. Statement 1 is wrong - ICMP reports on delivery failures but guarantees nothing, and IP remains best-effort. Statement 4 confuses the encapsulation: ICMP sits directly inside IP with protocol number 1, not inside UDP.

Answer: A

Correct answer: 1, 3 and 6

DHCP, SNMP and TFTP all use UDP. Each has a reason: DHCP must broadcast before the client has an address, SNMP sends small independent polls and traps where a lost one can simply be repeated, and TFTP handles its own acknowledgements. SMTP, FTP and HTTP all need the reliable ordered stream that TCP provides.

Answer: C

Correct answer: 10xxxxxx

The classes are distinguished by their leading bits: Class A begins with 0, Class B with 10, Class C with 110 and Class D with 1110. So a Class B address starts 10 in binary, which corresponds to a first octet of 128 to 191 in decimal - and reading it that way is a useful cross-check.

Answer: B

This seems like a hard question at first because it doesn't make sense. The listed answers are from the OSI model and the question asked about the TCP/IP protocol stack (DoD model). However, let's just look for what is wrong. First, the Session layer is not in the TCP/IP model; neither are the Data Link and Physical layers. This leaves us with the Transport layer (Host-to-Host in the DoD model), Internet layer (Network layer in the OSI), and Application layer (Application/Process in the DoD).

Answer: C

Dynamic Host Configuration Protocol (DHCP) is used to provide IP information to hosts on your network. DHCP can provide a lot of information, but the most common is IP address, subnet mask, default gateway, and DNS information.

Answer: D

Correct answer: 192.168.24.43

The private ranges defined by RFC 1918 are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16. Only the last option falls in one. The trap is 172.15.14.36, which is outside the private block - that range begins at 172.16 and ends at 172.31, so 172.15 and 172.32 are both public.

Answer: C

Correct answer: Proxy ARP

Proxy ARP lets a router answer an ARP request on behalf of a host on another network, supplying its own MAC address so that the sender forwards the frame to the router, which then routes it onward. It exists for hosts that have no default gateway configured or that believe the whole address range is local. RARP and inverse ARP work in the opposite direction, resolving an address from a known hardware or circuit identifier.