"Network Layer: IPv4, CIDR, Fragmentation, ARP, DHCP, ICMP & NAT" — Practice

Question 1

How many usable host addresses does a /26 block contain, and what is its mask?

Accepted Answer

It contains 2 to the 6, which is 64 addresses, of which 62 are usable. The mask is 255.255.255.192.

Question 2

What does the time to live field prevent, and what happens when it reaches zero?

Accepted Answer

It prevents a datagram circulating forever in a routing loop. A router decrementing it to zero discards the datagram and sends an ICMP time exceeded message to the source.

Question 3

Which three address ranges are reserved for private use?

Accepted Answer

10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16.

Question 4

Subnet 200.10.16.0/20 into 8 equal subnets and give the range of the third one.

Accepted Answer

Eight subnets need 3 borrowed bits, giving /23 with 512 addresses each. The third subnet spans 200.10.20.0 to 200.10.21.255, with usable hosts from 200.10.20.1 to 200.10.21.254 and mask 255.255.254.0.

Question 5

A table has 192.168.0.0/16 to A, 192.168.4.0/22 to B, 192.168.5.128/25 to C and a default route to D. Where does traffic for 192.168.5.130 go?

Accepted Answer

All four entries match, since 5 AND 252 is 4 for the /22 and 130 AND 128 is 128 for the /25. Longest prefix match selects the /25, so the datagram goes to interface C.

Question 6

Fragment a 4000-byte datagram with a 20-byte header for an MTU of 1500 bytes.

Accepted Answer

The payload is 3980 bytes and each fragment carries at most 1480, which is a multiple of 8. Fragment 1 carries bytes 0 to 1479 at offset 0, fragment 2 carries 1480 to 2959 at offset 185, and fragment 3 carries the remaining 1020 bytes at offset 370 with more-fragments clear.

Question 7

Can 200.10.8.0/24 through 200.10.11.0/24 be aggregated? What about 200.10.9.0/24 through 200.10.12.0/24?

Accepted Answer

The first four aggregate to 200.10.8.0/22, since the third octets 8 to 11 share their first six bits and 8 is aligned to a four-block boundary. The second set cannot, because 9 is not aligned; the shortest common prefix would be a /20 covering eleven blocks the organisation does not own.

Question 8

Host A at 10.0.1.5 sends to B at 10.0.2.9 via router R with interfaces 10.0.1.1 and 10.0.2.1. Give the source and destination IP and MAC in both frames.

Accepted Answer

In the first frame the IPs are 10.0.1.5 and 10.0.2.9, the source MAC is A's and the destination MAC is R's 10.0.1.1 interface. In the second frame the IPs are unchanged, the source MAC is R's 10.0.2.1 interface and the destination MAC is B's. IP addresses are end to end; MAC addresses change at every hop.

Question 9

Two internal hosts behind one NAT address each open a connection to the same server from source port 5000. Describe the translation and one thing NAT breaks.

Accepted Answer

NAT allocates distinct external ports, mapping the first to say port 40001 and the second to 40002, rewriting source address and port on the way out and matching on destination port on the way back, with checksums recomputed. It breaks unsolicited inbound connections, since no table entry exists until an internal host sends first, and it breaks protocols such as classic FTP that carry an IP address inside the payload.

Question 10

Explain why a single lost fragment destroys an entire datagram and how path MTU discovery avoids the problem.

Accepted Answer

Reassembly happens only at the destination and needs every fragment, so a missing one leaves the rest unusable until a reassembly timer expires and discards them all, with no notification of which fragment was lost. Path MTU discovery sets the don't-fragment flag so an oversized datagram is discarded with an ICMP message naming the MTU, letting the source send appropriately sized datagrams that are never fragmented.

Practice — "Network Layer: IPv4, CIDR, Fragmentation, ARP, DHCP, ICMP & NAT"

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