IPv4 subnetting
- Hosts per subnet
- 2(32 − prefix) − 2 (/31: 2, /32: 1)
- Block size
- 256 − (mask octet)
- Network address
- IP AND mask
- Broadcast address
- IP OR (NOT mask): all host bits set to 1
- Subnets from n borrowed bits
- 2n
Computer Networks: cram pack
Formula sheets, the key method steps for every topic and 6 practice problems, with full solutions at the back.
1Formulas and methods
Number systems
- Place values
- binary 2k, octal 8k, hex 16k
- To decimal
- Σ dk·bk
- From decimal
- divide by the base, read the remainders bottom-up
- Hex ↔ binary
- one hex digit = 4 bits
- Two’s complement
- invert the bits and add 1
2Practice problems
1.IPv4 subnettingMedium
The network 192.168.78.0/24 is divided into /25 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.78.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A router has routes to these networks: 10.246.40.0/24, 10.246.41.0/24, 10.246.42.0/24, 10.246.43.0/24, 10.246.44.0/24, 10.246.45.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.246.40.0/24 … 10.246.45.0/24
p =
3.Computer networksMedium
Which IPv4 header field stops packets from circulating forever in a routing loop?
- ATime to live (TTL)
- BType of service
- CHeader checksum
- DIdentification
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo compress the data more efficiently
- BTo encrypt the payload of each segment
- CTo select the best route to the destination for the segments of that connection
- DTo avoid confusion with old segments and make spoofed segments harder to inject
5.Number systemsMedium
Write −52 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −528 bit
U =
6.Number systemsHard
A system must give every one of 4734 items its own binary code of the same length. What is the smallest number of bits b needed?
4734
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 128
IPv4 subnetting: an address has 32 bits. The prefix /p says how many leading bits identify the network; the remaining 32 − p bits number the hosts. A subnet has 2³²⁻ᵖ addresses, but the first (network address) and last (broadcast address) cannot be given to hosts, so 2³²⁻ᵖ − 2 hosts fit. Exceptions: a /31 (point-to-point link, RFC 3021) uses both of its 2 addresses for hosts, and a /32 is one single address (a single host or a loopback).
Block size
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 128
2.IPv4 subnetting
Answer: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
40 = 00101000
45 = 00101101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.246.40.0/21 (it also covers 2 network(s) that are not in the list).
3.Computer networks
Answer: A · Time to live (TTL)
Why this is the answer
Every router decrements the TTL; at zero the packet is discarded and an ICMP Time Exceeded message is usually sent. IPv6 calls it hop limit.
Answer
A · Time to live (TTL)
4.Computer networks
Answer: D · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
D · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 204
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Invert and add one
52 = 0011 0100
inverted: 1100 1011
+1 → 1100 1100
Read it as unsigned
1100 1100₂ = 256 − 52 = 204
Answer
U = 204
6.Number systems
Answer: b = 13 bit
Positional number systems: each digit is worth its value times a power of the base. In binary the places are worth 1, 2, 4, 8, 16, …; in hexadecimal 1, 16, 256, …, with A–F standing for 10–15. In 8-bit two's complement the leftmost bit is worth −128 instead of +128, which is how negative numbers are stored.
Powers of two
212 = 4096 < 4734
213 = 8192 ≥ 4734
Answer
b = 13
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