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
A department needs 268 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
268 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.244.83/23. What is the third octet x of its network address?
172.20.244.83/23
x =
3.Computer networksMedium
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
- A10.0.0.0/8, because it covers more addresses than the /16
- BBoth routes, splitting the traffic equally
- C10.1.0.0/16, because it is the longest matching prefix
- D10.0.0.0/8, because it was added first
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- BIPv6 fragments at the transport layer inside TCP
- CRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- DIPv6 does not allow packets larger than 576 bytes
5.Number systemsMedium
Write −55 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −558 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 16 bits?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 23
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).
Host bits needed
28 − 2 = 254 < 268
29 − 2 = 510 ≥ 268 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 244
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 in the third octet
21 = 2
Subnet of this host
network: 172.20.244.0
broadcast: 172.20.245.255
Answer
x = 244
3.Computer networks
Answer: C · 10.1.0.0/16, because it is the longest matching prefix
Why this is the answer
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer
C · 10.1.0.0/16, because it is the longest matching prefix
4.Computer networks
Answer: C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 201
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
55 = 0011 0111
inverted: 1100 1000
+1 → 1100 1001
Read it as unsigned
1100 1001₂ = 256 − 55 = 201
Answer
U = 201
6.Number systems
Answer: N = 65535
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.
Largest value
216 − 1 = 65535
Answer
N = 65535
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