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 163 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
163 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.135.167/30. What is the fourth octet x of its network address?
10.20.135.167/30
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A62
- B64
- C30
- D126
4.Computer networksHard
Which routing protocol is a link-state protocol?
- AOSPF
- BRIP
- CARP
- DBGP
5.Number systemsMedium
The 8-bit pattern 1000 0100 stores a signed integer in two's complement. What is its value N in decimal?
1000 01008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0101 << 1. What is the result's value R in decimal?
1100 0101<< 1
R =
3Solutions
1.IPv4 subnetting
Answer: p = 24
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
27 − 2 = 126 < 163
28 − 2 = 254 ≥ 163 → h = 8
Prefix length
p = 32 − 8 = 24
Answer
p = /24
2.IPv4 subnetting
Answer: x = 164
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
22 = 4
Subnet of this host
network: ⌊167/4⌋·4 = 164
broadcast: 164 + 4 − 1 = 167
usable hosts: 165 … 166
Answer
x = 164
3.Computer networks
Answer: A · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
A · 62
4.Computer networks
Answer: A · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
A · OSPF
5.Number systems
Answer: N = −124
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.
Add the place values
The other bits keep their normal positive values.
−128 + 4 = −124
Answer
N = −124
6.Number systems
Answer: R = 138
Setting bits: each bit is worth a power of two (128, 64, …, 1). One hex digit is exactly four bits, and in 8-bit two's complement the leftmost bit is worth −128. Bitwise operations work on each bit position separately.
Bit by bit
Shifting left by 1 moves every bit 1 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 0101 << 1
= 1000 1010
Value of the result
1000 1010₂ = 138
Answer
R = 138
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