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 subnet uses the mask 255.255.255.224. How many usable host addresses H does it have?
255.255.255.224
H =
2.IPv4 subnettingHard
A host has the address 10.20.34.138/29. What is the fourth octet x of the last usable host address in its subnet?
10.20.34.138/29
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/32
- B/8
- C/24
- D/16
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 does not use addresses
- BIPv6 encrypts all addresses
- CIPv6 only works on local networks, so addresses never need to be shared globally
- DIts 128-bit address space is large enough to give every device a global address
5.Number systemsMedium
The 8-bit pattern 1110 1101 stores a signed integer in two's complement. What is its value N in decimal?
1110 11018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1011 << 2. What is the result's value R in decimal?
0101 1011<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Prefix length
255.255.255.224 = /27
Usable hosts
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 142
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
23 = 8
Subnet of this host
network: ⌊138/8⌋·8 = 136
broadcast: 136 + 8 − 1 = 143
usable hosts: 137 … 142
Answer
x = 142
3.Computer networks
Answer: C · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
C · /24
4.Computer networks
Answer: D · Its 128-bit address space is large enough to give every device a global address
Why this is the answer
IPv6 has about 3.4 × 10³⁸ addresses, so the scarcity that drove NAT in IPv4 disappears; firewalls still provide security.
Answer
D · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: N = −19
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 + 64 + 32 + 8 + 4 + 1 = −19
Answer
N = −19
6.Number systems
Answer: R = 108
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
0101 1011 << 2
= 0110 1100
Value of the result
0110 1100₂ = 108
Answer
R = 108
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