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 255.255.255.192. How many usable host addresses H does it have?
255.255.255.192
H =
2.IPv4 subnettingHard
A host has the address 10.20.145.106/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.145.106/28
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt is the switch port the host is connected to
- BIt is the router used for traffic to other networks
- CIt is the DNS server
- DIt assigns MAC addresses
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 encrypts all addresses
- BIPv6 does not use 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
Set the bits so they store −101 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1018 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 0110 AND 1000 0111. What is the result's value R in decimal?
0010 0110AND1000 0111
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
255.255.255.192 = /26
Usable hosts
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 110
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
24 = 16
Subnet of this host
network: ⌊106/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110
Answer
x = 110
3.Computer networks
Answer: B · It is the router used for traffic to other networks
Why this is the answer
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer
B · It is the router used for traffic to other networks
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: U = 155
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.
Invert and add one
101 = 0110 0101
inverted: 1001 1010
+1 → 1001 1011
Read it as unsigned
1001 1011₂ = 256 − 101 = 155
Answer
U = 155
6.Number systems
Answer: R = 6
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
AND gives 1 only where both bits are 1.
0010 0110
AND 1000 0111
= 0000 0110
Value of the result
0000 0110₂ = 6
Answer
R = 6
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