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 1641 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1641 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.168.144/30. What is the fourth octet x of the last usable host address in its subnet?
10.20.168.144/30
x =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.1.0/23
- B192.168.0.0/25
- C192.168.0.0/23
- D192.168.0.0/22
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AA router or layer 3 switch to route between the VLANs
- BThe same MAC address on both hosts
- CA hub connected to both VLANs
- DA longer cable between the hosts
5.Number systemsMedium
The 8-bit pattern 1011 1000 stores a signed integer in two's complement. What is its value N in decimal?
1011 10008 bit
N =
6.Number systemsHard
A system must give every one of 415 items its own binary code of the same length. What is the smallest number of bits b needed?
415
b = bit
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1641
211 − 2 = 2046 ≥ 1641 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 146
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: ⌊144/4⌋·4 = 144
broadcast: 144 + 4 − 1 = 147
usable hosts: 145 … 146
Answer
x = 146
3.Computer networks
Answer: C · 192.168.0.0/23
Why this is the answer
0 and 1 share all bits of the third octet except the last, so one bit less of prefix (/23) starting at 192.168.0.0 covers both exactly.
Answer
C · 192.168.0.0/23
4.Computer networks
Answer: A · A router or layer 3 switch to route between the VLANs
Why this is the answer
VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.
Answer
A · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: N = −72
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 + 32 + 16 + 8 = −72
Answer
N = −72
6.Number systems
Answer: b = 9 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
28 = 256 < 415
29 = 512 ≥ 415
Answer
b = 9
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