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 255.255.255.128. How many usable host addresses H does it have?
255.255.255.128
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.239.197.0/24, 10.239.198.0/24, 10.239.199.0/24, 10.239.200.0/24, 10.239.201.0/24, 10.239.202.0/24, 10.239.203.0/24, 10.239.204.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.239.197.0/24 … 10.239.204.0/24
p =
3.Computer networksMedium
Which of these is a private IPv4 address range?
- A192.168.0.0/16
- B200.1.1.0/24
- C100.0.0.0/8
- D8.8.8.0/24
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA hub connected to both VLANs
- CA longer cable between the hosts
- DA router or layer 3 switch to route between the VLANs
5.Number systemsMedium
Set the bits so they store −57 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −578 bit
U =
6.Number systemsHard
A system must give every one of 1532 items its own binary code of the same length. What is the smallest number of bits b needed?
1532
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 126
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
255.255.255.128 = /25
Usable hosts
27 − 2 = 126
Answer
H = 126
2.IPv4 subnetting
Answer: p = 20
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
197 = 11000101
204 = 11001100
common leading bits: 4
Prefix length
p = 16 + 4 = 20
Answer
p = /20
Summary route: 10.239.192.0/20 (it also covers 8 network(s) that are not in the list).
3.Computer networks
Answer: A · 192.168.0.0/16
Why this is the answer
Private ranges are 10.0.0.0/8, 172.16.0.0/12 and 192.168.0.0/16; they are not routed on the public internet.
Answer
A · 192.168.0.0/16
4.Computer networks
Answer: D · 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
D · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 199
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
57 = 0011 1001
inverted: 1100 0110
+1 → 1100 0111
Read it as unsigned
1100 0111₂ = 256 − 57 = 199
Answer
U = 199
6.Number systems
Answer: b = 11 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
210 = 1024 < 1532
211 = 2048 ≥ 1532
Answer
b = 11
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