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 865 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
865 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.199.238.0/24, 10.199.239.0/24, 10.199.240.0/24, 10.199.241.0/24, 10.199.242.0/24, 10.199.243.0/24, 10.199.244.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.199.238.0/24 … 10.199.244.0/24
p =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A0–65535
- B1024–49151
- C49152–65535
- D0–1023
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
- BA longer cable between the hosts
- CThe same MAC address on both hosts
- DA hub connected to both VLANs
5.Number systemsMedium
Set the bits so they store −25 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −258 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px32 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 22
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
29 − 2 = 510 < 865
210 − 2 = 1022 ≥ 865 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
2.IPv4 subnetting
Answer: p = 19
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
238 = 11101110
244 = 11110100
common leading bits: 3
Prefix length
p = 16 + 3 = 19
Answer
p = /19
Summary route: 10.199.224.0/19 (it also covers 25 network(s) that are not in the list).
3.Computer networks
Answer: D · 0–1023
Why this is the answer
Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.
Answer
D · 0–1023
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: U = 231
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
25 = 0001 1001
inverted: 1110 0110
+1 → 1110 0111
Read it as unsigned
1110 0111₂ = 256 − 25 = 231
Answer
U = 231
6.Number systems
Answer: M = 3072 KiB
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.
Bits and bytes
1024·768·32 = 25165824 bit = 3145728 B
Convert to KiB
3145728/1024 = 3072 KiB
Answer
M = 3072 KiB
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