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 255.255.255.248. How many usable host addresses H does it have?
255.255.255.248
H =
2.IPv4 subnettingHard
The network 192.168.250.0/24 is split with VLSM for these LANs: A: 43 hosts, B: 4 hosts, C: 13 hosts, D: 20 hosts, E: 21 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 43B: 4C: 13D: 20E: 21
x =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A0–65535
- B0–1023
- C49152–65535
- D1024–49151
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 only works on local networks, so addresses never need to be shared globally
- BIPv6 encrypts all addresses
- CIts 128-bit address space is large enough to give every device a global address
- DIPv6 does not use addresses
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
Set the bits to the 8-bit result of 1100 1001 << 3. What is the result's value R in decimal?
1100 1001<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
255.255.255.248 = /29
Usable hosts
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 96
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
A: 43 hosts → 26 = 64 (/26)
E: 21 hosts → 25 = 32 (/27)
D: 20 hosts → 25 = 32 (/27)
C: 13 hosts → 24 = 16 (/28)
B: 4 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .63
E: .64 – .95
D: .96 – .127
C: .128 – .143
B: .144 – .151
Answer
x = 96
LAN D: 192.168.250.96/27.
3.Computer networks
Answer: B · 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
B · 0–1023
4.Computer networks
Answer: C · 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
C · Its 128-bit address space is large enough to give every device a global address
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: R = 72
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 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 1001 << 3
= 0100 1000
Value of the result
0100 1000₂ = 72
Answer
R = 72
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