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
The network 192.168.75.0/24 is divided into /27 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 4?
192.168.75.0/24/27subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.12.0/24 is split with VLSM for these LANs: A: 3 hosts, B: 20 hosts, C: 9 hosts, D: 34 hosts, E: 14 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 E?
A: 3B: 20C: 9D: 34E: 14
x =
3.Computer networksMedium
Which single prefix summarises both 192.168.0.0/24 and 192.168.1.0/24?
- A192.168.0.0/22
- B192.168.0.0/25
- C192.168.0.0/23
- D192.168.1.0/23
4.Computer networksHard
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
- A/27
- B/28
- C/25
- D/26
5.Number systemsMedium
The 8-bit pattern 1000 1101 stores a signed integer in two's complement. What is its value N in decimal?
1000 11018 bit
N =
6.Number systemsHard
What is the largest unsigned integer N that fits in 10 bits?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 127
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
25 = 32
Subnet number 4
network: (4 − 1)·32 = 96
broadcast: 96 + 32 − 1 = 127
Answer
x = 127
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
D: 34 hosts → 26 = 64 (/26)
B: 20 hosts → 25 = 32 (/27)
E: 14 hosts → 24 = 16 (/28)
C: 9 hosts → 24 = 16 (/28)
A: 3 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .63
B: .64 – .95
E: .96 – .111
C: .112 – .127
A: .128 – .135
Answer
x = 96
LAN E: 192.168.12.96/28.
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 · /27
Why this is the answer
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
Answer
A · /27
5.Number systems
Answer: N = −115
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 + 8 + 4 + 1 = −115
Answer
N = −115
6.Number systems
Answer: N = 1023
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.
Largest value
210 − 1 = 1023
Answer
N = 1023
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