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 1646 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1646 hosts
p =
2.IPv4 subnettingHard
The network 192.168.243.0/24 is split with VLSM for these LANs: A: 7 hosts, B: 41 hosts, C: 5 hosts, D: 79 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 A?
A: 7B: 41C: 5D: 79
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo split one physical switched network into separate broadcast domains
- BTo increase the speed of every switch port
- CTo give each device a public IP address
- DTo encrypt all traffic between switches so other departments cannot read it
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- BIPv6 fragments at the transport layer inside TCP
- CRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- DIPv6 does not allow packets larger than 576 bytes
5.Number systemsMedium
The 8-bit pattern 0010 0111 stores a signed integer in two's complement. What is its value N in decimal?
0010 01118 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0011 XOR 0110 1110. What is the result's value R in decimal?
1010 0011XOR0110 1110
R =
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 < 1646
211 − 2 = 2046 ≥ 1646 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 192
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: 79 hosts → 27 = 128 (/25)
B: 41 hosts → 26 = 64 (/26)
A: 7 hosts → 24 = 16 (/28)
C: 5 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .127
B: .128 – .191
A: .192 – .207
C: .208 – .215
Answer
x = 192
LAN A: 192.168.243.192/28.
3.Computer networks
Answer: A · To split one physical switched network into separate broadcast domains
Why this is the answer
VLANs group ports logically, so broadcasts stay within each VLAN. This improves security and limits broadcast traffic.
Answer
A · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
Why this is the answer
An IPv6 router drops an oversized packet and returns ICMPv6 Packet Too Big. The source then sends smaller packets or uses a fragment extension header.
Answer
C · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: N = 39
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.
32 + 4 + 2 + 1 = 39
Answer
N = 39
6.Number systems
Answer: R = 205
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
XOR gives 1 where the two bits differ.
1010 0011
XOR 0110 1110
= 1100 1101
Value of the result
1100 1101₂ = 205
Answer
R = 205
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