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 255.255.248.0. How many usable host addresses H does it have?
255.255.248.0
H =
2.IPv4 subnettingHard
The network 192.168.114.0/24 is split with VLSM for these LANs: A: 91 hosts, B: 18 hosts, C: 3 hosts, D: 17 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 C?
A: 91B: 18C: 3D: 17
x =
3.Computer networksMedium
What is the main purpose of a VLAN?
- ATo increase the speed of every switch port
- BTo give each device a public IP address
- CTo split one physical switched network into separate broadcast domains
- 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
- BRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- CIPv6 fragments at the transport layer inside TCP
- DIPv6 does not allow packets larger than 576 bytes
5.Number systemsMedium
Set the bits so they store −103 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1038 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 0010 OR 0110 1101. What is the result's value R in decimal?
0100 0010OR0110 1101
R =
3Solutions
1.IPv4 subnetting
Answer: H = 2046
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
255.255.248.0 = /21
Usable hosts
211 − 2 = 2046
Answer
H = 2046
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
A: 91 hosts → 27 = 128 (/25)
B: 18 hosts → 25 = 32 (/27)
D: 17 hosts → 25 = 32 (/27)
C: 3 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .127
B: .128 – .159
D: .160 – .191
C: .192 – .199
Answer
x = 192
LAN C: 192.168.114.192/29.
3.Computer networks
Answer: C · 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
C · To split one physical switched network into separate broadcast domains
4.Computer networks
Answer: B · 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
B · Routers never fragment IPv6 packets; only the source can, using path MTU discovery
5.Number systems
Answer: U = 153
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
103 = 0110 0111
inverted: 1001 1000
+1 → 1001 1001
Read it as unsigned
1001 1001₂ = 256 − 103 = 153
Answer
U = 153
6.Number systems
Answer: R = 111
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
OR gives 1 where at least one bit is 1.
0100 0010
OR 0110 1101
= 0110 1111
Value of the result
0110 1111₂ = 111
Answer
R = 111
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