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 192.168.31.0/24 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 6?
192.168.31.0/24/27subnet 6 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.81.0/24 is split with VLSM for these LANs: A: 47 hosts, B: 12 hosts, C: 6 hosts, D: 123 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 B?
A: 47B: 12C: 6D: 123
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BCNAME
- CMX
- DAAAA
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA longer cable between the hosts
- CA router or layer 3 switch to route between the VLANs
- DA hub connected to both VLANs
5.Number systemsMedium
The 8-bit pattern 0100 0000 stores a signed integer in two's complement. What is its value N in decimal?
0100 00008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 1111 << 3. What is the result's value R in decimal?
0010 1111<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: x = 191
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 6
network: (6 − 1)·32 = 160
broadcast: 160 + 32 − 1 = 191
Answer
x = 191
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: 123 hosts → 27 = 128 (/25)
A: 47 hosts → 26 = 64 (/26)
B: 12 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .127
A: .128 – .191
B: .192 – .207
C: .208 – .215
Answer
x = 192
LAN B: 192.168.81.192/28.
3.Computer networks
Answer: D · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
D · AAAA
4.Computer networks
Answer: C · 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
C · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: N = 64
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.
64 = 64
Answer
N = 64
6.Number systems
Answer: R = 120
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.
0010 1111 << 3
= 0111 1000
Value of the result
0111 1000₂ = 120
Answer
R = 120
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