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 172.25.0.0/16 is divided into equal subnets with prefix /21. How many subnets N are there?
172.25.0.0/16/21
N =
2.IPv4 subnettingHard
The network 192.168.78.0/24 is split with VLSM for these LANs: A: 24 hosts, B: 22 hosts, C: 5 hosts, D: 13 hosts, E: 43 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: 24B: 22C: 5D: 13E: 43
x =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- APhysical (layer 1)
- BTransport (layer 4)
- CNetwork (layer 3)
- DData link (layer 2)
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIts 128-bit address space is large enough to give every device a global address
- BIPv6 encrypts all addresses
- CIPv6 does not use addresses
- DIPv6 only works on local networks, so addresses never need to be shared globally
5.Number systemsMedium
Set the bits so they store −107 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1078 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0110 1100 OR 1101 1100. What is the result's value R in decimal?
0110 1100OR1101 1100
R =
3Solutions
1.IPv4 subnetting
Answer: N = 32
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).
Borrowed bits
21 − 16 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /21 subnet has 211 − 2 = 2046 usable hosts.
2.IPv4 subnetting
Answer: x = 64
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
E: 43 hosts → 26 = 64 (/26)
A: 24 hosts → 25 = 32 (/27)
B: 22 hosts → 25 = 32 (/27)
D: 13 hosts → 24 = 16 (/28)
C: 5 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
A: .64 – .95
B: .96 – .127
D: .128 – .143
C: .144 – .151
Answer
x = 64
LAN A: 192.168.78.64/27.
3.Computer networks
Answer: D · Data link (layer 2)
Why this is the answer
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer
D · Data link (layer 2)
4.Computer networks
Answer: A · 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
A · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 149
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
107 = 0110 1011
inverted: 1001 0100
+1 → 1001 0101
Read it as unsigned
1001 0101₂ = 256 − 107 = 149
Answer
U = 149
6.Number systems
Answer: R = 252
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.
0110 1100
OR 1101 1100
= 1111 1100
Value of the result
1111 1100₂ = 252
Answer
R = 252
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