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 192.168.48.0/24 is divided into /25 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 2?
192.168.48.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.211.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 14 hosts, C: 27 hosts, D: 28 hosts, E: 39 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: 5B: 14C: 27D: 28E: 39
x =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- APhysical (layer 1)
- BNetwork (layer 3)
- CData link (layer 2)
- DTransport (layer 4)
4.Computer networksHard
Why does IPv6 not need NAT for address shortage?
- AIPv6 encrypts all addresses
- BIPv6 does not use addresses
- CIPv6 only works on local networks, so addresses never need to be shared globally
- DIts 128-bit address space is large enough to give every device a global address
5.Number systemsMedium
Write −65 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −658 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 20-bit two's complement?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 255
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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 255
2.IPv4 subnetting
Answer: x = 144
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: 39 hosts → 26 = 64 (/26)
D: 28 hosts → 25 = 32 (/27)
C: 27 hosts → 25 = 32 (/27)
B: 14 hosts → 24 = 16 (/28)
A: 5 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
D: .64 – .95
C: .96 – .127
B: .128 – .143
A: .144 – .151
Answer
x = 144
LAN A: 192.168.211.144/29.
3.Computer networks
Answer: C · 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
C · Data link (layer 2)
4.Computer networks
Answer: D · 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
D · Its 128-bit address space is large enough to give every device a global address
5.Number systems
Answer: U = 191
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.
Invert and add one
65 = 0100 0001
inverted: 1011 1110
+1 → 1011 1111
Read it as unsigned
1011 1111₂ = 256 − 65 = 191
Answer
U = 191
6.Number systems
Answer: N = −524288
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.
Most negative value
−219 = −524288
Answer
N = −524288
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