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 192.168.159.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 3?
192.168.159.0/24/26subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.150.0/24 is split with VLSM for these LANs: A: 5 hosts, B: 15 hosts, C: 30 hosts, D: 6 hosts, E: 13 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: 15C: 30D: 6E: 13
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AOnly faster loading
- BAnonymity of the user
- CProtection against all malware and phishing on the visited site
- DConfidentiality, integrity and server authentication
4.Computer networksHard
How does fragmentation differ in IPv6 compared with IPv4?
- AIPv6 does not allow packets larger than 576 bytes
- BRouters never fragment IPv6 packets; only the source can, using path MTU discovery
- CIPv6 routers fragment every packet larger than 64 bytes before forwarding it on the path
- DIPv6 fragments at the transport layer inside TCP
5.Number systemsMedium
Set the bits so they store −65 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −658 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0001 1111 >> 1. What is the result's value R in decimal?
0001 1111>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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
26 = 64
Subnet number 3
network: (3 − 1)·64 = 128
broadcast: 128 + 64 − 1 = 191
Answer
x = 128
2.IPv4 subnetting
Answer: x = 88
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
C: 30 hosts → 25 = 32 (/27)
B: 15 hosts → 25 = 32 (/27)
E: 13 hosts → 24 = 16 (/28)
D: 6 hosts → 23 = 8 (/29)
A: 5 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .31
B: .32 – .63
E: .64 – .79
D: .80 – .87
A: .88 – .95
Answer
x = 88
LAN A: 192.168.150.88/29.
3.Computer networks
Answer: D · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
D · Confidentiality, integrity and server authentication
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 = 191
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
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: R = 15
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 right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0001 1111 >> 1
= 0000 1111
Value of the result
0000 1111₂ = 15
Answer
R = 15
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