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 192.168.10.0/24 must be divided into at least 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
The network 192.168.65.0/24 is split with VLSM for these LANs: A: 43 hosts, B: 48 hosts, C: 8 hosts, D: 19 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: 43B: 48C: 8D: 19
x =
3.Computer networksMedium
What is the role of a default gateway on a host?
- AIt assigns MAC addresses
- BIt is the switch port the host is connected to
- CIt is the router used for traffic to other networks
- DIt is the DNS server
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe switch's MAC address table
- BThe server's table of half-open connections
- CThe DNS cache of the client
- DThe server's disk space
5.Number systemsMedium
Write −85 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −858 bit
U =
6.Number systemsHard
An uncompressed image is 512 × 512 pixels with 8 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
512 × 512 px8 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 30
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).
Bits to borrow
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 160
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
B: 48 hosts → 26 = 64 (/26)
A: 43 hosts → 26 = 64 (/26)
D: 19 hosts → 25 = 32 (/27)
C: 8 hosts → 24 = 16 (/28)
Allocation, largest first
B: .0 – .63
A: .64 – .127
D: .128 – .159
C: .160 – .175
Answer
x = 160
LAN C: 192.168.65.160/28.
3.Computer networks
Answer: C · It is the router used for traffic to other networks
Why this is the answer
The host sends packets for non-local destinations to the default gateway, usually its local router.
Answer
C · It is the router used for traffic to other networks
4.Computer networks
Answer: B · The server's table of half-open connections
Why this is the answer
Each SYN makes the server reserve state and wait for an ACK that never comes. SYN cookies avoid storing state until the handshake completes.
Answer
B · The server's table of half-open connections
5.Number systems
Answer: U = 171
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
85 = 0101 0101
inverted: 1010 1010
+1 → 1010 1011
Read it as unsigned
1010 1011₂ = 256 − 85 = 171
Answer
U = 171
6.Number systems
Answer: M = 256 KiB
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.
Bits and bytes
512·512·8 = 2097152 bit = 262144 B
Convert to KiB
262144/1024 = 256 KiB
Answer
M = 256 KiB
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