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
A host has the address 10.20.216.200/28. What is the fourth octet x of its network address?
10.20.216.200/28
x =
3.Computer networksMedium
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
- ABoth routes, splitting the traffic equally
- B10.0.0.0/8, because it covers more addresses than the /16
- C10.0.0.0/8, because it was added first
- D10.1.0.0/16, because it is the longest matching prefix
4.Computer networksHard
In a SYN flood attack, what resource is the attacker trying to exhaust?
- AThe switch's MAC address table
- BThe DNS cache of the client
- CThe server's disk space
- DThe server's table of half-open connections
5.Number systemsMedium
Write −113 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1138 bit
U =
6.Number systemsHard
A system must give every one of 1002 items its own binary code of the same length. What is the smallest number of bits b needed?
1002
b = bit
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 = 192
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
24 = 16
Subnet of this host
network: ⌊200/16⌋·16 = 192
broadcast: 192 + 16 − 1 = 207
usable hosts: 193 … 206
Answer
x = 192
3.Computer networks
Answer: D · 10.1.0.0/16, because it is the longest matching prefix
Why this is the answer
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer
D · 10.1.0.0/16, because it is the longest matching prefix
4.Computer networks
Answer: D · 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
D · The server's table of half-open connections
5.Number systems
Answer: U = 143
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
113 = 0111 0001
inverted: 1000 1110
+1 → 1000 1111
Read it as unsigned
1000 1111₂ = 256 − 113 = 143
Answer
U = 143
6.Number systems
Answer: b = 10 bit
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.
Powers of two
29 = 512 < 1002
210 = 1024 ≥ 1002
Answer
b = 10
ExamProblems · examproblems.com/courses/computer-networks