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 192.168.37.0/24 is divided into /28 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 11?
192.168.37.0/24/28subnet 11 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.192.173/28. What is the fourth octet x of the last usable host address in its subnet?
10.20.192.173/28
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- BIt only checks the MAC address of each frame
- CIt encrypts every packet it forwards
- DIt tracks connections and allows replies that belong to established sessions
4.Computer networksHard
You must divide 192.168.1.0/24 into at least 5 subnets with up to 25 hosts each. Which prefix length fits best?
- A/25
- B/27
- C/28
- D/26
5.Number systemsMedium
Set the bits so they store −49 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −498 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 1111 OR 1100 0111. What is the result's value R in decimal?
0010 1111OR1100 0111
R =
3Solutions
1.IPv4 subnetting
Answer: x = 160
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 number 11
network: (11 − 1)·16 = 160
broadcast: 160 + 16 − 1 = 175
Answer
x = 160
2.IPv4 subnetting
Answer: x = 174
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: ⌊173/16⌋·16 = 160
broadcast: 160 + 16 − 1 = 175
usable hosts: 161 … 174
Answer
x = 174
3.Computer networks
Answer: D · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
D · It tracks connections and allows replies that belong to established sessions
4.Computer networks
Answer: B · /27
Why this is the answer
/27 gives 8 subnets with 30 usable hosts each. /26 gives only 4 subnets, and /28 allows only 14 hosts per subnet.
Answer
B · /27
5.Number systems
Answer: U = 207
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
49 = 0011 0001
inverted: 1100 1110
+1 → 1100 1111
Read it as unsigned
1100 1111₂ = 256 − 49 = 207
Answer
U = 207
6.Number systems
Answer: R = 239
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.
0010 1111
OR 1100 0111
= 1110 1111
Value of the result
1110 1111₂ = 239
Answer
R = 239
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