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 192.168.239.0/24 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 2?
192.168.239.0/24/28subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.27.48/21. What is the third octet x of its subnet's broadcast address?
172.20.27.48/21
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/8
- B/16
- C/32
- D/24
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AThe physical cable
- BThe default gateway
- CName resolution (DNS)
- DThe network card driver
5.Number systemsMedium
Write −72 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −728 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 1010 << 2. What is the result's value R in decimal?
1110 1010<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: x = 16
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 2
network: (2 − 1)·16 = 16
broadcast: 16 + 16 − 1 = 31
Answer
x = 16
2.IPv4 subnetting
Answer: x = 31
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 in the third octet
23 = 8
Subnet of this host
network: 172.20.24.0
broadcast: 172.20.31.255
Answer
x = 31
3.Computer networks
Answer: D · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
D · /24
4.Computer networks
Answer: C · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
C · Name resolution (DNS)
5.Number systems
Answer: U = 184
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
72 = 0100 1000
inverted: 1011 0111
+1 → 1011 1000
Read it as unsigned
1011 1000₂ = 256 − 72 = 184
Answer
U = 184
6.Number systems
Answer: R = 168
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 left by 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1110 1010 << 2
= 1010 1000
Value of the result
1010 1000₂ = 168
Answer
R = 168
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