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
A subnet uses the mask 255.255.255.128. How many usable host addresses H does it have?
255.255.255.128
H =
2.IPv4 subnettingHard
A host has the address 10.20.166.14/27. What is the fourth octet x of the last usable host address in its subnet?
10.20.166.14/27
x =
3.Computer networksMedium
Which TCP flag signals that a sender has finished sending data?
- APSH
- BSYN
- CURG
- DFIN
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AThe same MAC address on both hosts
- BA hub connected to both VLANs
- CA router or layer 3 switch to route between the VLANs
- DA longer cable between the hosts
5.Number systemsMedium
Set the bits so they store −60 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −608 bit
U =
6.Number systemsHard
An uncompressed image is 640 × 480 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
640 × 480 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 126
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).
Prefix length
255.255.255.128 = /25
Usable hosts
27 − 2 = 126
Answer
H = 126
2.IPv4 subnetting
Answer: x = 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).
Block size
25 = 32
Subnet of this host
network: ⌊14/32⌋·32 = 0
broadcast: 0 + 32 − 1 = 31
usable hosts: 1 … 30
Answer
x = 30
3.Computer networks
Answer: D · FIN
Why this is the answer
Each side sends FIN when it has no more data, and the other side acknowledges it. SYN is for opening a connection; RST aborts one.
Answer
D · FIN
4.Computer networks
Answer: C · A router or layer 3 switch to route between the VLANs
Why this is the answer
VLANs are isolated at layer 2, so traffic between them must be routed, for example by a router on a trunk ("router on a stick") or a layer 3 switch.
Answer
C · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 196
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
60 = 0011 1100
inverted: 1100 0011
+1 → 1100 0100
Read it as unsigned
1100 0100₂ = 256 − 60 = 196
Answer
U = 196
6.Number systems
Answer: M = 600 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
640·480·16 = 4915200 bit = 614400 B
Convert to KiB
614400/1024 = 600 KiB
Answer
M = 600 KiB
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