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 department needs 589 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
589 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.206.47/20. What is the third octet x of its network address?
172.20.206.47/20
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt tracks connections and allows replies that belong to established sessions
- BIt encrypts every packet it forwards
- CIt only checks the MAC address of each frame
- DIt blocks all incoming traffic without exceptions, including replies to outgoing requests
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AA longer cable between the hosts
- BA router or layer 3 switch to route between the VLANs
- CThe same MAC address on both hosts
- DA hub connected to both VLANs
5.Number systemsMedium
Write −91 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −918 bit
U =
6.Number systemsHard
An uncompressed image is 1280 × 720 pixels with 24 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1280 × 720 px24 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: p = 22
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).
Host bits needed
29 − 2 = 510 < 589
210 − 2 = 1022 ≥ 589 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
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 in the third octet
24 = 16
Subnet of this host
network: 172.20.192.0
broadcast: 172.20.207.255
Answer
x = 192
3.Computer networks
Answer: A · 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
A · It tracks connections and allows replies that belong to established sessions
4.Computer networks
Answer: B · 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
B · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 165
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
91 = 0101 1011
inverted: 1010 0100
+1 → 1010 0101
Read it as unsigned
1010 0101₂ = 256 − 91 = 165
Answer
U = 165
6.Number systems
Answer: M = 2700 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
1280·720·24 = 22118400 bit = 2764800 B
Convert to KiB
2764800/1024 = 2700 KiB
Answer
M = 2700 KiB
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