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 12 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2412 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.103.146/20. What is the third octet x of its network address?
172.20.103.146/20
x =
3.Computer networksMedium
Which medium access method did classic shared (half-duplex) Ethernet use?
- ACSMA/CD
- BToken passing
- CCSMA/CA
- DTDMA
4.Computer networksHard
Hosts in VLAN 10 and VLAN 20 on the same switch cannot communicate. What is needed?
- AA router or layer 3 switch to route between the VLANs
- BThe same MAC address on both hosts
- CA hub connected to both VLANs
- DA longer cable between the hosts
5.Number systemsMedium
Set the bits so they store −65 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −658 bit
U =
6.Number systemsHard
An uncompressed image is 256 × 256 pixels with 16 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
256 × 256 px16 bit/px1 KiB = 1024 B
M = KiB
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 12 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
2.IPv4 subnetting
Answer: x = 96
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: 172.20.96.0
broadcast: 172.20.111.255
Answer
x = 96
3.Computer networks
Answer: A · CSMA/CD
Why this is the answer
Carrier sense multiple access with collision detection: stations stop when they detect a collision and retry after a random backoff. Full-duplex switched Ethernet has no collisions.
Answer
A · CSMA/CD
4.Computer networks
Answer: A · 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
A · A router or layer 3 switch to route between the VLANs
5.Number systems
Answer: U = 191
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
65 = 0100 0001
inverted: 1011 1110
+1 → 1011 1111
Read it as unsigned
1011 1111₂ = 256 − 65 = 191
Answer
U = 191
6.Number systems
Answer: M = 128 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
256·256·16 = 1048576 bit = 131072 B
Convert to KiB
131072/1024 = 128 KiB
Answer
M = 128 KiB
ExamProblems · examproblems.com/courses/computer-networks