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 855 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
855 hosts
p =
2.IPv4 subnettingHard
A host has the address 172.20.106.213/20. What is the third octet x of its network address?
172.20.106.213/20
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- Aff02::1
- B::1
- Cfe80::1
- D::
4.Computer networksHard
An 8-port switch (no VLANs) has one host on each port. How many collision domains and broadcast domains are there?
- A8 collision domains, 8 broadcast domains
- B1 collision domain, 8 broadcast domains
- C8 collision domains, 1 broadcast domain
- D1 collision domain, 1 broadcast domain
5.Number systemsMedium
Write −17 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −178 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 1111 << 2. What is the result's value R in decimal?
1001 1111<< 2
R =
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 < 855
210 − 2 = 1022 ≥ 855 → h = 10
Prefix length
p = 32 − 10 = 22
Answer
p = /22
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: B · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
B · ::1
4.Computer networks
Answer: C · 8 collision domains, 1 broadcast domain
Why this is the answer
Each switch port is its own collision domain, but broadcasts are forwarded to all ports, so the whole switch is one broadcast domain. A hub would give one of each.
Answer
C · 8 collision domains, 1 broadcast domain
5.Number systems
Answer: U = 239
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
17 = 0001 0001
inverted: 1110 1110
+1 → 1110 1111
Read it as unsigned
1110 1111₂ = 256 − 17 = 239
Answer
U = 239
6.Number systems
Answer: R = 124
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.
1001 1111 << 2
= 0111 1100
Value of the result
0111 1100₂ = 124
Answer
R = 124
ExamProblems · examproblems.com/courses/computer-networks