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 192.168.234.0/24 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 3?
192.168.234.0/24/28subnet 3 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.207.0/24 is split with VLSM for these LANs: A: 8 hosts, B: 38 hosts, C: 102 hosts, D: 3 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN D?
A: 8B: 38C: 102D: 3
x =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A4
- B0
- C2
- D1
4.Computer networksHard
Why does a browser warn when a site presents a self-signed certificate?
- AThe certificate is always expired
- BSelf-signed certificates use no encryption at all, so the traffic is sent in clear text
- CNo trusted certificate authority vouches that the key belongs to that site
- DThe site must be using HTTP instead of HTTPS
5.Number systemsMedium
Set the bits so they store −68 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −688 bit
U =
6.Number systemsHard
A system must give every one of 601 items its own binary code of the same length. What is the smallest number of bits b needed?
601
b = bit
3Solutions
1.IPv4 subnetting
Answer: x = 32
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 3
network: (3 − 1)·16 = 32
broadcast: 32 + 16 − 1 = 47
Answer
x = 32
2.IPv4 subnetting
Answer: x = 208
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 102 hosts → 27 = 128 (/25)
B: 38 hosts → 26 = 64 (/26)
A: 8 hosts → 24 = 16 (/28)
D: 3 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
B: .128 – .191
A: .192 – .207
D: .208 – .215
Answer
x = 208
LAN D: 192.168.207.208/29.
3.Computer networks
Answer: B · 0
Why this is the answer
1011001 contains four 1-bits, already an even number, so the parity bit is 0. With odd parity it would be 1.
Answer
B · 0
4.Computer networks
Answer: C · No trusted certificate authority vouches that the key belongs to that site
Why this is the answer
Browsers trust certificates signed, directly or through a chain, by CAs in their trust store. A self-signed certificate could be made by anyone, such as a man-in-the-middle.
Answer
C · No trusted certificate authority vouches that the key belongs to that site
5.Number systems
Answer: U = 188
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
68 = 0100 0100
inverted: 1011 1011
+1 → 1011 1100
Read it as unsigned
1011 1100₂ = 256 − 68 = 188
Answer
U = 188
6.Number systems
Answer: b = 10 bit
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.
Powers of two
29 = 512 < 601
210 = 1024 ≥ 601
Answer
b = 10
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