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 192.168.79.0/24 is divided into /27 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 4?
192.168.79.0/24/27subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.128.112/21. What is the third octet x of its network address?
172.20.128.112/21
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AProtection against all malware and phishing on the visited site
- BConfidentiality, integrity and server authentication
- CAnonymity of the user
- DOnly faster loading
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AThe network card driver
- BThe physical cable
- CThe default gateway
- DName resolution (DNS)
5.Number systemsMedium
Set the bits so they store −61 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −618 bit
U =
6.Number systemsHard
A system must give every one of 1732 items its own binary code of the same length. What is the smallest number of bits b needed?
1732
b = bit
3Solutions
1.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
25 = 32
Subnet number 4
network: (4 − 1)·32 = 96
broadcast: 96 + 32 − 1 = 127
Answer
x = 96
2.IPv4 subnetting
Answer: x = 128
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
23 = 8
Subnet of this host
network: 172.20.128.0
broadcast: 172.20.135.255
Answer
x = 128
3.Computer networks
Answer: B · Confidentiality, integrity and server authentication
Why this is the answer
TLS encrypts traffic, detects tampering and authenticates the server with certificates (optionally the client too).
Answer
B · Confidentiality, integrity and server authentication
4.Computer networks
Answer: D · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
D · Name resolution (DNS)
5.Number systems
Answer: U = 195
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
61 = 0011 1101
inverted: 1100 0010
+1 → 1100 0011
Read it as unsigned
1100 0011₂ = 256 − 61 = 195
Answer
U = 195
6.Number systems
Answer: b = 11 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
210 = 1024 < 1732
211 = 2048 ≥ 1732
Answer
b = 11
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