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 30 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2430 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.16.221/25. What is the fourth octet x of its network address?
10.20.16.221/25
x =
3.Computer networksMedium
TLS provides which security properties for HTTPS?
- AOnly faster loading
- BProtection against all malware and phishing on the visited site
- CAnonymity of the user
- DConfidentiality, integrity and server authentication
4.Computer networksHard
The count-to-infinity problem occurs in which kind of routing protocol, and what helps against it?
- ADistance-vector; split horizon and poison reverse
- BLink-state; flooding link updates
- CPath-vector; longer AS paths
- DStatic routing; adding default routes
5.Number systemsMedium
Write −114 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −1148 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0011 1111 AND 1111 0000. What is the result's value R in decimal?
0011 1111AND1111 0000
R =
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 30 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
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
27 = 128
Subnet of this host
network: ⌊221/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 128
3.Computer networks
Answer: D · 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
D · Confidentiality, integrity and server authentication
4.Computer networks
Answer: A · Distance-vector; split horizon and poison reverse
Why this is the answer
After a failure, distance-vector routers can keep advertising stale routes to each other, raising the metric slowly. Split horizon, poison reverse and a maximum metric limit this.
Answer
A · Distance-vector; split horizon and poison reverse
5.Number systems
Answer: U = 142
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
114 = 0111 0010
inverted: 1000 1101
+1 → 1000 1110
Read it as unsigned
1000 1110₂ = 256 − 114 = 142
Answer
U = 142
6.Number systems
Answer: R = 48
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
AND gives 1 only where both bits are 1.
0011 1111
AND 1111 0000
= 0011 0000
Value of the result
0011 0000₂ = 48
Answer
R = 48
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