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 1025 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1025 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.146.74.0/24, 10.146.75.0/24, 10.146.76.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.146.74.0/24 … 10.146.76.0/24
p =
3.Computer networksMedium
Using even parity, which parity bit is added to the data bits 1011001?
- A1
- B0
- C4
- D2
4.Computer networksHard
Which routing protocol is a link-state protocol?
- ARIP
- BOSPF
- CBGP
- DARP
5.Number systemsMedium
Set the bits so they store −98 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −988 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 20-bit two's complement?
20 bit
N =
3Solutions
1.IPv4 subnetting
Answer: p = 21
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
210 − 2 = 1022 < 1025
211 − 2 = 2046 ≥ 1025 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: p = 21
Route summarisation: a summary route a.b.c.0/p covers all networks whose first p bits match. Write the differing octet of the first and last network in binary; the number of common leading bits, plus the bits of the octets before it, is the longest summary prefix.
Compare the bits
74 = 01001010
76 = 01001100
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.146.72.0/21 (it also covers 5 network(s) that are not in the list).
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: B · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
B · OSPF
5.Number systems
Answer: U = 158
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
98 = 0110 0010
inverted: 1001 1101
+1 → 1001 1110
Read it as unsigned
1001 1110₂ = 256 − 98 = 158
Answer
U = 158
6.Number systems
Answer: N = −524288
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.
Most negative value
−219 = −524288
Answer
N = −524288
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