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 1350 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1350 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.192.83.0/24, 10.192.84.0/24, 10.192.85.0/24, 10.192.86.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.192.83.0/24 … 10.192.86.0/24
p =
3.Computer networksMedium
What characterises a distributed denial-of-service (DDoS) attack?
- AOne attacker guesses a user's password
- BMany compromised machines flood a target with traffic
- CAn attacker reads unencrypted Wi-Fi traffic from a nearby car
- DA virus deletes files on a single PC
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
- BPath-vector; longer AS paths
- CStatic routing; adding default routes
- DLink-state; flooding link updates
5.Number systemsMedium
Set the bits so they store −45 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −458 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 12-bit two's complement?
12 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 < 1350
211 − 2 = 2046 ≥ 1350 → 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
83 = 01010011
86 = 01010110
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.192.80.0/21 (it also covers 4 network(s) that are not in the list).
3.Computer networks
Answer: B · Many compromised machines flood a target with traffic
Why this is the answer
In a DDoS attack many sources overwhelm a server or link so legitimate users cannot be served. The many sources make simple blocking hard.
Answer
B · Many compromised machines flood a target with traffic
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 = 211
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
45 = 0010 1101
inverted: 1101 0010
+1 → 1101 0011
Read it as unsigned
1101 0011₂ = 256 − 45 = 211
Answer
U = 211
6.Number systems
Answer: N = −2048
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
−211 = −2048
Answer
N = −2048
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