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 1377 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1377 hosts
p =
2.IPv4 subnettingHard
A router has routes to these networks: 10.161.153.0/24, 10.161.154.0/24, 10.161.155.0/24, 10.161.156.0/24, 10.161.157.0/24, 10.161.158.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.161.153.0/24 … 10.161.158.0/24
p =
3.Computer networksMedium
What is the IPv6 loopback address?
- A::
- Bfe80::1
- Cff02::1
- D::1
4.Computer networksHard
A firewall blocks all ICMP. Small web requests work, but large downloads over some paths hang. What is the likely cause?
- ADNS needs ICMP to resolve names
- BTCP cannot open connections without ICMP
- CPath MTU discovery fails because "fragmentation needed" messages are blocked
- DTLS certificates are sent over ICMP
5.Number systemsMedium
The 8-bit pattern 1010 1100 stores a signed integer in two's complement. What is its value N in decimal?
1010 11008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1110 0011 >> 1. What is the result's value R in decimal?
1110 0011>> 1
R =
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 < 1377
211 − 2 = 2046 ≥ 1377 → 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
153 = 10011001
158 = 10011110
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.161.152.0/21 (it also covers 2 network(s) that are not in the list).
3.Computer networks
Answer: D · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
D · ::1
4.Computer networks
Answer: C · Path MTU discovery fails because "fragmentation needed" messages are blocked
Why this is the answer
PMTUD relies on ICMP messages to learn the smaller MTU. If they are dropped, large packets vanish silently, creating a "black hole" connection.
Answer
C · Path MTU discovery fails because "fragmentation needed" messages are blocked
5.Number systems
Answer: N = −84
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.
Add the place values
The other bits keep their normal positive values.
−128 + 32 + 8 + 4 = −84
Answer
N = −84
6.Number systems
Answer: R = 113
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
Shifting right by 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1110 0011 >> 1
= 0111 0001
Value of the result
0111 0001₂ = 113
Answer
R = 113
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