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 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
A router has routes to these networks: 10.11.184.0/24, 10.11.185.0/24, 10.11.186.0/24, 10.11.187.0/24, 10.11.188.0/24, 10.11.189.0/24. What is the longest prefix length p of a single summary route that covers all of them?
10.11.184.0/24 … 10.11.189.0/24
p =
3.Computer networksMedium
What is the IPv6 loopback address?
- Afe80::1
- B::
- C::1
- Dff02::1
4.Computer networksHard
Why does the side that closes a TCP connection first stay in the TIME_WAIT state for twice the maximum segment lifetime?
- ATo keep the encryption keys for later reuse
- BTo let the routers along the path update their routing tables for the next connection
- CTo wait for the DNS cache to expire
- DSo delayed segments die out and the final ACK can be resent if it was lost
5.Number systemsMedium
Write −43 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −438 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0010 0111 >> 1. What is the result's value R in decimal?
0010 0111>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
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
184 = 10111000
189 = 10111101
common leading bits: 5
Prefix length
p = 16 + 5 = 21
Answer
p = /21
Summary route: 10.11.184.0/21 (it also covers 2 network(s) that are not in the list).
3.Computer networks
Answer: C · ::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
C · ::1
4.Computer networks
Answer: D · So delayed segments die out and the final ACK can be resent if it was lost
Why this is the answer
TIME_WAIT ensures a lost final ACK can be repeated and old duplicate segments expire before the same address and port pair is reused.
Answer
D · So delayed segments die out and the final ACK can be resent if it was lost
5.Number systems
Answer: U = 213
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
43 = 0010 1011
inverted: 1101 0100
+1 → 1101 0101
Read it as unsigned
1101 0101₂ = 256 − 43 = 213
Answer
U = 213
6.Number systems
Answer: R = 19
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.
0010 0111 >> 1
= 0001 0011
Value of the result
0001 0011₂ = 19
Answer
R = 19
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