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 5 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/245 subnets
H =
2.IPv4 subnettingHard
A host has the address 10.20.207.96/28. What is the fourth octet x of its network address?
10.20.207.96/28
x =
3.Computer networksMedium
Which DNS record type maps a host name to an IPv6 address?
- AA
- BCNAME
- CMX
- DAAAA
4.Computer networksHard
Why was the TCP window scale option introduced?
- AWithout it, TCP could not detect lost segments on long-distance links
- BThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- CUDP needed a way to control its sending rate on congested paths
- DThe 32-bit window field was too large and wasted header space in every TCP segment sent
5.Number systemsMedium
Set the bits so they store −12 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −128 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 1110 >> 3. What is the result's value R in decimal?
1100 1110>> 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 30
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
23 = 8 ≥ 5 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 96
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
24 = 16
Subnet of this host
network: ⌊96/16⌋·16 = 96
broadcast: 96 + 16 − 1 = 111
usable hosts: 97 … 110
Answer
x = 96
3.Computer networks
Answer: D · AAAA
Why this is the answer
An A record holds an IPv4 address, AAAA an IPv6 address. MX names mail servers and CNAME gives an alias.
Answer
D · AAAA
4.Computer networks
Answer: B · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
B · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 244
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
12 = 0000 1100
inverted: 1111 0011
+1 → 1111 0100
Read it as unsigned
1111 0100₂ = 256 − 12 = 244
Answer
U = 244
6.Number systems
Answer: R = 25
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 3 moves every bit 3 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1100 1110 >> 3
= 0001 1001
Value of the result
0001 1001₂ = 25
Answer
R = 25
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