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 275 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
275 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.166.77/29. What is the fourth octet x of its network address?
10.20.166.77/29
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A12 ms
- B0.125 ms
- C8 ms
- D1 ms
4.Computer networksHard
A student runs a game server on a PC behind a home NAT router. Friends on the internet cannot connect. What is the usual fix?
- AGive the PC a different MAC address so the router can find it from outside
- BChange the PC's subnet mask to /32
- CTurn off the PC's DNS client
- DConfigure port forwarding on the router to the PC's private address
5.Number systemsMedium
Set the bits so they store −68 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −688 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 1011 XOR 0011 1001. What is the result's value R in decimal?
1010 1011XOR0011 1001
R =
3Solutions
1.IPv4 subnetting
Answer: p = 23
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
28 − 2 = 254 < 275
29 − 2 = 510 ≥ 275 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.IPv4 subnetting
Answer: x = 72
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
23 = 8
Subnet of this host
network: ⌊77/8⌋·8 = 72
broadcast: 72 + 8 − 1 = 79
usable hosts: 73 … 78
Answer
x = 72
3.Computer networks
Answer: D · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
D · 1 ms
4.Computer networks
Answer: D · Configure port forwarding on the router to the PC's private address
Why this is the answer
NAT creates mappings only for connections started from inside. A port-forwarding rule maps a public port to the internal host so inbound connections can reach it.
Answer
D · Configure port forwarding on the router to the PC's private address
5.Number systems
Answer: U = 188
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
68 = 0100 0100
inverted: 1011 1011
+1 → 1011 1100
Read it as unsigned
1011 1100₂ = 256 − 68 = 188
Answer
U = 188
6.Number systems
Answer: R = 146
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
XOR gives 1 where the two bits differ.
1010 1011
XOR 0011 1001
= 1001 0010
Value of the result
1001 0010₂ = 146
Answer
R = 146
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