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 1157 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1157 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.204.161/25. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.204.161/25
x =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- AData link (layer 2)
- BPhysical (layer 1)
- CNetwork (layer 3)
- DTransport (layer 4)
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo select the best route to the destination for the segments of that connection
- BTo compress the data more efficiently
- CTo avoid confusion with old segments and make spoofed segments harder to inject
- DTo encrypt the payload of each segment
5.Number systemsMedium
Set the bits so they store −8 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −88 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1001 OR 1000 1110. What is the result's value R in decimal?
0101 1001OR1000 1110
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 < 1157
211 − 2 = 2046 ≥ 1157 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 255
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
27 = 128
Subnet of this host
network: ⌊161/128⌋·128 = 128
broadcast: 128 + 128 − 1 = 255
usable hosts: 129 … 254
Answer
x = 255
3.Computer networks
Answer: A · Data link (layer 2)
Why this is the answer
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer
A · Data link (layer 2)
4.Computer networks
Answer: C · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
C · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 248
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
8 = 0000 1000
inverted: 1111 0111
+1 → 1111 1000
Read it as unsigned
1111 1000₂ = 256 − 8 = 248
Answer
U = 248
6.Number systems
Answer: R = 223
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
OR gives 1 where at least one bit is 1.
0101 1001
OR 1000 1110
= 1101 1111
Value of the result
1101 1111₂ = 223
Answer
R = 223
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