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 subnet uses the mask 255.255.255.224. How many usable host addresses H does it have?
255.255.255.224
H =
2.IPv4 subnettingHard
The network 192.168.157.0/24 is split with VLSM for these LANs: A: 10 hosts, B: 11 hosts, C: 53 hosts, D: 5 hosts, E: 19 hosts. Subnets are allocated from .0 upwards, the LAN with the most hosts first, each with the smallest subnet that fits. What is the fourth octet x of the network address of LAN A?
A: 10B: 11C: 53D: 5E: 19
x =
3.Computer networksMedium
Which metric does RIP use to choose routes?
- AHop count
- BBusiness policy
- CLink bandwidth
- DMeasured delay
4.Computer networksHard
Stations A and C both reach access point B but cannot hear each other, so their frames collide at B. How does 802.11 help?
- ABy routing A's frames through C
- BOptional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
- CBy making A and C use collision detection so they stop sending when frames overlap
- DBy increasing the TTL of the frames
5.Number systemsMedium
Set the bits so they store −23 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −238 bit
U =
6.Number systemsHard
An uncompressed image is 1024 × 768 pixels with 32 bits per pixel. How much memory M does it take, in KiB (1 KiB = 1024 bytes)?
1024 × 768 px32 bit/px1 KiB = 1024 B
M = KiB
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).
Prefix length
255.255.255.224 = /27
Usable hosts
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 112
VLSM: each LAN gets the smallest subnet with 2ʰ − 2 ≥ its host count, i.e. prefix 32 − h. Allocating the largest subnets first keeps every block aligned on a multiple of its own size, so the network addresses follow by adding up the block sizes.
Block size per LAN
C: 53 hosts → 26 = 64 (/26)
E: 19 hosts → 25 = 32 (/27)
B: 11 hosts → 24 = 16 (/28)
A: 10 hosts → 24 = 16 (/28)
D: 5 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .63
E: .64 – .95
B: .96 – .111
A: .112 – .127
D: .128 – .135
Answer
x = 112
LAN A: 192.168.157.112/28.
3.Computer networks
Answer: A · Hop count
Why this is the answer
RIP is a distance-vector protocol that counts hops, with 16 meaning unreachable. This limits it to small networks.
Answer
A · Hop count
4.Computer networks
Answer: B · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
Why this is the answer
A station sends Request to Send; the AP answers Clear to Send, which every station in its range hears, so hidden stations defer for the reserved time.
Answer
B · Optional RTS/CTS exchanges: the AP's CTS tells all nearby stations to wait
5.Number systems
Answer: U = 233
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
23 = 0001 0111
inverted: 1110 1000
+1 → 1110 1001
Read it as unsigned
1110 1001₂ = 256 − 23 = 233
Answer
U = 233
6.Number systems
Answer: M = 3072 KiB
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.
Bits and bytes
1024·768·32 = 25165824 bit = 3145728 B
Convert to KiB
3145728/1024 = 3072 KiB
Answer
M = 3072 KiB
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