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 192.168.82.0/24 is split with VLSM for these LANs: A: 28 hosts, B: 11 hosts, C: 9 hosts, D: 47 hosts, E: 4 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 C?
A: 28B: 11C: 9D: 47E: 4
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe frame has a wrong IP checksum and must be checked by every host
- BThe frame is too large for one port
- CThe switch is acting as a router
- DThe destination MAC address is not yet in its address table
4.Computer networksHard
A PC can reach printers on its own subnet but nothing beyond it. Its IP address and mask are correct. What is the most likely fault?
- AA duplicate MAC address on the internet
- BA missing or wrong default gateway
- CA faulty DNS server setting
- DThe TCP window is too small
5.Number systemsMedium
Set the bits so they store −31 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −318 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1000 << 3. What is the result's value R in decimal?
1011 1000<< 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).
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
D: 47 hosts → 26 = 64 (/26)
A: 28 hosts → 25 = 32 (/27)
B: 11 hosts → 24 = 16 (/28)
C: 9 hosts → 24 = 16 (/28)
E: 4 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .63
A: .64 – .95
B: .96 – .111
C: .112 – .127
E: .128 – .135
Answer
x = 112
LAN C: 192.168.82.112/28.
3.Computer networks
Answer: D · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
D · The destination MAC address is not yet in its address table
4.Computer networks
Answer: B · A missing or wrong default gateway
Why this is the answer
Hosts send off-subnet packets to the default gateway. If it is wrong or missing, local traffic still works but nothing outside is reachable, even by IP address.
Answer
B · A missing or wrong default gateway
5.Number systems
Answer: U = 225
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
31 = 0001 1111
inverted: 1110 0000
+1 → 1110 0001
Read it as unsigned
1110 0001₂ = 256 − 31 = 225
Answer
U = 225
6.Number systems
Answer: R = 192
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 left by 3 moves every bit 3 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1011 1000 << 3
= 1100 0000
Value of the result
1100 0000₂ = 192
Answer
R = 192
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