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 3 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/243 subnets
H =
2.IPv4 subnettingHard
The network 192.168.203.0/24 is split with VLSM for these LANs: A: 59 hosts, B: 117 hosts, C: 6 hosts, D: 8 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: 59B: 117C: 6D: 8
x =
3.Computer networksMedium
Which host is in the same /16 subnet as 10.1.2.3?
- A10.0.2.3
- B10.1.200.5
- C10.2.2.3
- D11.1.2.3
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AName resolution (DNS)
- BThe physical cable
- CThe network card driver
- DThe default gateway
5.Number systemsMedium
The 8-bit pattern 1111 1110 stores a signed integer in two's complement. What is its value N in decimal?
1111 11108 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 0101 >> 1. What is the result's value R in decimal?
0101 0101>> 1
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 3 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 208
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
B: 117 hosts → 27 = 128 (/25)
A: 59 hosts → 26 = 64 (/26)
D: 8 hosts → 24 = 16 (/28)
C: 6 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
A: .128 – .191
D: .192 – .207
C: .208 – .215
Answer
x = 208
LAN C: 192.168.203.208/29.
3.Computer networks
Answer: B · 10.1.200.5
Why this is the answer
With /16 the first 16 bits (10.1) are the network part. Only 10.1.200.5 shares them, so it is on 10.1.0.0/16.
Answer
B · 10.1.200.5
4.Computer networks
Answer: A · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
A · Name resolution (DNS)
5.Number systems
Answer: N = −2
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.
Add the place values
The other bits keep their normal positive values.
−128 + 64 + 32 + 16 + 8 + 4 + 2 = −2
Answer
N = −2
6.Number systems
Answer: R = 42
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 1 moves every bit 1 places to the right; bits that fall off the right end are lost and zeros come in on the left.
0101 0101 >> 1
= 0010 1010
Value of the result
0010 1010₂ = 42
Answer
R = 42
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