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 1757 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
1757 hosts
p =
2.IPv4 subnettingHard
The network 192.168.67.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 29 hosts, C: 3 hosts, D: 26 hosts, E: 62 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 B?
A: 13B: 29C: 3D: 26E: 62
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A62
- B64
- C126
- D30
4.Computer networksHard
Which routing protocol is a link-state protocol?
- AOSPF
- BARP
- CRIP
- DBGP
5.Number systemsMedium
Write −11 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −118 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0110 << 2. What is the result's value R in decimal?
1100 0110<< 2
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 < 1757
211 − 2 = 2046 ≥ 1757 → h = 11
Prefix length
p = 32 − 11 = 21
Answer
p = /21
2.IPv4 subnetting
Answer: x = 64
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
E: 62 hosts → 26 = 64 (/26)
B: 29 hosts → 25 = 32 (/27)
D: 26 hosts → 25 = 32 (/27)
A: 13 hosts → 24 = 16 (/28)
C: 3 hosts → 23 = 8 (/29)
Allocation, largest first
E: .0 – .63
B: .64 – .95
D: .96 – .127
A: .128 – .143
C: .144 – .151
Answer
x = 64
LAN B: 192.168.67.64/27.
3.Computer networks
Answer: A · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
A · 62
4.Computer networks
Answer: A · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
A · OSPF
5.Number systems
Answer: U = 245
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.
Invert and add one
11 = 0000 1011
inverted: 1111 0100
+1 → 1111 0101
Read it as unsigned
1111 0101₂ = 256 − 11 = 245
Answer
U = 245
6.Number systems
Answer: R = 24
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1100 0110 << 2
= 0001 1000
Value of the result
0001 1000₂ = 24
Answer
R = 24
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