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 10.85.220.0/22 is divided into equal subnets with prefix /27. How many subnets N are there?
10.85.220.0/22/27
N =
2.IPv4 subnettingHard
The network 192.168.8.0/24 is split with VLSM for these LANs: A: 53 hosts, B: 11 hosts, C: 15 hosts, D: 22 hosts, E: 6 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: 53B: 11C: 15D: 22E: 6
x =
3.Computer networksMedium
Which range of TCP/UDP port numbers is reserved as "well-known" ports?
- A0–1023
- B0–65535
- C1024–49151
- D49152–65535
4.Computer networksHard
A path has 100 Mbit/s bandwidth and a 40 ms round-trip time. How much data must be in flight to keep it fully used?
- A500 kB
- B2.5 MB
- C50 kB
- D4 kB
5.Number systemsMedium
Write −51 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −518 bit
U =
6.Number systemsHard
What is the largest unsigned integer N that fits in 16 bits?
16 bit
N =
3Solutions
1.IPv4 subnetting
Answer: N = 32
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).
Borrowed bits
27 − 22 = 5
Number of subnets
N = 25 = 32
Answer
N = 32
Each /27 subnet has 25 − 2 = 30 usable hosts.
2.IPv4 subnetting
Answer: x = 128
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
A: 53 hosts → 26 = 64 (/26)
D: 22 hosts → 25 = 32 (/27)
C: 15 hosts → 25 = 32 (/27)
B: 11 hosts → 24 = 16 (/28)
E: 6 hosts → 23 = 8 (/29)
Allocation, largest first
A: .0 – .63
D: .64 – .95
C: .96 – .127
B: .128 – .143
E: .144 – .151
Answer
x = 128
LAN B: 192.168.8.128/28.
3.Computer networks
Answer: A · 0–1023
Why this is the answer
Ports 0–1023 are well-known ports for standard services. 1024–49151 are registered ports and 49152–65535 are dynamic or private ports.
Answer
A · 0–1023
4.Computer networks
Answer: A · 500 kB
Why this is the answer
The bandwidth-delay product is 100 × 10⁶ bit/s × 0.04 s = 4 × 10⁶ bits = 500 kB. The sender's window must be at least this large.
Answer
A · 500 kB
5.Number systems
Answer: U = 205
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
51 = 0011 0011
inverted: 1100 1100
+1 → 1100 1101
Read it as unsigned
1100 1101₂ = 256 − 51 = 205
Answer
U = 205
6.Number systems
Answer: N = 65535
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.
Largest value
216 − 1 = 65535
Answer
N = 65535
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