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 6 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/246 subnets
H =
2.IPv4 subnettingHard
The network 192.168.173.0/24 is split with VLSM for these LANs: A: 15 hosts, B: 113 hosts, C: 29 hosts, D: 3 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 D?
A: 15B: 113C: 29D: 3
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A30
- B64
- C62
- D126
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 640 Mbit/s
- BAbout 0.5 Mbit/s
- CAbout 50 Mbit/s
- DAbout 5 Mbit/s
5.Number systemsMedium
Set the bits so they store −22 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −228 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 20-bit two's complement?
20 bit
N =
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).
Bits to borrow
23 = 8 ≥ 6 → b = 3, /27
Hosts per subnet
25 − 2 = 30
Answer
H = 30
2.IPv4 subnetting
Answer: x = 192
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: 113 hosts → 27 = 128 (/25)
C: 29 hosts → 25 = 32 (/27)
A: 15 hosts → 25 = 32 (/27)
D: 3 hosts → 23 = 8 (/29)
Allocation, largest first
B: .0 – .127
C: .128 – .159
A: .160 – .191
D: .192 – .199
Answer
x = 192
LAN D: 192.168.173.192/29.
3.Computer networks
Answer: C · 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
C · 62
4.Computer networks
Answer: D · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
D · About 5 Mbit/s
5.Number systems
Answer: U = 234
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
22 = 0001 0110
inverted: 1110 1001
+1 → 1110 1010
Read it as unsigned
1110 1010₂ = 256 − 22 = 234
Answer
U = 234
6.Number systems
Answer: N = −524288
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.
Most negative value
−219 = −524288
Answer
N = −524288
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