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 192.168.77.0/24 is split with VLSM for these LANs: A: 15 hosts, B: 10 hosts, C: 61 hosts, D: 73 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 A?
A: 15B: 10C: 61D: 73
x =
3.Computer networksMedium
What does a subnet mask of 255.255.255.0 correspond to in CIDR notation?
- A/24
- B/32
- C/8
- D/16
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 5 Mbit/s
- BAbout 0.5 Mbit/s
- CAbout 50 Mbit/s
- DAbout 640 Mbit/s
5.Number systemsMedium
The 8-bit pattern 1000 0101 stores a signed integer in two's complement. What is its value N in decimal?
1000 01018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1001 0000 << 3. What is the result's value R in decimal?
1001 0000<< 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).
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
D: 73 hosts → 27 = 128 (/25)
C: 61 hosts → 26 = 64 (/26)
A: 15 hosts → 25 = 32 (/27)
B: 10 hosts → 24 = 16 (/28)
Allocation, largest first
D: .0 – .127
C: .128 – .191
A: .192 – .223
B: .224 – .239
Answer
x = 192
LAN A: 192.168.77.192/27.
3.Computer networks
Answer: A · /24
Why this is the answer
Three octets of 255 are 24 one-bits, so 255.255.255.0 = /24.
Answer
A · /24
4.Computer networks
Answer: A · 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
A · About 5 Mbit/s
5.Number systems
Answer: N = −123
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 + 4 + 1 = −123
Answer
N = −123
6.Number systems
Answer: R = 128
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.
1001 0000 << 3
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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