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 10 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2410 subnets
H =
2.IPv4 subnettingHard
The network 192.168.112.0/24 is split with VLSM for these LANs: A: 13 hosts, B: 5 hosts, C: 66 hosts, D: 45 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: 13B: 5C: 66D: 45
x =
3.Computer networksMedium
How long does it take to put a 1500-byte packet onto a 12 Mbit/s link (transmission delay)?
- A12 ms
- B8 ms
- C1 ms
- D0.125 ms
4.Computer networksHard
In the TCP/IP model, which layer adds port numbers?
- ALink layer
- BApplication layer
- CInternet layer
- DTransport layer
5.Number systemsMedium
The 8-bit pattern 1010 1011 stores a signed integer in two's complement. What is its value N in decimal?
1010 10118 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1111 0100 >> 2. What is the result's value R in decimal?
1111 0100>> 2
R =
3Solutions
1.IPv4 subnetting
Answer: H = 14
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
24 = 16 ≥ 10 → b = 4, /28
Hosts per subnet
24 − 2 = 14
Answer
H = 14
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
C: 66 hosts → 27 = 128 (/25)
D: 45 hosts → 26 = 64 (/26)
A: 13 hosts → 24 = 16 (/28)
B: 5 hosts → 23 = 8 (/29)
Allocation, largest first
C: .0 – .127
D: .128 – .191
A: .192 – .207
B: .208 – .215
Answer
x = 128
LAN D: 192.168.112.128/26.
3.Computer networks
Answer: C · 1 ms
Why this is the answer
1500 bytes = 12,000 bits. 12,000 / 12,000,000 bit/s = 0.001 s = 1 ms. Transmission delay depends on packet size and link rate, not distance.
Answer
C · 1 ms
4.Computer networks
Answer: D · Transport layer
Why this is the answer
TCP and UDP headers contain source and destination ports; the internet layer (IP) adds addresses.
Answer
D · Transport layer
5.Number systems
Answer: N = −85
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 + 32 + 8 + 2 + 1 = −85
Answer
N = −85
6.Number systems
Answer: R = 61
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 2 moves every bit 2 places to the right; bits that fall off the right end are lost and zeros come in on the left.
1111 0100 >> 2
= 0011 1101
Value of the result
0011 1101₂ = 61
Answer
R = 61
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