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 4 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/244 subnets
H =
2.IPv4 subnettingHard
The network 192.168.109.0/24 is split with VLSM for these LANs: A: 25 hosts, B: 22 hosts, C: 4 hosts, D: 114 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: 25B: 22C: 4D: 114
x =
3.Computer networksMedium
NAT (network address translation) is mainly used to:
- ATranslate domain names into IP addresses for devices on a local network
- BIncrease the speed of a network cable
- CEncrypt traffic between two sites
- DLet many devices with private addresses share one public IP address
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?
- A2.5 MB
- B500 kB
- C50 kB
- D4 kB
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 0111 0000 << 3. What is the result's value R in decimal?
0111 0000<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: H = 62
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
22 = 4 ≥ 4 → b = 2, /26
Hosts per subnet
26 − 2 = 62
Answer
H = 62
2.IPv4 subnetting
Answer: x = 160
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: 114 hosts → 27 = 128 (/25)
A: 25 hosts → 25 = 32 (/27)
B: 22 hosts → 25 = 32 (/27)
C: 4 hosts → 23 = 8 (/29)
Allocation, largest first
D: .0 – .127
A: .128 – .159
B: .160 – .191
C: .192 – .199
Answer
x = 160
LAN B: 192.168.109.160/27.
3.Computer networks
Answer: D · Let many devices with private addresses share one public IP address
Why this is the answer
NAT rewrites addresses (and ports) so private networks can access the internet through a smaller number of public addresses.
Answer
D · Let many devices with private addresses share one public IP address
4.Computer networks
Answer: B · 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
B · 500 kB
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 = 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.
0111 0000 << 3
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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