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 192.168.227.0/24 is divided into /28 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the broadcast address of subnet 12?
192.168.227.0/24/28subnet 12 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.50.0/24 is split with VLSM for these LANs: A: 48 hosts, B: 16 hosts, C: 51 hosts, D: 8 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: 48B: 16C: 51D: 8
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- A::1
- Bfe80::1
- Cff02::1
- D::
4.Computer networksHard
Why was the TCP window scale option introduced?
- AThe 32-bit window field was too large and wasted header space in every TCP segment sent
- BUDP needed a way to control its sending rate on congested paths
- CWithout it, TCP could not detect lost segments on long-distance links
- DThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systemsMedium
Write −89 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −898 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0100 0000 XOR 0110 1010. What is the result's value R in decimal?
0100 0000XOR0110 1010
R =
3Solutions
1.IPv4 subnetting
Answer: x = 191
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).
Block size
24 = 16
Subnet number 12
network: (12 − 1)·16 = 176
broadcast: 176 + 16 − 1 = 191
Answer
x = 191
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
C: 51 hosts → 26 = 64 (/26)
A: 48 hosts → 26 = 64 (/26)
B: 16 hosts → 25 = 32 (/27)
D: 8 hosts → 24 = 16 (/28)
Allocation, largest first
C: .0 – .63
A: .64 – .127
B: .128 – .159
D: .160 – .175
Answer
x = 160
LAN D: 192.168.50.160/28.
3.Computer networks
Answer: A · ::1
Why this is the answer
::1 is the loopback address. :: is the unspecified address, fe80::/10 is link-local and ff02::1 is the all-nodes multicast group.
Answer
A · ::1
4.Computer networks
Answer: D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
Why this is the answer
The original header allows at most 65,535 bytes of unacknowledged data. Window scaling multiplies this by a power of two so high bandwidth-delay paths can be filled.
Answer
D · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 167
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
89 = 0101 1001
inverted: 1010 0110
+1 → 1010 0111
Read it as unsigned
1010 0111₂ = 256 − 89 = 167
Answer
U = 167
6.Number systems
Answer: R = 42
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
XOR gives 1 where the two bits differ.
0100 0000
XOR 0110 1010
= 0010 1010
Value of the result
0010 1010₂ = 42
Answer
R = 42
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