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 192.168.145.0/24 is divided into /26 subnets, numbered 1, 2, 3, … from the lowest address (subnet 1 starts at .0). What is the fourth octet x of the network address of subnet 2?
192.168.145.0/24/26subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
The network 192.168.159.0/24 is split with VLSM for these LANs: A: 37 hosts, B: 29 hosts, C: 54 hosts, D: 12 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: 37B: 29C: 54D: 12
x =
3.Computer networksMedium
What is the IPv6 loopback address?
- A::1
- Bff02::1
- Cfe80::1
- D::
4.Computer networksHard
Why was the TCP window scale option introduced?
- AThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- 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 32-bit window field was too large and wasted header space in every TCP segment sent
5.Number systemsMedium
Set the bits so they store −19 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −198 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 1001 << 3. What is the result's value R in decimal?
1011 1001<< 3
R =
3Solutions
1.IPv4 subnetting
Answer: x = 64
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
26 = 64
Subnet number 2
network: (2 − 1)·64 = 64
broadcast: 64 + 64 − 1 = 127
Answer
x = 64
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: 54 hosts → 26 = 64 (/26)
A: 37 hosts → 26 = 64 (/26)
B: 29 hosts → 25 = 32 (/27)
D: 12 hosts → 24 = 16 (/28)
Allocation, largest first
C: .0 – .63
A: .64 – .127
B: .128 – .159
D: .160 – .175
Answer
x = 160
LAN D: 192.168.159.160/28.
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: A · 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
A · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 237
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
19 = 0001 0011
inverted: 1110 1100
+1 → 1110 1101
Read it as unsigned
1110 1101₂ = 256 − 19 = 237
Answer
U = 237
6.Number systems
Answer: R = 200
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.
1011 1001 << 3
= 1100 1000
Value of the result
1100 1000₂ = 200
Answer
R = 200
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