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
A host has the address 172.20.83.42/21. What is the third octet x of its network address?
172.20.83.42/21
x =
3.Computer networksMedium
How many usable host addresses does a /26 IPv4 subnet have?
- A64
- B126
- C62
- D30
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
- BWithout it, TCP could not detect lost segments on long-distance links
- CThe 32-bit window field was too large and wasted header space in every TCP segment sent
- DUDP needed a way to control its sending rate on congested paths
5.Number systemsMedium
The 8-bit pattern 0100 0000 stores a signed integer in two's complement. What is its value N in decimal?
0100 00008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1100 0110 >> 2. What is the result's value R in decimal?
1100 0110>> 2
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 = 80
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 in the third octet
23 = 8
Subnet of this host
network: 172.20.80.0
broadcast: 172.20.87.255
Answer
x = 80
3.Computer networks
Answer: C · 62
Why this is the answer
A /26 has 6 host bits: 2⁶ = 64 addresses, minus the network and broadcast addresses, gives 62 usable hosts.
Answer
C · 62
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: N = 64
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.
64 = 64
Answer
N = 64
6.Number systems
Answer: R = 49
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.
1100 0110 >> 2
= 0011 0001
Value of the result
0011 0001₂ = 49
Answer
R = 49
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