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
A department needs 483 host addresses in a single subnet. What is the longest prefix length p (smallest subnet) that is big enough?
483 hosts
p =
2.IPv4 subnettingHard
A host has the address 10.20.53.180/26. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.53.180/26
x =
3.Computer networksMedium
Which protocol does the "ping" command use?
- AHTTP
- BTCP
- CUDP
- DICMP
4.Computer networksHard
Why was the TCP window scale option introduced?
- AUDP needed a way to control its sending rate on congested paths
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- 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 −27 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −278 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 1000 0001 << 2. What is the result's value R in decimal?
1000 0001<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: p = 23
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).
Host bits needed
28 − 2 = 254 < 483
29 − 2 = 510 ≥ 483 → h = 9
Prefix length
p = 32 − 9 = 23
Answer
p = /23
2.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
26 = 64
Subnet of this host
network: ⌊180/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190
Answer
x = 191
3.Computer networks
Answer: D · ICMP
Why this is the answer
Ping sends ICMP echo requests and measures the time until echo replies return.
Answer
D · ICMP
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 = 229
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
27 = 0001 1011
inverted: 1110 0100
+1 → 1110 0101
Read it as unsigned
1110 0101₂ = 256 − 27 = 229
Answer
U = 229
6.Number systems
Answer: R = 4
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 2 moves every bit 2 places to the left; bits that fall off the left end are lost and zeros come in on the right.
1000 0001 << 2
= 0000 0100
Value of the result
0000 0100₂ = 4
Answer
R = 4
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