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 20 equal subnets. If you borrow as few bits as possible, how many usable hosts H does each subnet have?
192.168.10.0/2420 subnets
H =
2.IPv4 subnettingHard
A host has the address 172.20.238.130/20. What is the third octet x of its subnet's broadcast address?
172.20.238.130/20
x =
3.Computer networksMedium
Why does a switch sometimes send a frame out of all ports except the one it came in on?
- AThe frame has a wrong IP checksum and must be checked by every host
- BThe destination MAC address is not yet in its address table
- CThe switch is acting as a router
- DThe frame is too large for one port
4.Computer networksHard
Why does TCP choose an unpredictable initial sequence number for each connection?
- ATo avoid confusion with old segments and make spoofed segments harder to inject
- BTo select the best route to the destination for the segments of that connection
- CTo compress the data more efficiently
- DTo encrypt the payload of each segment
5.Number systemsMedium
Write −49 as an 8-bit two's complement pattern. Read the pattern as an unsigned binary number: what is its value U in decimal?
N = −498 bit
U =
6.Number systemsHard
A system must give every one of 3537 items its own binary code of the same length. What is the smallest number of bits b needed?
3537
b = bit
3Solutions
1.IPv4 subnetting
Answer: H = 6
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
25 = 32 ≥ 20 → b = 5, /29
Hosts per subnet
23 − 2 = 6
Answer
H = 6
2.IPv4 subnetting
Answer: x = 239
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
24 = 16
Subnet of this host
network: 172.20.224.0
broadcast: 172.20.239.255
Answer
x = 239
3.Computer networks
Answer: B · The destination MAC address is not yet in its address table
Why this is the answer
For an unknown unicast address the switch floods the frame. When the destination replies, the switch learns its port from the source MAC.
Answer
B · The destination MAC address is not yet in its address table
4.Computer networks
Answer: A · To avoid confusion with old segments and make spoofed segments harder to inject
Why this is the answer
Random initial sequence numbers stop delayed segments from old connections being accepted and make it hard for off-path attackers to forge valid segments.
Answer
A · To avoid confusion with old segments and make spoofed segments harder to inject
5.Number systems
Answer: U = 207
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
49 = 0011 0001
inverted: 1100 1110
+1 → 1100 1111
Read it as unsigned
1100 1111₂ = 256 − 49 = 207
Answer
U = 207
6.Number systems
Answer: b = 12 bit
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.
Powers of two
211 = 2048 < 3537
212 = 4096 ≥ 3537
Answer
b = 12
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