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 192.168.206.0/24 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 network address of subnet 2?
192.168.206.0/24/28subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 172.20.241.33/19. What is the third octet x of its subnet's broadcast address?
172.20.241.33/19
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 is too large for one port
- BThe frame has a wrong IP checksum and must be checked by every host
- CThe switch is acting as a router
- DThe destination MAC address is not yet in its address table
4.Computer networksHard
A TCP connection has a 64 KB window and a 100 ms round-trip time, and no window scaling. What is its approximate maximum throughput?
- AAbout 5 Mbit/s
- BAbout 640 Mbit/s
- CAbout 50 Mbit/s
- DAbout 0.5 Mbit/s
5.Number systemsMedium
Set the bits so they store −62 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −628 bit
U =
6.Number systemsHard
What is the most negative integer N that can be stored in 10-bit two's complement?
10 bit
N =
3Solutions
1.IPv4 subnetting
Answer: x = 16
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 2
network: (2 − 1)·16 = 16
broadcast: 16 + 16 − 1 = 31
Answer
x = 16
2.IPv4 subnetting
Answer: x = 255
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
25 = 32
Subnet of this host
network: 172.20.224.0
broadcast: 172.20.255.255
Answer
x = 255
3.Computer networks
Answer: D · 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
D · The destination MAC address is not yet in its address table
4.Computer networks
Answer: A · About 5 Mbit/s
Why this is the answer
Throughput ≤ window / RTT = 65,536 × 8 bits / 0.1 s ≈ 5.2 Mbit/s, however fast the link is.
Answer
A · About 5 Mbit/s
5.Number systems
Answer: U = 194
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
62 = 0011 1110
inverted: 1100 0001
+1 → 1100 0010
Read it as unsigned
1100 0010₂ = 256 − 62 = 194
Answer
U = 194
6.Number systems
Answer: N = −512
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.
Most negative value
−29 = −512
Answer
N = −512
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