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 192.168.189.0/24 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 broadcast address of subnet 4?
192.168.189.0/24/28subnet 4 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.76.216/27. What is the fourth octet x of its network address?
10.20.76.216/27
x =
3.Computer networksMedium
How does a stateful firewall differ from a simple packet filter?
- AIt blocks all incoming traffic without exceptions, including replies to outgoing requests
- BIt tracks connections and allows replies that belong to established sessions
- CIt encrypts every packet it forwards
- DIt only checks the MAC address of each frame
4.Computer networksHard
A user can ping 8.8.8.8 but cannot open any website by name. Where is the problem most likely?
- AThe physical cable
- BName resolution (DNS)
- CThe network card driver
- DThe default gateway
5.Number systemsMedium
The 8-bit pattern 1010 1001 stores a signed integer in two's complement. What is its value N in decimal?
1010 10018 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1011 0110 AND 0011 0000. What is the result's value R in decimal?
1011 0110AND0011 0000
R =
3Solutions
1.IPv4 subnetting
Answer: x = 63
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 4
network: (4 − 1)·16 = 48
broadcast: 48 + 16 − 1 = 63
Answer
x = 63
2.IPv4 subnetting
Answer: x = 192
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
25 = 32
Subnet of this host
network: ⌊216/32⌋·32 = 192
broadcast: 192 + 32 − 1 = 223
usable hosts: 193 … 222
Answer
x = 192
3.Computer networks
Answer: B · It tracks connections and allows replies that belong to established sessions
Why this is the answer
A stateful firewall keeps a table of active connections, so return traffic is allowed automatically. A stateless filter judges each packet in isolation.
Answer
B · It tracks connections and allows replies that belong to established sessions
4.Computer networks
Answer: B · Name resolution (DNS)
Why this is the answer
A successful ping to an external IP proves cabling, addressing and routing work. Failing names point to a DNS server or resolver setting problem.
Answer
B · Name resolution (DNS)
5.Number systems
Answer: N = −87
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.
−128 + 32 + 8 + 1 = −87
Answer
N = −87
6.Number systems
Answer: R = 48
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
AND gives 1 only where both bits are 1.
1011 0110
AND 0011 0000
= 0011 0000
Value of the result
0011 0000₂ = 48
Answer
R = 48
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