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 192.168.99.0/24 is divided into /27 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 5?
192.168.99.0/24/27subnet 5 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.122.134/26. What is the fourth octet x of the last usable host address in its subnet?
10.20.122.134/26
x =
3.Computer networksMedium
At which OSI layer does a switch normally operate?
- ATransport (layer 4)
- BNetwork (layer 3)
- CData link (layer 2)
- DPhysical (layer 1)
4.Computer networksHard
Which routing protocol is a link-state protocol?
- AOSPF
- BBGP
- CARP
- DRIP
5.Number systemsMedium
The 8-bit pattern 1101 1100 stores a signed integer in two's complement. What is its value N in decimal?
1101 11008 bit
N =
6.Number systemsHard
Set the bits to the 8-bit result of 1010 0000 << 2. What is the result's value R in decimal?
1010 0000<< 2
R =
3Solutions
1.IPv4 subnetting
Answer: x = 128
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 number 5
network: (5 − 1)·32 = 128
broadcast: 128 + 32 − 1 = 159
Answer
x = 128
2.IPv4 subnetting
Answer: x = 190
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: ⌊134/64⌋·64 = 128
broadcast: 128 + 64 − 1 = 191
usable hosts: 129 … 190
Answer
x = 190
3.Computer networks
Answer: C · Data link (layer 2)
Why this is the answer
A standard switch learns MAC addresses and forwards frames at layer 2. Layer 3 switches also route.
Answer
C · Data link (layer 2)
4.Computer networks
Answer: A · OSPF
Why this is the answer
OSPF routers build a full map of the area and compute shortest paths with Dijkstra's algorithm. RIP is distance-vector; BGP is path-vector.
Answer
A · OSPF
5.Number systems
Answer: N = −36
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 + 64 + 16 + 8 + 4 = −36
Answer
N = −36
6.Number systems
Answer: R = 128
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.
1010 0000 << 2
= 1000 0000
Value of the result
1000 0000₂ = 128
Answer
R = 128
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