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 192.168.141.0/24 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 8?
192.168.141.0/24/27subnet 8 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.172.152/30. What is the last number (fourth octet) x of its subnet's broadcast address?
10.20.172.152/30
x =
3.Computer networksMedium
What is the main difference between the HTTP methods GET and POST?
- AGET is encrypted, POST is not
- BGET uses UDP, POST uses TCP
- CGET deletes a resource, POST reads it from the server's database
- DGET retrieves a resource, POST submits data in the request body
4.Computer networksHard
Why was the TCP window scale option introduced?
- AWithout it, TCP could not detect lost segments on long-distance links
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- CThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- DUDP needed a way to control its sending rate on congested paths
5.Number systemsMedium
Set the bits so they store −109 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1098 bit
U =
6.Number systemsHard
Set the bits to the 8-bit result of 0101 1100 XOR 1010 0101. What is the result's value R in decimal?
0101 1100XOR1010 0101
R =
3Solutions
1.IPv4 subnetting
Answer: x = 224
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 8
network: (8 − 1)·32 = 224
broadcast: 224 + 32 − 1 = 255
Answer
x = 224
2.IPv4 subnetting
Answer: x = 155
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
22 = 4
Subnet of this host
network: ⌊152/4⌋·4 = 152
broadcast: 152 + 4 − 1 = 155
usable hosts: 153 … 154
Answer
x = 155
3.Computer networks
Answer: D · GET retrieves a resource, POST submits data in the request body
Why this is the answer
GET asks for a resource and should not change server state. POST sends data, such as a form, in the body for the server to process.
Answer
D · GET retrieves a resource, POST submits data in the request body
4.Computer networks
Answer: C · 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
C · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 147
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
109 = 0110 1101
inverted: 1001 0010
+1 → 1001 0011
Read it as unsigned
1001 0011₂ = 256 − 109 = 147
Answer
U = 147
6.Number systems
Answer: R = 249
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
XOR gives 1 where the two bits differ.
0101 1100
XOR 1010 0101
= 1111 1001
Value of the result
1111 1001₂ = 249
Answer
R = 249
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