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 192.168.111.0/24 is divided into /25 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 2?
192.168.111.0/24/25subnet 2 (subnet 1 starts at .0)
x =
2.IPv4 subnettingHard
A host has the address 10.20.137.244/27. What is the fourth octet x of its network address?
10.20.137.244/27
x =
3.Computer networksMedium
A router has routes for 10.0.0.0/8 and 10.1.0.0/16. Which route does it use for a packet to 10.1.5.5?
- A10.0.0.0/8, because it was added first
- B10.0.0.0/8, because it covers more addresses than the /16
- CBoth routes, splitting the traffic equally
- D10.1.0.0/16, because it is the longest matching prefix
4.Computer networksHard
Why was the TCP window scale option introduced?
- AThe 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
- BThe 32-bit window field was too large and wasted header space in every TCP segment sent
- CWithout it, TCP could not detect lost segments on long-distance links
- DUDP needed a way to control its sending rate on congested paths
5.Number systemsMedium
Set the bits so they store −105 in 8-bit two's complement. Read as an unsigned number, what is the pattern's value U?
N = −1058 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 = 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
27 = 128
Subnet number 2
network: (2 − 1)·128 = 128
broadcast: 128 + 128 − 1 = 255
Answer
x = 255
2.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 of this host
network: ⌊244/32⌋·32 = 224
broadcast: 224 + 32 − 1 = 255
usable hosts: 225 … 254
Answer
x = 224
3.Computer networks
Answer: D · 10.1.0.0/16, because it is the longest matching prefix
Why this is the answer
IP forwarding uses longest prefix match: among all routes that match the destination, the one with the longest mask is chosen.
Answer
D · 10.1.0.0/16, because it is the longest matching prefix
4.Computer networks
Answer: A · 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
A · The 16-bit window field limits the window to 65,535 bytes, too small for fast long-distance paths
5.Number systems
Answer: U = 151
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
105 = 0110 1001
inverted: 1001 0110
+1 → 1001 0111
Read it as unsigned
1001 0111₂ = 256 − 105 = 151
Answer
U = 151
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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