Question

Два узла находятся в одной сети. IP1=211.115.61.154 IP2=211.115.59.37 Найти наибольшее возможное значение третьего слева байта маски, адрес сети, количество компьютеров в этой сети и номера компьютеров с указанными IP-адресами 10 класс

Answer 1

Ответ:

211.115.61.154

211.115.59.37

61 = 0011 1101

59 = 0011 1011

3-й байт маски

  = 1100 0000 = 192

маска: 255.255.192.0

или так: 8 + 8 + 2 => /18 (CIDR запись маски)

адрес сети: 211.115.0.0

мин адрес: 211.115.0.1

макс адрес: 211.115.63.254

кол-во хостов: 2^(32 - 18) - 2 = 16382

2 хоста используются для обратной петли и broadcast, поэтому -2

номер IP1: ...|...|0011 1101|1001 1010 => 15770

номер IP2: ...|...|0011 1011|0010 0101 => 15141

Answer 2

Finding the Subnet Mask, Network Address, Number of Computers, and Computer Numbers

To find the subnet mask, network address, number of computers, and computer numbers for the given IP addresses, we need to determine the network portion of the IP addresses and calculate the necessary values.

Let's break down the given IP addresses:

- IP1: 211.115.61.154 - IP2: 211.115.59.37

Determining the Network Portion

To determine the network portion of the IP addresses, we need to compare the binary representation of the IP addresses. The network portion is the part of the IP address that remains the same for all devices within the same network.

Let's convert the given IP addresses to binary:

- IP1: 11010011.01110011.00111101.10011010 - IP2: 11010011.01110011.00111011.00100101

By comparing the binary representation, we can determine that the network portion is the first 21 bits (from left to right) for both IP addresses.

Finding the Subnet Mask

The subnet mask is used to divide an IP address into a network portion and a host portion. It is represented by a series of 1s followed by a series of 0s. The number of 1s in the subnet mask determines the size of the network.

In this case, since the network portion consists of the first 21 bits, the subnet mask will have 21 1s followed by 11 0s.

Converting the subnet mask to decimal, we get: 255.255.248.0

Finding the Network Address

The network address is obtained by performing a bitwise AND operation between the IP address and the subnet mask. This operation sets all the bits in the host portion to 0, effectively isolating the network portion.

Performing the bitwise AND operation for IP1 and the subnet mask:

- IP1: 11010011.01110011.00111101.10011010 - Subnet Mask: 11111111.11111111.11111000.00000000 - Network Address: 11010011.01110011.00111000.00000000

Converting the network address to decimal, we get: 211.115.56.0

Finding the Number of Computers

The number of computers in a network is determined by the number of available host addresses. In this case, since the network portion consists of 21 bits, the host portion consists of 11 bits.

The number of available host addresses can be calculated using the formula 2^n - 2, where n is the number of bits in the host portion.

Calculating the number of available host addresses:

- Number of bits in the host portion: 11 - Number of available host addresses: 2^11 - 2 = 2046

Therefore, there are 2046 available host addresses in this network.

Finding the Computer Numbers

To find the computer numbers for the given IP addresses, we need to convert the host portion of the IP addresses to decimal.

Converting the host portion of IP1 to decimal:

- Host portion of IP1: 00111101.10011010 - Computer number for IP1: 61.154

Converting the host portion of IP2 to decimal:

- Host portion of IP2: 00111011.00100101 - Computer number for IP2: 59.37

Therefore, the computer numbers for the given IP addresses are 61.154 and 59.37.

Summary

Based on the given IP addresses (IP1 = 211.115.61.154 and IP2 = 211.115.59.37), the largest possible value for the third byte of the subnet mask is 248. The network address is 211.115.56.0, and there are 2046 available host addresses in this network. The computer numbers for the given IP addresses are 61.154 and 59.37.

Please let me know if there's anything else I can help you with!