Solve the basics of the network by the masters

You're still putting it... I'll give you a copy too

1.(2.Subnet mask, which should be.

3.(1) Subnet mask,. (2) Subnet number. (3) Main engine range 2-24, 2-24.

I'm a real-world experience.

3.First Subnet: Mask: Subnet Number:

Host Scope: Second Subnet: Mask: Subnet Number: Host Scope:

1.In the CRC checksum. The generative polynomial is known to be g(x)=x4+x3+1. Requires the CRC check digit of the information 1011001 to be written. Solution:

The polynomial g(x)=11001 is generated, which is 5 bits, and the balance of the check is 4 digits, and the calculation process is as follows according to modulo 2 division

11001 011110 11001 011100 11001 1010 Remainder r(x) = 1010

CRC checksum = 1011001 1010

2.The two parties use the CRC cyclic check code to communicate, and it is known that the generated polynomial is x4+x3+x+1, and the received codeword is 10111010011. Determine whether the information is incorrect. Solution:

According to the question, the polynomial g(x)=11011 is generated, and if the information is correct, the remainder of modulo 2 division should be 0

11011 11100 11011 11111 11011 100 Result: The remainder r(x) = 100 is not zero, so the result is wrong.

In a channel with a bandwidth of 3kHz and no noise, the limit value of the code rate that can be achieved is 6kbps The code rate is the limit of the channel's ability to transmit data, Nyquist first gave the relationship between the limit value of the code rate and the channel bandwidth in the case of no noise: b=2h (baud) where h is the bandwidth of the channel, also known as the frequency range, that is, the difference between the upper and lower frequencies that the channel can transmit. From this, the Nyquist formula, which characterizes the data transmission capability of the channel, can be derived

c=2 h log2n (bps) for a particular channel, the symbol rate cannot exceed 2 times the channel bandwidth, but the data transmission rate can be increased exponentially if the number of discrete values that can be taken for each symbol can be increased. For example, if the bandwidth of an ordinary ** line is about 3kHz, the limit value of its symbol rate is 6kbaud. If the number of discrete values that can be obtained for each symbol is 32 (i.e., n=32), the maximum data transmission rate can reach c=2*3k*log2 32=30kbps.

Shannon further studied the situation of the channel disturbed by random noise, and gave Shannon's formula for calculating the channel capacity: c=h*log2(1+s n) (bps) where s is the signal power and n is the noise power, so it can be seen that the maximum data transmission rate of the channel can be increased as long as the signal-to-noise ratio of the channel is increased.

Hope it helps.

How much does the data rate of a digital signal being transmitted over a 3kHz channel with a signal-to-noise ratio of 20dB be exceeded, with and without thermal noise, when thermal noise is taken into account?

1. Equal range division.

1) Based on host bits.

Let the required host bit be x and y be the network bit, i.e., y=8-x; (Note: This setting is based on the re-question setting).

2^x>=19 → x=5，y=3；

As shown above, you need to borrow 3 bits from the original host bit as the network bit;

Subnet mask: The network address is the first IP address of the appeal, the broadcast address is the last address, and the available range is the range of excluding the network address and the broadcast address; Taking the first subnet as an example, the network address is, the broadcast address is, and the available range is -

2) Network bits-based.

As you know, if you need to assign IP addresses to four departments, you need to borrow 2 bits.

Subnet mask: The network address is the first IP address of the appeal, the broadcast address is the last address, and the available range is the range of excluding the network address and the broadcast address; Taking the first subnet as an example, the network address is, the broadcast address is, and the available range is -

2. Non-equal range division.

This method is actually the same as the previous "hostbit-based" partition, except that (1) uses an equal range, i.e., the same number of available addresses for each subnet. The "non-equal range division" is based on the idea of the "host-based" method, but the division continues after the original subnet division.

As can be seen from the question, the technical department consists of 18 people, and the required host bit is set as x, and y is the network bit, that is, y=8-x;

2^x>=18 → x=5，y=3；

Subnet mask: 19 people in the sales department:

2^x>=19 → x=5，y=3；

Subnet mask: 12 people in the personnel department:

2^x>=12 → x=4，y=4；

Subnet mask: 11 people from the finance department:

2^x>=11 → x=4，y=4；

Subnet Mask: