Number of hosts per network: The default subnet mask of class A is 255.0.0.08. The IP address is 32 bits. In this problem, we take 4 bits from the host and add it to network bits. So, 8 + 4 = 12 bits. Therefore, the mask bit is = 12. Consider each octet has 8 bits In this, we take 8 bits from the 1 st octet and 4 bit from the 2 nd octet so it can be used in 11 departments. The remaining bits are 4 bits+ 8 bits+ 8 bits= 20 bits...

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Answer 1

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Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

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Answer two JAVA OOP problems.

Answer 2

Expert Solution & Answer

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Expert Solution & Answer

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Check out a sample textbook solution

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Answer 3

Expert Solution & Answer

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Answer 4

Expert Solution & Answer

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Expert Solution & Answer

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Check out a sample textbook solution

See solution

Answer 5

Expert Solution & Answer

Answer 6

Expert Solution & Answer

Answer 7

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Answer 8

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Answer 9

Chapter 12, Problem 1DE

Explanation of Solution

Number of hosts per network:

The default subnet mask of class A is 255.0.0.08.

The IP address is 32 bits.

In this problem, we take 4 bits from the host and add it to network bits.

So,

8 + 4 = 12 bits.

Therefore, the mask bit is = 12.

Consider each octet has 8 bits

In this, we take 8 bits from the 1^st octet and 4 bit from the 2^nd octet so it can be used in 11 departments.

The remaining bits are

4 bits+ 8 bits+ 8 bits= 20 bits...

Answer 10

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Answer 11

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Answer 12

Expert Solution & Answer

Answer 13

Expert Solution & Answer

Answer 14

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Check out a sample textbook solution

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Answer 15

Ch. 12 - Prob. 11RQ Ch. 12 - Prob. 1DE

Solution Summary: The author explains that the default subnet mask of class A is 255.0.0.08. The IP address is 32 bits.

Concept explainers

Explanation of Solution

Explanation of Solution

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...

Want to see the full answer?

Chapter 12 Solutions

SUBNET	RANGE OF HOST	SUBNET SIZE	BROADCAST
100.0.0.0	100.0.0.1 TO 100.15.255.254	1048574	100.15.255.255
100...

Solution Summary: The author explains that the default subnet mask of class A is 255.0.0.08. The IP address is 32 bits.

Concept explainers

Explanation of Solution

Explanation of Solution

Five range of addresses: The first 5 range of addresses that can be assigned to different departments are, SUBNETRANGE OF HOSTSUBNET SIZEBROADCAST100.0.0.0100.0.0.1 TO 100.15.255.2541048574100.15.255.255100...

Want to see the full answer?

Chapter 12 Solutions

Five range of addresses:

The first 5 range of addresses that can be assigned to different departments are,

SUBNET RANGE OF HOST SUBNET SIZE BROADCAST
100.0.0.0 100.0.0.1 TO 100.15.255.254 1048574 100.15.255.255
100...