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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Discrete Mathematics for Computer Engineering

this module is java 731 . make sure my answers are 1005 correct and the layout and structure is perfect and also include all comments etc. thank you i have attached question 1 (40 marks) and question 2 (30 marks ) this is question 3: Question3: (30 MARKS) Passenger Rail Agency for South Africa Train Scheduling System Problem Statement Design and implement a train scheduling system for Prasa railway network. The system should handle the following functionalities: 1. Scheduling trains: Allow the addition of train schedules, ensuring that no two trains use the same platform at the same time at any station. 2. Dynamic updates: Enable adding new train schedules and canceling existing ones. 3. Real-time simulation: Use multithreading to simulate the operation of trains (e.g., arriving, departing). 4. Data management: Use ArrayList to manage train schedules and…

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

See solution

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

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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

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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

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Answer 13

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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...