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3 years ago

Anyone can help me with this?

Computers and Technology

1 answer:

dybincka [34]3 years ago

5 0

Answer:

Find Largest and Smallest Number in an Array Example

This Java Example shows how to find largest and smallest number in an

array.

public class FindLargestSmallestNumber {

public static void main(String[] args) {

//array of 10 numbers

int numbers[] = new int[]{32,43,53,54,32,65,63,98,43,23};

//assign first element of an array to largest and smallest

int smallest = numbers[0];

int largetst = numbers[0];

for(int i=1; i< numbers.length; i++)

{

if(numbers[i] > largetst)

largetst = numbers[i];

else if (numbers[i] < smallest)

smallest = numbers[i];

}

System.out.println("Largest Number is : " + largetst);

System.out.println("Smallest Number is : " + smallest);

}

Output of this program would be

Largest Number is : 98

Smallest Number is : 23

Explanation:

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Direct Mapped Cache. Memory is byte addressable. Fill in the missing fields based upon the properties of a direct-mapped cache.

xz_007 [3.2K]

The question given is incomplete and by finding it on internet i found the complete question as follows:

Direct Mapped Cache.

Memory is byte addressable

Fill in the missing fields based upon the properties of a direct-mapped cache. Click on "Select" to access the list of possible answers Main Memory Size Cache Size Block Size Number of Tag Bits 3 1) 16 KiB 128 KiB 256 B 20 2) 32 GiB 32 KiB 1 KiB 3) 64 MiB 512 KiB 1 KiB Select] 4 KiB 4) 16 GiB 10 Select ] Select ] 5) 10 64 MiB [ Select ] 6) Select] 512 KiB 7

For convenience, the table form of the question is attached in the image below.

Answers of blanks:

1. 3 bits

2. 20 bits

3. 64 MB

4. 16 MB

5. 64 KB

6. 64 MB

Explanation:

Following is the solution for question step-by-step:

Part 1:

No. of Tag bits = No. of bits to represent

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 128 KB = $2^7 * 2^{10} = 2^{17}$

Cache Memory = 16 KB = $2^4 * 2^{10}= 2^{14}$

Putting values in A:

Tag bits = 17 - 14 = 3 bits

Part 2:

Tag bits = Main memory - cache size bits -------- (A)

Given:

Main memory = 32 GB = $2^5 * 2^{30} = 2^{35}$

Cache Memory = 16 KB = $2^5 * 2^{10}= 2^{15}$

Putting values in A:

Tag bits = 35 - 15 = 20 bits

Part 3:

Given:

Tag bits = 7

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

So from equation A

7 = Main Memory size - 19

Main Memory = 7 + 19

Main memory = 26

Main Memory = $2^6 * 2^{20} = 64 MB$

Part 4:

Given that:

Main Memory Size = $2^4 * 2^{30} = 2^{34}$

Tag bits = 10

Cache Memory Bits = 34 - 10 = 24

Cache Memory Size = $2^4 * 2^{20} = 16 MB$

Part 5:

Given that:

Main Memory Size = 64 MB = $2^6 * 2^{20}$

Tag bits = 10

Cache Memory Bits = 26 - 10 = 16

Cache Memory Size = $2^{16} = 2^6 * 2^{10} = 64 KB$

Part 6:

Cache Memory = 512 KB = $2^9 * 2^{10} = 2^{19}$

Tag Bits = 7

Main Memory Bits = 19 + 7 = 26

Main Memory size = $2^{26} = 2^6 * 2^20 = 64 MB$

i hope it will help you!

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To improve readability,______ use text on a dark green background.

Olenka [21]

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You have been instructed to connect a computer to a group of hosts that have been segmented from the regular network. What kind

san4es73 [151]

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VLAN also has the ability to provide the segmentation in the network by dividing the physical network into segments and making logical network. Router is the source for maintaining the communication in the VLAN.

Thus according to the requirement mentioned in the question ,VLAN is the network device that should be used.

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Anastaziya [24]

Answer:

Explanation:

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