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luda_lava [24]
3 years ago
15

You have been hired by an educational software company to create a program that automatically calculates the sum of each place-v

alue of two numbers. You can assume that the user of this program will only enter positive whole numbers greater than zero and less than 10,000.

Computers and Technology
1 answer:
Vinvika [58]3 years ago
4 0

Hi, you haven't provided the programing language, therefore, we will use python but you can extend it to any programing language by reading the code and the explanation.

Answer:

n1 = int(input("First numeber: "))

n2 = int(input("Second numeber: "))

for i in range(5):

   r1 = n1%10

   r2 = n2%10

   print(r1+r2)

   n1 = n1//10

   n2 = n2//10

 

Explanation:

  1. First, we ask for the user input n1 and n2
  2. We create a for-loop to calculate the sum of each place-value of two numbers
  3. We obtain the last number in n1 by using the mod operator (%) an the number ten this way we can always take the last value, we make the same for n2
  4. Then we print the result of adding the last two numbers (place value)
  5. Finally, we get rid of the last value and overwrite n1 and n2 to continue with the process

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Answer:

False

Explanation:

It is rather the other way round, the intuitive and analytical approach are more internally focused while the benchmark approach is more externally focused.

Hence the statement from the question is false.

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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.
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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:

<u>Part 1:</u>

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

<u>Part 2:</u>

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

<u>Part 3:</u>

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

OR

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

<u>Part 4:</u>

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

<u>Part 5:</u>

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

<u>Part 6:</u>

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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