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mixas84 [53]
2 years ago
15

Find the error in the following pseudocode.

Computers and Technology
1 answer:
Ber [7]2 years ago
5 0

Answer:

Following are the error in the given program are explain in the explanation part .

Explanation:

The main() function call the function getMileage() function after that the control moves to the getMileage() definition of function. In this firstly declared the  "mileage" variable after that it taking the input and module is finished we see that the "mileage" variable is local variable i,e it is declared inside the  " Module getMileage() "  and we prints in the main module that are not possible to correct these we used two technique.

Technique 1: Declared the mileage variable as the global variable that is accessible anywhere in the program.

Technique 2: We print the value of  "mileage" inside the Module getMileage() in the program .

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MAR bit

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The MPC computers area set of software and hardware standards that was developed by the consortium of computer firms which is led by Microsoft. It contains the address of a next microcode for the Mic1 emulator foe execution.

The part of the executing microcode instruction that determines the value which is placed in the MOC is the MAR bit. The MAR is the memory address register in the CPU which stores the memory address or such addresses to which some data will be sent and also stored.

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Describe the ways in which a computer-aided design (CAD) drawing makes the details of an image easier to understand.
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Answer: CAD is mainly used for detailed engineering of 3D models or 2D drawings of physical components, but it is also used throughout the engineering process from conceptual design and layout of products, through strength and dynamic analysis of assemblies to definition of manufacturing methods of components.

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Given six memory partitions of 300 KB, 600 KB, 350 KB, 200 KB, 750 KB, and 125 KB (in order), how would the first-fit, best-fit,
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Answer:

In terms of efficient use of memory: Best-fit is the best (it still have a free memory space of 777KB and all process is completely assigned) followed by First-fit (which have free space of 777KB but available in smaller partition) and then worst-fit (which have free space of 1152KB but a process cannot be assigned). See the detail in the explanation section.

Explanation:

We have six free memory partition: 300KB (F1), 600KB (F2), 350KB (F3), 200KB (F4), 750KB (F5) and 125KB (F6) (in order).

Using First-fit

First-fit means you assign the first available memory that can fit a process to it.

  • 115KB will fit into the first partition. So, F1 will have a remaining free space of 185KB (300 - 115).
  • 500KB will fit into the second partition. So, F2 will have a remaining free space of  100KB (600 - 500)
  • 358KB will fit into the fifth partition. So, F5 will have a remaining free space of 392KB (750 - 358)
  • 200KB will fit into the third partition. So, F3 will have a remaining free space of 150KB (350 -200)
  • 375KB will fit into the remaining partition of F5. So, F5 will a remaining free space of 17KB (392 - 375)

Using Best-fit

Best-fit means you assign the best memory available that can fit a process to the process.

  • 115KB will best fit into the last partition (F6). So, F6 will now have a free remaining space of 10KB (125 - 115)
  • 500KB will best fit into second partition. So, F2 will now have a free remaining space of 100KB (600 - 500)
  • 358KB will best fit into the fifth partition. So, F5 will now have a free remaining space of 392KB (750 - 358)
  • 200KB will best fit into the fourth partition and it will occupy the entire space with no remaining space (200 - 200 = 0)
  • 375KB will best fit into the remaining space of the fifth partition. So, F5 will now have a free space of 17KB (392 - 375)

Using Worst-fit

Worst-fit means that you assign the largest available memory space to a process.

  • 115KB will be fitted into the fifth partition. So, F5 will now have a free remaining space of 635KB (750 - 115)
  • 500KB will be fitted also into the remaining space of the fifth partition. So, F5 will now have a free remaining space of 135KB (635 - 500)
  • 358KB will be fitted into the second partition. So, F2 will now have a free remaining space of 242KB (600 - 358)
  • 200KB will be fitted into the third partition. So, F3 will now have a free remaining space of 150KB (350 - 200)
  • 375KB will not be assigned to any available memory space because none of the available space can contain the 375KB process.
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