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ehidna [41]
4 years ago
9

Suppose you have a 16-bit machine with a page size of 16B. Assume that any unsigned 16-bit integer can be a memory address. Also

, assume that the machine can support up to only 32KB of physical memory. (1K = 210)
Required:
For paging-based memory allocation, how many page table entries do you need to map the virtual address space of a process, with pure paging alone?
Computers and Technology
1 answer:
irinina [24]4 years ago
7 0

Answer:

2^11

Explanation:

Physical Memory Size = 32 KB = 32 x 2^10 B

Virtual Address space = 216 B

Page size is always equal to frame size.

Page size = 16 B. Therefore, Frame size = 16 B

If there is a restriction, the number of bits is calculated like this:  

number of page entries = 2^[log2(physical memory size) - log2(n bit machine)]

where

physical memory size = 32KB  which is the restriction

n bit machine = frame size = 16

Hence, we have page entries = 2^[log2(32*2^10) - log2(16)] = 2ˆ[15 - 4 ] = 2ˆ11

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Students who finish their homework after school are meeting a. intrapersonal and short-term goals b. normative and short-term go
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Answer: a. intrapersonal and short-term goals

Explanation:

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6 0
3 years ago
Read 2 more answers
Find the double word-length 2's complement representation of each of the following decimal numbers:a. 3874
miss Akunina [59]

Answer:

-3874₁₀ = 1111 1111 1111 1111 1111 1111 1101 1110₂

Explanation:

2's complement is a way for us to represent negative numbers in binary.

To get 2's complement:

1. Invert all the bits

2. Add 1 to the inverted bits

Summary: 2's complement = -N = ~N + 1

1. Inverting the number

3874₁₀ = 1111 0010 0010₂

~3874₁₀ = 0000 1101 1101₂

2. Add 1 to your inverted bits

~3874₁₀ + 1 = 0000 1101 1101₂ + 1

= 0000 1101 1110₂

You can pad the most signigicant bits with 1's if you're planning on using more bits.

so,

12 bits                          16 bits

0000 1101 1110₂  = 1111 0000 1101 1110₂

They asked for double word-length (a fancy term for 32-bits), so pad the left-most side with 1s' until you get a total of 32 bits.

           32 bits

= 1111 1111 1111 1111 1111 1111 1101 1110

7 0
3 years ago
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