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dybincka [34]
3 years ago
15

What is the best summation of the main idea in Andrew Blum’s speech, "Discover the Physical Side of the Internet"? Question 10 o

ptions: a) Being online too much is bad for one’s health. b) More jobs are needed in the technology sector. c) We should be aware of the network and geographic presence behind our internet access rather than taking it for granted.
Computers and Technology
1 answer:
rewona [7]3 years ago
6 0

Answer:

c) We should be aware of the network and geographic presence behind our internet access rather than taking it for granted.

Explanation:

Andrew Blum’s speech is about the physical world and the size of the internet, he spent more time on the computer and cell phone than in the real world, and he had an issue with his internet, and the technician told him, squirrel has bittern your internet.

This can mean that the internet can be something complex and a huge thing, but can be damage for a squirrel, can be something simple.

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Explanation:

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For the following questions, consider a paged memory system that has a physical main memory size of 32KB (215) and a page frame
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Complete Question

For the following questions, consider a paged memory system that has a physical main memory size of 32KB (215) and a page frame size of 8KB (213). Consider a process P whose logical address space is 64KB (216). Important Note: If an answer requires an exponent, use the ^ character. For example: 216 would be entered as 2^16.

Question 1: How many physical page frames are there in the above paged memory system? ________(Your answer should be in exponential form)

How many bits are needed to represent a physical page frame number in the system? ____________

Question 2: How many logical pages are there for P? _____________(Your answer should be in exponential form. )

How many bits are needed to represent a logical page number for P? _____________

Answer:

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1b) The number of bits that can be used to represent the physical page frame = 5 bits

2a) Number of logical pages = 2³

2b) Number of logical frame bits = 3 bits

Explanation:

Size of physical main memory = 32 kB = 2¹⁵

Space for Logical address = 64 KB = 2¹⁶

Page frame size = 8 KB = 2¹³

1a) The number of physical page frames in the  paged memory system

Number of page frames = physical main memory size / page frame size

 Number of page frames  = 2¹⁵ / 2¹³

Number of page frames = 2⁵

1b)The number of bits that can be used to represent the physical page frame

physical memory size = log2(2²⁰) = 20 bits.

Offset bits = log2(2¹⁵)  = 15 bits

Frame bits = physical memory size – offset bits

Frame bits = 20 – 15

Frame bits = 5 bits

2a)  Number of logical pages = logical address space / page frame size

Number of logical pages = 2¹⁶/2¹³

Number of logical pages = 2³

2b) Logical address space = log2(2¹⁶) = 16 bits

Offset bits = log2(2¹³)

Offset bits = 13 bits

Frame bits = logical address space bits – offset bits

Frame bits   = 16 – 13

Frame bits  = 3 bits

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