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ehidna [41]
3 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]3 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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Host B on subnet Teachers transmits a packet to host D on subnet Students. Which Layer 2 and Layer 3 addresses are contained in
mr Goodwill [35]

Answer:

Layer 2 destination address = 00-00-0c-94-36-ab

Layer 2 source address = 00-00-0c-94-36-bb

Layer 3 destination address = 172.16.20.200

Layer 3 source address = 172.16.10.200

Explanation:

To properly understand the given answer, let us define some terms used in the narrative of the question:

Subnet: The sub network or as popularly called subnet  is a logical division of the internet protocol of a network into many different and smaller network. It is typically used to create partitions or subdivisions of large networks into bits.

PDU: The PDU can be said to be an abbreviation for the word power distribution unit or protocol data unit.  It can also be refereed to as an IT device that possess many output ports or terminals. They are mostly used to distribute or transmit power or data to other operating devices with the telecommunication system.

Now referring back to the question asked and the image attached below, we can see that host a and b send their signal through subnet teachers while host c and d send their signal through subnet students thus,

Layer 2 destination address = 00-00-0c-94-36-ab

Layer 2 source address = 00-00-0c-94-36-bb

Layer 3 destination address = 172.16.20.200

Layer 3 source address = 172.16.10.200 are the correct destination.

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3 years ago
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Answer:

The pseudocode is as follows

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