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timama [110]
3 years ago
10

A computer’s memory is composed of 8K words of 32 bits each. How many bits are required for memory addressing if the smallest ad

dressable memory unit is a word?
13

8

10

6

32
Computers and Technology
1 answer:
victus00 [196]3 years ago
7 0

Answer:

The minimum number of bits necessary to address 8K words is 13.

Explanation:

You have the number of words to address that is 8000 words, a word is the smallest addressable memory unit.  

8000 words can be addressed with 2^{n}  units. Now you have to find the value of n that approximates to the number of words.  

2^2=4\\2^4=16\\2^7=128\\2^{10}=1024\\2^{12}=4096\\2^{13}=8192

So you can see that 13 bits are needed to address 8K words.

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3 years ago
Consider an unpipelined or single-stage processor design like the one discussed in slide 6 of lecture 17. At the start of a cycl
pantera1 [17]

Answer:

a. Clock Speed of Processor = 0.5 GHz

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c. Throughput = 1 billion Instruction per Second

Explanation

Given Data

Time Take to complete the single Cycle = 2000ps = 2000 x 10⁻¹²

To Find

a. Clock Speed of Processor = ?

b. Cycles per Instruction (CPI) = ?

c. Throughput = ?

Solution:

a. Clock Speed of Processor = ?

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                      =  0.0005 x 10¹²  Hz

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                      = 0.5 GHz

b. Cycles per Instruction (CPI) = ?

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for above mentioned processor.

c. Throughput = ?

Throughput = CPI x Clock Speed

where

CPI = 1 cycle per instruction

Clock Speed = 1 billion Instructions per Second

as

Clock Speed = 1 billion Cycles per Second

Throughput = 1 cycle per instruction x 1 billion Cycles per Second

Throughput = 1 billion Instruction per Second

                         

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Answer:

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