Determining Number of Bits in MAR

  • Thread starter SpiffWilkie
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In summary, for a memory system with 64K x 32 capacity, the MBR would require 32 bits and the MAR would require 16 bits for word-addressable memory, and 18 bits for byte-addressable memory (since each unit is 4 bytes). The unit size is irrelevant for word-addressable memory.
  • #1
SpiffWilkie
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I'm trying to determine how many bits are required in the memory buffer register and in the memory address register given certain memory systems.

For example, given the below system, how many bits are needed in MAR, and MBR if the memory is word addressable and how many bits if the memory is byte-addressable.

64K x 32


From what I understand the bits in the MBR are equal to the number of bits in the memory unit, or 32 for this example. I'm having a hard time grasping the MAR concept, though.
If I'm using word address memory, would it just be 16 bits (26 x 210 = 64K)? I guess, I'm asking if the unit size is irrelevant in that case.
If the memory is byte-addressable, would it be 18 bits? 216 x 22 (since each unit is 4 bytes?)

Thanks for any insight.
 
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  • #2
SpiffWilkie said:
I'm trying to determine how many bits are required in the memory buffer register and in the memory address register given certain memory systems.

For example, given the below system, how many bits are needed in MAR, and MBR if the memory is word addressable and how many bits if the memory is byte-addressable.

64K x 32


From what I understand the bits in the MBR are equal to the number of bits in the memory unit, or 32 for this example. I'm having a hard time grasping the MAR concept, though.
If I'm using word address memory, would it just be 16 bits (26 x 210 = 64K)? I guess, I'm asking if the unit size is irrelevant in that case.
If the memory is byte-addressable, would it be 18 bits? 216 x 22 (since each unit is 4 bytes?)

Thanks for any insight.

All of your numbers look correct to me.
 

1. How do you calculate the number of bits in MAR?

To calculate the number of bits in MAR (Memory Address Register), you need to know the total number of memory locations in the computer's memory. Then, you can use the following formula: log2(memory locations) = number of bits in MAR. For example, if the computer has 64K (64,000) memory locations, the number of bits in MAR would be log2(64,000) = 16.

2. Why is it important to determine the number of bits in MAR?

Knowing the number of bits in MAR is important because it determines the maximum amount of memory that can be addressed by the computer. If the number of bits in MAR is too small, the computer will not be able to access and use all of the available memory, which can limit its capabilities and performance.

3. Can the number of bits in MAR be different for different computers?

Yes, the number of bits in MAR can vary for different computers. It depends on the computer's architecture and the size of its memory. For example, a 32-bit computer can address up to 4GB (4 billion) of memory, while a 64-bit computer can address much larger amounts of memory.

4. What is the relationship between the number of bits in MAR and the computer's word size?

The number of bits in MAR is directly related to the computer's word size. The word size is the number of bits that the computer can process at one time. The size of MAR must be at least equal to the word size in order for the computer to be able to access and use all of the available memory.

5. How does the number of bits in MAR affect a computer's speed?

The number of bits in MAR does not directly affect a computer's speed. However, if the number of bits is too small and the computer cannot access all of the available memory, it may cause slower performance due to frequent data transfers between the memory and the processor.

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