Determining Number of Bits in MAR

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In a memory system of 64K x 32, the Memory Buffer Register (MBR) requires 32 bits, as it corresponds to the size of the memory unit. For a word-addressable memory, the Memory Address Register (MAR) needs 16 bits, calculated from the total addressable words (64K). If the memory is byte-addressable, the MAR would require 18 bits, accounting for the 4 bytes per word. The calculations confirm that the unit size is relevant in determining the MAR for byte-addressable memory. Overall, understanding the relationship between memory size and addressing is crucial for accurate bit determination in MAR and MBR.
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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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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.
 

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