You need to show detail work on how you get the answer. Dont use answers from other websites please! Please type your answer not picture 3. Suppose that a 32M x 16 main memory is built using 512K × 8 RAM chips and memory is word-addressable. a) How many RAM chips are necessary? Number of RAM Chips = Memory Size/ Chip Size =225x 16/219x 8 =27 128 RAM chips Needed b) If we were accessing one full word, how many chips would be involved? As given RAM chips holds 8 bits ( 512K × 8)for a word. Main memory holds 16 bits (32M * 16 ). So that design needs 16 bit word which means two byes. And chip hold only 1 byte. So that 2 chips would be involved to access one word c) How many address bits are needed for each RAM chip? For 512K × 8 chip, bits required are 512K =>512 x 1024 = 29x 210 = 219 219 bits required. And 19 address bits are required for each chip.
You need to show detail work on how you get the answer.
Dont use answers from other websites please!
Please type your answer not picture
3. Suppose that a 32M x 16 main memory is built using 512K × 8 RAM chips and memory is word-addressable.
a) How many RAM chips are necessary?
Number of RAM Chips = Memory Size/ Chip Size
=225x 16/219x 8
=27
128 RAM chips Needed
b) If we were accessing one full word, how many chips would be involved?
- As given RAM chips holds 8 bits ( 512K × 8)for a word. Main memory holds 16 bits (32M * 16 ).
- So that design needs 16 bit word which means two byes. And chip hold only 1 byte.
- So that 2 chips would be involved to access one word
c) How many address bits are needed for each RAM chip?
For 512K × 8 chip, bits required are 512K
=>512 x 1024 = 29x 210 = 219
219 bits required.
And 19 address bits are required for each chip.
d) How many banks will this memory have?
The number of banks = main memory addressable / chip addressable items
So that, no. of Banks = 225 / 2 19
=26 = 64 Banks
Therefore, the number of banks equals 64
e) How many address bits are needed for all of the memory?
The number of bits required for memory is 32 M = 25x 210x 210
=225
Therefore, 25 bits are needed to address the main memory.
f) If high-order interleaving is used, where would address 32(base 10) be located? (Your answer should be "Bank#, Offset#")
g) Repeat (f) for low-order interleaving.
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Draw a diagram of the 128M x 16 chip built from 32M x 8 chips. Include the range of
addresses for each chip.
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