(14 points) Suppose that a 2M x 16 main memory is built using 256K × 8 RAM chips
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Question
(14 points) Suppose that a 2M x 16 main memory is built using 256K × 8 RAM chips and memory is word-addressable.
How many RAM chips are necessary?
If we were accessing one full word, how many chips would be involved?
How many address bits are needed for each RAM chip?
How many banks will this memory have?
How many address bits are needed for all of memory?
If high-order interleaving is used, where would address 14 (which is E in hex) be located?
s. 14 points) Suppose that a 2M x 16 main memory is built using 256K x 8 RAM chips and memory is word-addressable a. How many RAM chips are necessary? b. If we were accessing one full word, how many chips would be involved? c. How many address bits are needed for each RAM chip? d. How many banks will this memory have!? e. How many address bits are needed for all of memory? f Ifhigh-order interleaving is used, where would address 14 (which is E in hex) be located? g. Repeat Exercise f) for low-order interleaving.Explanation / Answer
How many RAM chips are necessary?
Answer:
a) Amount of memory = 2M = 2 x 220words
The RAM chips are 8 bits wide and contain 256K of memory. The number needed to make upthe 2M x 16 memory would be:
Number of chips = (2M x 2 bytes/word) / (256K)
= (2 x 2^20x 2) / (28x 2^10)
= 2^22/ 2^18
= 2^4
= 16 RAM chips
b)If we were addressing one full word, how many chips would be involved?
Answer:
Divide the number of bits at each address in the composite memory by the number of bits at each address in the component memory chips.
n=16/8
=2
2 RAM chips are there per memory word
c) How many address lines are needed for each RAM chip?
Answer:
Find the log2 M, where M is the number of address cells in the component RAM chips.
A=log2 256K=log2 (2^8*2^10)
=log2 2^18
=18
256K = 2^18, so 18 bits
d) How many banks will this memory have?
Answer:
The number of banks is equal to the number of component chips divided by the number of chips required to access a full word at a single address
K=16/2=8
8 banks this memory have.
e) How many address lines are needed for the composite memory?
Answer:
Find the log2 M, where M is the number of address cells in the composite memory.
A=log2 2M
=log2(2*2^20)
=21 bits
f) If high-order interleaving is used, where would address 14 (which is E in hex) be located?
Answer:
Address 14 = 00000E
Bank select is 000
g) Repeat exercise 10f for low-order interleaving.
Answer:
Address 14 = 00000E
Bank select is 110=6
Bank 6 (110) if counting from 0, Bank 7 if counting from 1
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