For Boo closed Architecture a 4G times 16 Bit RAM chip What is the size of each

Question

For Boo closed Architecture a 4G times 16 Bit RAM chip What is the size of each memory addresses? How many memory addresses are there? Assuming the memory addressing starts from 0, what is the highest memory address? If the FM broadcast band ranges from approximately 87.9 MHz to 107.9 MHz, what is the clock Cycle Time at each end of this range? Suppose a computer using fully-associative cache has 222 words of main memory and a cache of 32 blocks, where each cache block contains 64 words How many blocks of main memory are there? Assuming memory addressing starts from zero, what is the highest memory address? How many bits are needed to represent all of the memory addresses? What is the format of a memory address as seen by the cache, that is, what are the sizes of the tag and word fields? To which cache block will the memory reference E872_16 (assume as many leading zeros as needed per your answer to part c)?

Explanation / Answer

Here given, 4G*16bit it means that you have :

Memory address of size 16 bit each and

4G locations ,4G=4*1024 M=4096M=4096*1024 B

There are 4G of memory addresses.

Since each memory address requires 2 bytes=16 bit as given.

In 1 MB of memory you can address 512 memory locations

So we can calculate last or highest address by,

512*4096=2097152 .

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