Is it correct to assume that the number of addresses for 32k of ram is 2^15 addresses? Then for the peripherals is it 2^6 addresses?
closed as unclear what you're asking by Chris Stratton, PeterJ, Nick Alexeev♦ Feb 15 '16 at 6:39
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As posed this question is not answerable. It reeks of an oversimplified homework question that can only be answered in the context of the course it goes with and the assumpetions that course has taught.
A computer system has RAM of 32K bytes
The word "byte" has two different definitions. The original defintion was "the smallest addressable unit of memory" but sometimes it is uses as a synonym of "octet". If the latter meaning is used then the number of addresses needed will depend on the size of the smallest addressable unit of data (which is not nessacerally the same as the data bus width).
If we assume that "byte" does reffer to the smallest addressable unit of memory (either because that is the definition the asker is using or because your system has 8 bits as it's smallest addresable unit) then yes you will need 215 addresses to address your ram. Since most computers nowadays do indeed use 8 bits as their smallest addressable unit this answer is probablly right.
and 64 peripherals.
This is where it gets really messy. If we assume the perhiperals use one address each then your answer would be correct but most real perhiperals use more than that. How many more varies wildly.
What is the number of distinct addresses required?
In addition to the above there was no mention of rom in your question, a real computer system is almost certain to have some rom.
And finally note that it would be very weird for a computer system to have a non power of two address space size. It's not entirely clear what is meant by "required" in your question but it would not surprise me if your course expects you to round up to the next power of 2