hw 3-24

[Joseph Lee]

A computer with a 32-bit wide data bus uses 1M × 1 dynamic RAM memory chips. What is the smallest memory (in bytes) that this computer can have?

I tried to look for this in the textbook and found nothing about it. I forgot to ask it during the lecture today too and I'm worried it'll show up on the exam. Where do we learn this and the same to other problems that aren't mentioned in textbook and lecture?

[Frank (sjsu) Lin]

so, as a minimum, how many 1M x 1 do we need to build 32-bit wide data bus?

On Mon, Mar 4, 2013 at 3:00 PM, Joseph Lee <joseph...@gmail.com> wrote:

A computer with a 32-bit wide data bus uses 1M × 1 dynamic RAM memory chips. What is the smallest memory (in bytes) that this computer can have?

I tried to look for this in the textbook and found nothing about it. I forgot to ask it during the lecture today too and I'm worried it'll show up on the exam. Where do we learn this and the same to other problems that aren't mentioned in textbook and lecture?

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[Joseph Lee]

My first issue was what does the 1M mean? 

[Frank (sjsu) Lin]

1 mega.  in decimal terms, it is 1K x 1K (i.e. 1000 times 1000).  In binary logic, it means 1024 times 1024.

[Joseph Lee]

32?

[Joseph Lee]

Oh wait, is the answer 4 MB?

[Frank (sjsu) Lin]

Joseph,

My suggestion:  tell us why you think that way (i.e. how did you derive the answer)..  In many cases, you will be clear about the answer.  If there is doubt in your reasoning, air it in your own explanation.

for your consideration,
frank

[Joseph Lee]

Well there is a 32-bit bus that is the capacity which uses a 1M x 1 RAM. To maximize it's capacity, we need 32 of those chips since each chip is 1-bit. Now to find the minimum, we multiply using one chip, so 32 * 1M * 1 = 32. But the question asks for bytes so we have to divide by 8, which makes it 4 MB. Is that right?

[Frank (sjsu) Lin]

So, do you have any doubt about your reasoning?  If none, then it is most likely the right answer. 
In my experience, many things in real life do not have "correct" answers.  What I like to learn is the method of reasoning such that I have "no doubt" about the conclusion I derived.

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[Joseph Lee]

That is understandable, afterall, nothing in life is perfect and exact. Thank you!

[Enrique Padilla]

Well there is a 32-bit bus that is the capacity which uses a 1M x 1 RAM. To maximize it's capacity, we need 32 of those chips since each chip is 1-bit. Now to find the minimum, we multiply using one chip, so 32 * 1M * 1 = 32. But the question asks for bytes so we have to divide by 8, which makes it 4 MB. Is that right?

Joseph, 

What do you mean you need 32 of those chips? 

If you had 8-bit bus then it would 8*1*1(the thrid one is because theres only one chip) = 8/8 = 1MB 

[Enrique Padilla]

 A computer with 8-bit wide data bus uses 1M x 1 memory chips.  The smallest memory size that this computer can have is __________________ (how many) bytes.  In 10 words or less, explain why?

Pretty much that question... 

Hmmm im not really understanding the question,im just doing math which doesnt help :/ 

[joseph lee]

Ah the question I asked said it had a 32 bit bus. This is on the hw, not the midterm practice.

[Laimonas Turauskas]

I do not understand this question as well.

Since, the communication is 32 bit wide and memory has 1Mx1, isn't smallest memory size is 1024*1024 bits. Because, 1Mx1 is 1024^2 words of 1 bit each. You can have 1024 * 1024 / 32 spaces available for such data. 

Then, I'm not sure what is meant by bytes because commonly 1 byte represents 8 bits, but it isn't always the case. So are we going by this standard or for example is 1 byte = 1 bit, since we could say 1 word = 1 byte and 1 word = 1 bit. 

[Frank (sjsu) Lin]

1 byte is 8 bits.

Laim, what is the meaning of 1Mx1 memory chip?  How many data pins are there for a 32 bit wide bus?

[Lance Barrett]

Ok... from page 179 of the text, "To build a memory with a 32-bit word from 4096K x 1 chips requires 32 chips in parrallel."
So my understanding is that we can think of the data bus "width" as equivalent to the required width of the memory chip array. Since we're dealing with a 1-bit wide memory chip, we would need 32 parrallel chips in order to handle the bus. (Thus the "inconvenience" of 1-bit wide memory chips).

So we need 32 chips - which is 32 * 1M total storage. It asks for bytes... Memory is usually sized in bits, so I assume the 1M is 1 Megabit.
That is 32 * 2^20. In Megabytes, it would be (32/8) * 2^20 = 4 MB.

So I think the minimum possible memory is 4 MB - which agrees with Joseph's conclusion and Frank's dodgey pseudo-confirmation. =P

[Laimonas Turauskas]

1Mx1 memory chip is 1Mbit chip. I think 32 bit wide bus has 32 data pins. Each pin for 1 bit. So this means we need 32, 1Mx1 chips. 

32 * 2 ^ 20 / 8 = 4 MB

On Tuesday, March 5, 2013 9:51:41 PM UTC-8, Frank (sjsu) Lin wrote:

1 byte is 8 bits.

Laim, what is the meaning of 1Mx1 memory chip?  How many data pins are there for a 32 bit wide bus?

On Tuesday, March 5, 2013, Laimonas Turauskas wrote:

I do not understand this question as well.

Since, the communication is 32 bit wide and memory has 1Mx1, isn't smallest memory size is 1024*1024 bits. Because, 1Mx1 is 1024^2 words of 1 bit each. You can have 1024 * 1024 / 32 spaces available for such data. 

Then, I'm not sure what is meant by bytes because commonly 1 byte represents 8 bits, but it isn't always the case. So are we going by this standard or for example is 1 byte = 1 bit, since we could say 1 word = 1 byte and 1 word = 1 bit. 

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