1. How many bits would you need to address a 32G X 32 memory...

1. How many bits would you need to address a 32G X 32 memory if \begin{itemize} \item The memory is word-addressable(nothing else is given, what will be the default value here) \item The memory is word-addressable (word size = 16)? \item The memory is byte-addressable? \end{itemize} 2. Convert the following numbers. \begin{enumerate} \item $(3AF.8C)_{16} = (\quad)_2$ \item $(1110010001)_2 = (\quad)_{16}$ \item $(100111001.11011)_2 = (\quad)_{16}$ \end{enumerate} 3. Suppose a 512M x 32 word-addressable memory built using 128K x 8 RAM chips \begin{enumerate} \item How many RAM chips are required to build an entire memory? \item How many address bits are needed for a single RAM chip addressing? \item How many banks will this memory have? \item How many address bits are needed for the entire memory? \end{enumerate}

Transcript

00:01 Hello students in this question to determine the first part to determine the memory capacity in megabytes that can be accessed using 10 bits memory for addressing.

00:15 We need to calculate the total number of memory locations that can be addressed.

00:19 So the memory location can be an 8 bit which is 1 byte.

00:24 So we can express the total number of terms which is 2 times 10 2 raised to 10.

00:30 So that is equal to 1024 since each memory location has a capacity of 1 by the total memory capacity can be calculated as times this divided by so 10 1024 divided by 2 raised to 10 kilobytes.

00:49 So that is equal to 1 k kilobyte.

00:54 Okay.

00:55 So now converting kb into megabyte.

00:57 What you will get is 1 by 1 24.

01:01 So you will get 0 .0009766 megabytes.

01:13 So that is therefore that using 10 bits of addressing allows us to access approximately 0 .0009766 megabytes of memory.

01:24 Now, if you have 264 megabytes of available memory and we need to determine the number of bits required for addressing that memory.

01:36 We can start let's convert megabytes into bytes.

01:42 So that will be equal to me.

01:44 This will be equal to 264 times 2 raised to 20 bytes that is equal to 2827 05942 bytes.

01:57 Now we need to find out the minimum number of bytes required to represent this number that minimum number of bits required to represent this number in bytes.

02:09 So we can calculate the logarithm of this function to the base of 2 so that will give you the minimum number.

02:18 So that is log to the base of 2 2 8 2 7 0 5 9 4 2 so that is approximately equal to 28 .15.

02:28 So we cannot have fraction so we can round up to the up next big integers of that is 29.

02:36 So we need at least we need a minimum of 29 bits to address a 269 megabits of memory.

02:49 Okay.

02:50 Now the c part what is a ram so ram stands for random access memory.

02:58 So it is a type of computer memory that allows you to read and write operation.

03:10 So it is a volatile memory.

03:12 So it means that its contents are lost when the power is turned off.

03:16 It's volatile...

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