Binary Bits — Largest and Smallest Values Possible

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This here is a list of the largest and smallest possible values in each n-bit type of numbers which are possible to be represented as 2s-complements :

2-bits (-2 : 1)

3-bits (-4 : 3)

4-bits (-8 : 7)

5-bits (-16 : 15)

6-bits (-32 : 31)

7-bits (-64, 63)

8-bits (-128 : 127)

9-bits (-256 : 255)

10-bits (-512 : 511)

11-bits (-1024 : 1023)

12-bits (-2048 : 2047)

16-bits (-32,768 : 32,767)

32-bits (-2,147,483,648 : 2,147,483,647)

64-bits (-9,223,372,036,854,775,808 : 9,223,372,036,854,775,807)

70-bits (-590,295,810,358,705,651,712 : 590,295,810,358,705,651,711)

2-bits: smallest 2 bit integer is -2, largest 2 bit integer is 1

3-bits: smallest 3 bit integer is -4, largest 3 bit integer is 3

4-bits: smallest 4 bit integer is -8, largest 4 bit integer is 7

5-bits: smallest 5 bit integer is -16, largest 5 bit integer is 15

6-bits: smallest 6 bit integer is -32, largest 6 bit integer is 31

7-bits: smallest 7 bit integer is -64, largest 7 bit integer is 63

8-bits: smallest 8 bit integer is -128, largest 8 bit integer is 127

9-bits: smallest 9 bit integer is -256, largest 9 bit integer is 255

10-bits: smallest 10 bit integer is -512, largest 10 bit integer is 511

11-bits: smallest 11 bit integer is -1024, largest 11 bit integer is 1023

12-bits: smallest 12 bit integer is -2048, largest 12 bit integer is 2047

16-bits: smallest 16 bit integer is -32,768, largest 16 bit integer is 32,767

32-bits: smallest 32 bit integer is -2,147,483,648, largest 32 bit integer is 2,147,483,647

64-bits: smallest 64 bit integer is -9,223,372,036,854,775,808, largest 64 bit integer is 9,223,372,036,854,775,807

70-bits: smallest 70 bit integer is -590,295,810,358,705,651,712, largest 70 bit integer is 590,295,810,358,705,651,711

Thus, it is evident that for n-bits, the smallest possible value will be -(2^n) and the largest possible value would be ((2^n)-1).

Currently, our modern computers have operating systems running up to 64-bits. Exceeding 70-bits is going to cause a problem known as a ‘floating point precision error’ or ‘floating point bug’. You can search that term up for better and detailed explanations, if you are curious.

When referring to a memory location, if the size of a ‘word’ is 8 bits (a byte), then it is conventionally called “ byte addressable ”. It’s counterpart — word addressable — exists when the size of a word is greater than 8 bits (i.e. 16 bits, 32 bits, 64 bits), the location is then “ word addressable ”.

That’s easy (kind of). For illustrative purposes, let’s take a 1 GiB (gibibyte) module of RAM, that is composed of several smaller 64 MiB (mebibyte) pieces.

Firstly, we need to convert the memory to bytes for convenience. So, 1 GiB = 1073741824 bytes, 64 MiB = 67108864 bytes. Done! (Conversion processes will be explained later)

Now you take the logarithm of the main memory module’s (the total memory value) value — in bytes — to base 2. You would get :

log(2, 1073741824) = 30

That means the 1 GiB memory module will be using 30-bit memory registers.

Now, since the whole 1 GiB memory module is actually made up of 64 MiB pieces — just like smaller pieces of chocolate are manufactured joined up together to form a large chocolate bar — we need to find exactly “ how many” of those 64 MiB tiny modules/RAM chips we need to add up to the total of 1 GiB.

So we calculate it like this (simple unitary method) :

1073741824/67108864 = 16

or if the large and overly-detailed numbers is uncomfortable for you, you can rely on good old powers of 2 for all your calculations :

2³⁰/2²⁶ = 16

Phew. Feeling tired already? Hang on in there!

We know 2⁴ = 16, so the first 4 bits of the whole 30-bit memory register needs to be used for selecting which ‘chip’ among the sixteen chips used to build up the 1 GiB module. And the remaining 26 bits is used for the memory addressing.

Done.

Tough question. (Just joking)

Just remember this for now :

Regardless of how many bits the memory has, the lowest address for it will always be zero. Dewan M.I. Mukto

Regardless of how many bits the memory has, the lowest address for it will always be zero.

That answers 50% of your question.

As for calculating the highest possible memory address for a defined byte range of memory, you can rely on this trend :

Consider there are 2ᵡ (2^x) bytes in memory.

So, if the word size is 8 bits, then the memory is byte addressable (as explained above) . The highest memory address would be = 2^x-1

If the word size is 16 bits, then the highest memory address would be = 2^(x-1)-1, because 16 bits is a 2-byte (2¹) word. And you’d be reducing the total number of available addresses by that amount. Thus, 1 is subtracted from the total x-bits.

If the word size is 32 bites, then the highest memory address would be = 2^(x-2)-1. The explanation is same as above. 32 bits is a 4-byte (2²) word. So the amount is subtracted from x-bits.

Once more, if the word size is 64 bits, the highest possible address is 2^(x-3)-1, as 64 is a 8-byte (2³) word.

This trend continues for as many increments of powers of 2 you need.

Voila! You’ve now understood the concept, I presume.

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