Suppose that I have a computer with an address register of size 16 bits (MAR, for example). The smallest addressable unit in this computer is a word and each word is of size 2 bytes. What is the maximum memory size (in bytes) this system can support?
I thought it would be 2^16 = 65536 bytes, but the part about the smallest addressable unit implies that this is not the way to solve it.
Thanks in advance
There is no direct correlation to the maximum amount of memory a system can support, and the size of address registers.
16bit computers 30 years ago could very well support more than 64 kilobytes. On the other hand, modern 64bit processors typilcally only have lanes for 52 bits (or less), but even so a typical computer cannot nearly support 2^52 bytes of memory.
Typical 64bit computers today could in theory address 16 exibytes, but present-time CPUs only support 4 petabytes of phyisical and 256 terabytes of per-process virtual memory. Typical desktop mainboards support 128GiB maximum, if you buy extra expensive DIMMS. With affordable DIMMS, you're limited to about half as much (there are only so and so many slots).
Operating systems typically allow for main memory sizes in the hundreds of gigabytes only (e.g. 512 GiB for Windows 8 enterprise/professional, and 128GiB otherwise, or as little as 16GiB for Windows 7 Home Premium)
Generally the smallest addressable size is one byte, as you have calculated it, if it were one byte it would be 2^16*1 = 65536 bytes. However, because on this system there are two bytes per address, it is actually 2^16*2 = 131072 bytes.
Related
My understanding of word size and memory addresses is as follows. An 8bit machine will have an address bus of size 8bit and have 256 memory addresses. A memory address is the location of each address so this machine can make use of 256 bytes of RAM. Now a 32 bit machine has an 32 bit word size ie 4bytes. At this point I get confused. In terms of usable memory, online tells me 4gb but if each memory address is 4bytes in size then surely it only has 1 million total memory addresses available ie 1gb of ram can be used? What am I mixing up here?
This question already has answers here:
Size of a Word and addressing
(2 answers)
How does 32-bit address 4GB if 2³² bits = 4 Billion bits not Bytes?
(4 answers)
Closed 2 years ago.
I have a question regarding (RAM) memory units:
for a cpu architecture x32,
we will have 32bit size cpu registers, as well as data bus to ram of 32 wire and address bus of 32 wire.
so the maximum memory address unit is 2^32 = 4,294,967,296
on other words we have 4,294,967,296 memory units, and for each memory unit, data bus (32) size should be writable so for each memory unit its size should be 32bit to handle the data bus
if i concluded right which i doubt, (Ram) total memory size should be = no of memory units * size each =
4,294,967,296 * 32 = 137,438,953,472 bit. which is not true
after research, i found out, rams unit memory are standardized to be 8 bit per each memory unit,
so if this is the case, how come single memory unit (8bit) can store (32bit data bus) ?
Indeed what is true, is that if you won't have 32 bit data, what this tells you is that the memory can hold 2^32 addresses for memory. However for each memory what you have is a byte, or eight bits, then you will have about 4 GiB = 4*2^30 Bytes = 2^32 bits.
This may be a duplicate and I apologies if that is so but I really want a definitive answer as that seems to change depending upon where I look.
Is it acceptable to say that a gigabyte is 1024 megabytes or should it be said that it is 1000 megabytes? I am taking computer science at GCSE and a typical exam question could be how many bytes in a kilobyte and I believe the exam board, AQA, has the answer for such a question as 1024 not 1000. How is this? Are both correct? Which one should I go with?
Thanks in advance- this has got me rather bamboozled!
The sad fact is that it depends on who you ask. But computer terminology is slowly being aligned with normal terminology, in which kilo is 103 (1,000), mega is 106 (1,000,000), and giga is 109 (1,000,000,000).
This is reflected in the International System of Quantities and the International Electrotechnical Commission, which define gigabyte as 109 and use gibibyte for the computer-specific 1024 x 1024 x 1024 value.
The reason it "depends who you ask," is that for many years, specifically in relation to "bytes" of storage, the prefixes kilo, mega, and giga meant 1024, 10242, and 10243. But that flies in the face of normal convention with regard to these prefixes. So again, computer terminology is being aligned with non-computer terminology.
The term gigabyte is commonly used to mean either 10003 bytes or 10243 bytes depending on the context. Disk manufacturers prefer the decimal term while memory manufacturers use the binary.
Decimal definition
1 GB = 1,000,000,000 bytes (= 10003 B = 109 B)
Based on powers of 10, this definition uses the prefix as defined in the International System of Units (SI). This is the recommended definition by the International Electrotechnical Commission (IEC). This definition is used in networking contexts and most storage media, particularly hard drives, flash-based storage, and DVDs, and is also consistent with the other uses of the SI prefix in computing, such as CPU clock speeds or measures of performance.
Binary definition
1 GiB = 1,073,741,824 bytes (= 10243 B = 230 B).
The binary definition uses powers of the base 2, as is the architectural principle of binary computers. This usage is widely promulgated by some operating systems, such as Microsoft Windows in reference to computer memory (e.g., RAM). This definition is synonymous with the unambiguous unit gibibyte.
The difference between units based on decimal and binary prefixes increases as a semi-logarithmic (linear-log) function—for example, the decimal kilobyte value is nearly 98% of the kibibyte, a megabyte is under 96% of a mebibyte, and a gigabyte is just over 93% of a gibibyte value. This means that a 300 GB (279 GiB) hard disk might be indicated variously as 300 GB, 279 GB or 279 GiB, depending on the operating system.
The Wikipedia article https://en.wikipedia.org/wiki/Gigabyte has a good writeup of the confusion surrounding the usage of the term
Why is the amount of RAM always a power of 2?
512, 1024, etc.
Specifically, what is the difference between using 512, 768, and 1024 RAM for an Android emulator?
Memory is closely tied to the CPU, so making their size a power of two
means that multiple modules can be packed requiring a minimum of logic
in order to switch between them; only a few bits from the end need to
be checked (since the binary representation of the size is 1000...0000
regardless of its size) instead of many more bits were it not a power
of two.
Hard drives are not tied to the CPU and not packed in the same manner,
so exactness of their size is not required.
from https://superuser.com/questions/235030/why-are-ram-size-usually-in-powers-of-2-512-mb-1-2-4-8-gb
as referenced by BrajeshKumar in the comments on the OP. Thanks Brajesh!
Because computers deal with binary values such as 0 and 1, because registers are on(1) or off(0)
So if you use powers of 2, your hardware will use 100% of the registers.
If computers used ternary values in their circuits, then we'd have memory, processors and anything else in powers of 3.
I think, it is related with the number of bits in an address bus (or bits used to select between address spaces). n bits can address 2^n bytes, so whenever the number of address bits increases to n+1, automatically the space increases by a factor of 2. The manufacturers use their maximum address capacity when including memory chips to the design.
In Android emulator, the increase in RAM may make your program more efficient, because when your application exceeds the RAM, a part of ROM (non-volatile memory) and it is slower.
Does anyone know what aspect ratio mean in memories? and how it's different in block ram and distributed ram in Xilinx FPGA?
Thanks
"Aspect ratio" refers to the number of address bits and data bits when accessing memory.
For example, say you have memory that holds 65,536 bits. If you use 8 data bits per address, you have 8,192 addresses (65,536 / 8 = 8,192), which means 13 address bits (2^13 = 8,192). So one aspect ratio for accessing 65,536 bits is 8 data bits and 13 address bits.
But say you want 16 data bits per address. Then you can only have 4,096 addresses since 4,096 * 16 is 65,536, and that's all the memory you have. In that case, you can only have 12 address bits. So another aspect ratio for accessing 65,536 bits of memory is 16 data bits and 12 address bits.
Aspect Ratio is another term that is used for the data organization of a RAM block.
In Xilinx FPGAs, a Block RAM is a dedicated two-port memory containing several kilobits of RAM. The FPGA contains several (or many) of these blocks.
Inside of each small logic block is a configurable lookup table. It is normally used for logic functions, but you can reconfigure it as a few bits of RAM. You can combine several (or many) of them into a larger RAM. This is distributed RAM.
Both types of RAM can be initialized with data, or used as ROM.
More Information can be found here:
http://www.xilinx.com/support/index.htm#nav=sd-nav-link-182711&tab=tab-sd