If there are 9,000 address lines, it implies that the system can address (2^{9000}) different memory locations. However, this is an impractically large number since the addressable space would be astronomically high. Instead, if you meant 9 kilobytes (kB), then the memory size would be 9,000 bytes, which is equivalent to 9 kB. For a more precise answer, clarifying the context of "9k" would be helpful.
To determine the number of address lines required for 1 GB of memory, we can use the formula (2^n = \text{Memory Size}), where (n) is the number of address lines. Since 1 GB equals (2^{30}) bytes, we need (30) address lines to uniquely address each byte in 1 GB of memory. Therefore, (30) address lines are required for 1 GB.
Let N be the number of addresses line 2 megabyte = 2*1024 =2048 N = log (size in bytes) /log 2 N= log 2048/log 2 N=11
It depends on how wide the data buses are on each chip, and how they're connected. If they're one byte wide, you could need over 256 million addresses, one for each byte. if they're wider, and connected to show an even wider combined data bus, it could be much less; around 32 million.
8086 has 20 address lines. Therefore it can address 220 bits or 1,048,576 bits of memory, or roughly 1 MB (mega byte).
A memory with a 16 bit address bus can address 216 or 65536 distinct items. If each item is 32 bits in size, then the item is 4 bytes. The size of this memory is then 262144 bytes. (256Kb)
14 lines can address 16,384 locations. The 8085, however, has 16 lines, and can address 65,536 locations.The system design, of course, may limit that to 14, so 16,384 is the answer in that case.
The size of the address bus affects the maximum amount of memory a computer can directly access. Specifically, it determines the number of unique memory addresses that can be generated, which is calculated as 2 raised to the power of the address bus size (in bits). For example, a 32-bit address bus can address up to 4 GB of memory, while a 64-bit address bus can theoretically access 16 exabytes. Thus, a larger address bus allows for greater memory capacity and can enhance overall system performance.
The minimum size of an address bus depends on the amount of memory the system needs to access. To calculate the minimum size, you can use the formula (2^n), where (n) is the number of bits in the address bus and determines the number of unique addresses it can generate. For example, a 32-bit address bus can address (2^{32}) locations, or 4 GB of memory. Therefore, the minimum size of the address bus must be large enough to accommodate the maximum memory requirement of the system.
With 48-bit virtual addresses, the machine can address up to 256 terabytes of memory. This large address space allows for more efficient memory management, as it can accommodate a greater number of processes and data. However, the increased address size may also lead to higher memory overhead and potential performance issues due to the larger memory footprint. Overall, the implications of a 48-bit address size on memory management and system performance include improved scalability but potential trade-offs in memory efficiency and performance.
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You don't need a specific size memory card to play the GameCube; any GameCube memory card will do. Different games take up a different amount of blocks on a memory card, so what size you'll need depends on what games you want to play.
consider a RAM of 64 words with a size of 16 bits.Assume that this memory have a cache memory of 8 Blocks with block size of 32 bits.Draw a diagram to show the address mapping of RAM and Cache, if 4-way set associative memory scheme is used.