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.
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.
2N
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.
The 8085 microprocessor is an 8-bit processor with a 16-bit address bus. This means it can access a maximum of 64 KB (2^16) of memory. The 8085 can address memory locations from 0000H to FFFFH, totaling 64 KB of memory space. This limitation is due to the 16-bit address bus, which can only address up to 64 KB of memory.
Awais khanLOC(LA[K]) =base(LA)+w(k-lowerbound)
To determine the memory size of a device, you can calculate it by multiplying the number of memory cells by the size of each memory cell. This will give you the total memory size of the device.