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You can address 2^20 (or 1,048,576) locations. Each location will consist of 16 bits (1 or 0) which is commonly called a Word (though that isn't always the case).
I think what you're really looking for is the answer in bytes. Each data bus line holds a bit, 8 bits is a byte, so you're data line is 2 bytes. Since each location holds two bytes, this would be 2,097,152 bytes, or 2MB
What else can I help you with?
What is 31-bit addressing on a mainframe?
31 bit addressing is a way of accessing the virtual memory. Based on the memory available the bit in the address can change. Or viceversa. for 31 bit the memory available is gb and for 24 bit its 16mb.
What are the implications of having a machine with 48-bit virtual addresses and 32-bit physical addresses?
Having a machine with 48-bit virtual addresses and 32-bit physical addresses means that the system can address a larger amount of virtual memory than physical memory. This can lead to potential issues with memory management, such as increased overhead for address translation and the possibility of running out of physical memory space. It may also impact the efficiency and performance of the system, as the mismatch between virtual and physical memory sizes can result in slower data access times.
How much is 10mb of virtual memory?
no just a bit sweg
How much Ram is there on a virtual Windows XP?
However much is assigned to it by the virtual machine. Virtual or not, the 32-bit version of WinXP will not be able to address more than 4GB of memory in any combination of video and system ram. Dial back to 4GB instead of allocating excess...you gain nothing and just take away memory from the host. This presumes that your host is a 64-bit OS. If the host is 32-bit, also, you want to limit your virtual machine to half the available memory (2GB) or your host will suffer performance issues.
Suppose that a machine has 48-bit virtual addresses, what are the implications of this address size on memory management and system performance?
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.
What factors determine how much physical memory an 80286 can address?
An 80286 has a 24 bit address bus. As such, it can address 224, or 16,777,216, or 16 MB of memory.
If your address bus is 20-bits wide, how much memory can your computer address approximately?
With a 20-bit address bus, a computer can address approximately 1,048,576 memory locations, which is equivalent to 1 megabyte of memory.
What is the difference between a physical address and a mailing address?
Real memory uses Physical addresses.These are the members that the memory chips react to on the bus. Virtual addresses are the logical addresses that nrefer to a process' address space. Thus, a machine with a 16-bit word can generate virtual addresses upto 64K, regardless of whether the machine has more or less memory than 64 KB
What is memory interfacing in 8086 microprocessor?
The 8086/8088 is a 16 bit computer running on a 20 bit address bus. Processes use a segmented memory architecture to access one of four 64kb memory segments from a physical space of 1mb.
How to calculate virtual memory in 80286 microprocessor?
Virtual memory does not exist physically but it is available in the systemThe procedure of fetching the chosen pgm. Segments or data from the secondary storage into the physical memory is called swappingSo it can address 1GBPhysical address calculation in PVAMIt uses 16-bit content of a segment register as a selector to address a descriptor stored in physical memoryThe descriptor is a block of contiguous memory locations containing information of a segmentPrivilege levels prevent unauthorized accessesMaximum segment size will be of 64kb
If a memory had 16 bit address bus and a 32 bit data bus what is the largest size of the memory?
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)
Why microcomputer system based on 8 bit microprocessor have 64K memory?
Even though the 8085 is an 8 bit microprocessor, it can address 64K memory, because it has a 16 bit address bus.
