As segment Size = 4096 Byte so 12 bits will be used to represent the offset
Logical address or Virtual address
| segment No | offset |
Given Virutal address 2DAC0H
12 bit are used to represent offset so offset is AC0H (H for Hexadecimal)
Segment N0 = 2DH
Physical Address
| segment no | offset+Base |
Base corresponding to 2DH is BEEFH
so BEEFH + AC0 = C9AFH
=> overall physical address is 2DC9AFH
A VM system uses segmentation with segment tables having maximum 256 entries and a maximum segment...
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1. Assume that your machine uses 16-bit virtual address and 14-bit physical address with 4KB page size. To get full credit, show your work. a. Show bits of the virtual address and physical address. b. What is the maximum program size that can be run in this machine? c. What is the maximum physical memory that this machine can have? d. Show the numbers of virtual pages and page frames. e. What is the table size if a page entry...
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