Paging examples
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The document discusses paging examples with a 15-bit logical address space and 1K page size. It explains that with these parameters: - There are 32 pages in the system (2^5) - Pages are 4K in size (2^12) since 3 bits are needed for the 8 logical pages - To find the physical address for logical address 1025: convert it to binary, identify the page number and offset, use the page number to index the page table, and concatenate the physical page number and offset. The physical address for 1025 is byte 1.
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- A list of textbooks and online resources for the course.
- The course outcomes, which are to identify digital logic concepts, simplify Boolean expressions, design data processing circuits, and design sequential circuits.
- An outline of the course units, which cover basic logic circuits, number systems, Boolean algebra, circuit implementation, and sequential circuits.
The document provides an overview of the topics, resources, and goals of the DLD course.
Data ordering and Addressing Standards - Big Endian and Small Endian
The document discusses data ordering and addressing standards in computers. It explains that memory is organized as an array, with each data element having a unique address. It describes byte-addressable memory where each element is 1 byte in size. It also discusses data ordering standards like big endian and little endian and how they determine if the least significant or most significant byte is stored at the lowest memory address. The document is presented by Kaushik Ghosh from the computer science department.
More on number system
there are different number system such as binary, decimal, octal and hexadecimal. binary has 2 digits 0 & 1. decimal has 0 to 9 digits. octal has 0 to 7 digits. and hexadecimal number system has 0 to 9 digits and 10 to 15 are denoted by alphabets. such as A=10, B=11 etc.
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Paging examples
- 1. Paging Examples Assume a page size of 1K and a 15-bit logical address space. How many pages are in the system?
- 2. • Answer: 2^5 = 32. • Assuming a 15-bit address space with 8 logical pages. How large are the pages?
- 3. • Answer: 2^5 = 32. • Assuming a 15-bit address space with 8 logical pages. How large are the pages? • Answer: 2^12 = 4K. It takes 3 bits to reference 8 logical pages (2^3 = 8). This leaves 12 bits for the page size thus pages are 2^12.
- 4. Consider logical address 1025 and the following page table for some process P0. Assume a 15-bit address space with a page size of 1K. What is the physical address to which logical address 1025 will be mapped? 8 0 2
- 5. Consider logical address 1025 and the following page table for some process P0. Assume a 15bit address space with a page size of 1K. What is the physical address to which logical address 1025 maps? 8 Step 1. Convert to binary: 0 000010000000001 2
- 6. Consider logical address 1025 and the following page table for some process P0. Assume a 15bit address space with a page size of 1K. What is the physical address to which logical address 1025 maps? 8 0 2 Step2. Determine the logical page number: Since there are 5-bits allocated to the logical page, the address is broken up as follows: 00001 0000000001 Logical page number offset within page
- 7. Consider logical address 1025 and the following page table for some process P0. What is the physical address? 00001 0 Step 3. Use logical page number as an index into the page table. 2 00001 0000000001 8
- 8. Consider logical address 1025 and the following page table for some process P0. What is the physical address? 00001 8 0 Take the physical page number from the page table and concatenate the offset. The physical address is byte 1. 2 000000000000001