Part I Consider a computer that you have used or now use. (You may make an assum

Question

Part I Consider a computer that you have used or now use. (You may make an assumption on word size if needed.) How large is the memory of the computer? How long is a word in the computer above? How many bits are required to address the bytes in that memory? How many data lines are required to read data from the computer memory?

Part II Does your computer have cache? If so how big is it? How does your cache size compare with the size of main memory? Do you know what kind of cache it is? How did you find out about your computer's cache? Make an assumption that your cache is either: Fully associative Direct mapped Two-way set-associative Four-way set-associative Using the relationships in Part I above, determine: the size of the Tag and Word for Associative cache; OR the size of the Tag, Line, and Word for Direct-Mapped Cache ; Or the size of Tag, Set, and Word for K-Way Set-Associative Cache. You may make any assumptions necessary including the number of Words in each block (recommend 2 or 4 or 8)

Part III Given the following: Logical Memory size of 1000 Physical Memory size of 2000 Page (and frame) size of 100 Block A contains data for a program Select Block A’s size and its starting point in both memories. Then write the page table for Block A based on your selections. See the example above for the layout of both memories and an example of Block A of size 200.

Part IV Discuss the pros and cons of paging.

Explanation / Answer

1,My laptop is dell inspiron 1545.The memory size is RAM IS 3GB.Thee word size is 64 bits.Bits required to access the memory are 3Gb=2.4e+10 bits

2.Yes it has L2 cache of size 1MB.It is smaller than Ram MEMORY WHICH is 3 gb.

for direct mapped caching.we will consider each block is 32 bytes (8 words), so we need 5 offset bits to determine which byte in each block - Direct-mapped => number of sets = number of blocks = 4096 =>2^12=> we need 12 index bits .

=> tag bit = block bits-index bits=32 - 12 - 5 = 15bits

3

Logical Memory Physical Memory
location/ page location/frame
0 to 99/ 0 0 to 99/ 0
100 to 199 /1 Block A 100 to 199/ 1
200 to 299/ 2 Block A 200 to 299/ 2
300 to 399/ 3 300 to 399/ 3
400 to 499/ 4 400 to 499/ 4
500 to 599/ 5 500 to 599/ 5
600 to 699/ 6 600 to 699/ 6
700 to 799/ 7 700 to 799/ 7
800 to 899/ 8 800 to 899/ 8
900 to 999/ 9 900 to 999/ 9
1000 to 1099/ 10
1100 to 1199/ 11
1200 to 1299/ 12
1300 to 1399/ 13 Block A
1400 to 1499/ 14 Block A
1500 to 1599/ 15
1600 to 1699/ 16
1700 to 1799/ 17
1800 to 1899/ 18
1900 to 1999/ 19

4.Pros of paging are:

a)It allows demand pagign and prepaging.

b)It helps in more efficient swapping.

c)It helps to allocate memory in an easy way.

d)This helps to prevent external fragmentation.

Cons of paging are:

a)It has longer memory access times here.

b)It has guarded and inverted page tables.

c)Page Table Length Register (PTLR) to limit virtual memory size

d)It leads to internal fragmentation creating a big problem

Related Questions

Navigate

• Browse (All)
• Browse P
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.