1. ## CS604-Assignment 5

Calculate the following

1. Calculate the swap out + swap in time of a process which has 2MB of size having transfer rate of 10MB/sec and whose average latency is 20ms. [5 marks]

2. The logical address space of the process is 16 pages and the size of each page is 2 Kb. There are 128 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer the following questions: [2.5 * 4 = 10 marks]

• Calculate the size of Logical Address?
• Find the size of the Logical Memory?
• Calculate the size of Physical Address?
• Find the size of the Physical Memory?

3. Consider a 64-bit linear address having 16k page size. Calculate maximum number of pages in a process address space, number of bits needed for d and for p. [5 marks]

2. please mention the duedate of it

3. mention the duedate of assignment is 21 july(today)

4. ok thanks. we are trying our best to provide you solution

5. IDEA SOLUTION

(1) Calculate the swap out + swap in time of a process which has 2MB of size having transfer rate of 10MB/sec and whose average latency is 20ms.

swap out time = process size / transfer rate
net swap out time= latency time swap out time
and then
swap out + swap in = net swap out time + net swap out time
by using this method my answer is
swap out + swap in = 440 ms

(2) The logical address space of the process is 16 pages and the size of each page is 2 Kb. There are 128 frames in the main memory. The size of a frame is equivalent to the size of a page. With the given information, answer the following questions:

• Calculate the size of Logical Address?
• Find the size of the Logical Memory?
• Calculate the size of Physical Address?
• Find the size of the Physical Memory?

Size of logical address= |p| + |d|=15 bits
Size of logical memory= No. of pages * page size= 64
Size of physical address= |f|+ |d|=18 bits
Size of physical memory= No. of frames * size of each frame= 256

(3) Consider a 64-bit linear address having 16k page size. Calculate maximum number of pages in a process address space, number of bits needed for d and for p.

Page size = 16 k

Maximum pages in a process address space = 2^16 / 16k = 4
No. of bits needed for d =|d| = log 2 4k = 12 bits
No. of bits needed for p = |p|= 16 – 12 = 3 bits

6. That's good you provide the solution for just idea.

7. hmmmm that is really smart of u ......thankx

8. Thanks alot

9. you are welcome dear. Just post your assignment on very first day.