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Memoery Managment Problem

http://www.doc.ic.ac.uk/~eedwards/compsys/index.html


1. Given five memory partitions of 100Kb, 500Kb, 200Kb, 300Kb, 600Kb (in order), how would the first-fit, best-fit, and worst-fit algorithms place
   processes of 212 Kb, 417 Kb, 112 Kb, and 426 Kb (in order)? Which algorithm makes the most efficient use of memory?

 First-fit:
 212K is put in 500K partition
 417K is put in 600K partition
 112K is put in 288K partition (new partition 288K = 500K - 212K)
 426K must wait

 Best-fit:
 212K is put in 300K partition
 417K is put in 500K partition
 112K is put in 200K partition
 426K is put in 600K partition

 Worst-fit:
 212K is put in 600K partition
 417K is put in 500K partition
 112K is put in 388K partition
 426K must wait

 In this example, best-fit turns out to be the best.

2. Assuming a 1 KB page size, what are the page numbers and offsets for
the following address references (provided as decimal numbers):
a. 2375
b. 19366
c. 30000
d. 256
e. 16385

Answer:
a. page = 1; offset = 327
b. page = 18; offset = 934
c. page = 29; offset = 304
d. page = 0; offset = 256
e. page = 16; offset = 1
 
3. Consider  a logical address space of 64 pages of 1024 words each, mapped onto a physical memory of 32 frames.

a. How many bits are there in the logical address?
b. How many bits are there in the physical address?


Answer:
a. Logical address: 16 bits
b. Physical address: 15 bits


4. Consider a logical address space of 32 pages with 1024 words per page;
mapped onto a physical memory of 16 frames.
a. How many bits are required in the logical address?
b. How many bits are required in the physical address?

Answer:
a. 2^5  + 2^10  = 15 bits.

b. 2^4  + 2^10  = 14 bits.


5. Consider a paging system with the page table stored in memory.

a. If a memory reference takes 200 nanoseconds, how long does a
paged memory reference take?

b. If we add associative registers, and 75 percent of all page-table references
are found in the associative registers, what is the effective
memory reference time? (Assume that finding a page-table entry
in the associative registers takes zero time if the entry is there.)


Answer:

a. 400 nanoseconds: 200 nanoseconds to access the page table and 200
nanoseconds to access the word in memory.

b. Effective access time = 0.75 ? (200 nanoseconds) + 0.25 ? (400
nanoseconds) = 250 nanoseconds.


6. Consider the following segment table:
Segment Base   Length
0 219 600
1 2300 14
2     90 100
3 1327 580
4 1952 96
What are the physical addresses for the following logical addresses?
a. 0,430
b. 1,10
c. 2,500
d. 3,400
e. 4,112


Answer:
a. 219 + 430 = 649
b. 2300 + 10 = 2310
c. illegal reference, trap to operating system
d. 1327 + 400 = 1727
e. illegal reference, trap to operating system

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