## Memory-Management

 Question 1
Consider six memory partitions of sizes 200 KB, 400 KB, 600 KB, 500 KB, 300 KB and 250 KB, where KB refers to kilobyte. These partitions need to be allotted to four processes of sizes 357 KB, 210KB, 468 KB and 491 KB in that order. If the best fit algorithm is used, which partitions are NOT allotted to any process?
 A 200KBand 300 KB B 200KBand 250 KB C 250KBand 300 KB D 300KBand 400 KB
Engineering-Mathematics       Memory-Management       GATE 2015 -(Set-2)
Question 1 Explanation:

Since Best fit algorithm is used. So, process of size,
357KB will occupy 400KB
210KB will occupy 250KB
468KB will occupy 500KB
491KB will occupy 600KB
So, partitions 200KB and 300KB are NOT alloted to any process.
 Question 2
Consider a paging hardware with a TLB. Assume that the entire page table and all the pages are in the physical memory. It takes 10 milliseconds to search the TLB and 80 milliseconds to access the physical memory. If the TLB hit ratio is 0.6, the effective memory access time (in milliseconds) is  _________.
 A 122 B 123 C 124 D 125
Operating-Systems       Memory-Management       Gate 2014 Set -03
Question 2 Explanation:
Tavg = TLB access time + miss ratio of TLB × memory access time + memory access time
= 10 + 0.4 × 80 + 80
= 10 + 32 + 80
= 122 ms
 Question 3
The memory access time is 1 nanosecond for a read operation with a hit in cache, 5 nanoseconds for a read operation with a miss in cache, 2 nanoseconds for a write operation with a hit in cache and 10 nanoseconds for a write operation with a miss in cache. Execution of a sequence of instructions involves 100 instruction fetch operations, 60 memory operand read operations and 40 memory operand write operations. The cache hit-ratio is 0.9.  The average memory access time (in nanoseconds) in executing the sequence of instructions is   __________.
 A 1.68 B 1.69 C 1.7 D 1.71
Operating-Systems       Memory-Management       Gate 2014 Set -03
Question 3 Explanation:
Total instruction=100 instruction fetch operation +60 memory operand read operation +40 memory operand write op
=200 instructions (operations)
Time taken for fetching 100 instructions (equivalent to read) = 90*1ns + 10*5ns = 140ns
Memory operand Read operations = 90% (60)*1ns + 10% (60) × 5ns = 54ns + 30 ns = 84ns
Memory operands Write operations time = 90% (40)*2ns + 10% (40)*10ns
= 72ns + 40ns = 112ns
Total time taken for executing 200 instructions = 140 + 84 + 112 = 336ns
∴ Average memory access time =336 ns/200=1.68ns
 Question 4
 A 2 B 4 C 8 D 16
Operating-Systems       Memory-Management       Gate 2013
Question 4 Explanation:
Let the size of page is = 2p B
So the no. of entries in one page is 2p/4, where 4 is the page table entry size given in question.
So we know that process size or virtual address space size is equal to
No. of entries × Page size
So total no. of entries for 3 level page table is,
(2p/4)×(2p/4)×(2p/4)
So, No. of entries × Page size = VAS
(2p/4)×(2p/4)×(2p/4)× (2p) = 246
24p = 252
4p = 52
p = 13
∴ Page size = 213
 Question 5
 A 2 B 4 C 8 D 16
Operating-Systems       Memory-Management       Gate 2013
Question 5 Explanation:
Architecture of physically indexed cache: Architecture of virtual indexed physically tagged (VIPT): VIPT cache and aliasing effect and synonym.
Alias: Same physical address can be mapped to multiple virtual addresses.
Synonym: Different virtual addresses mapped to same physical address (for data sharing).
So these synonyms should be in same set to avoid write-update problems.
In our problem VA = 46bits We are using 16bits for indexing into cache.
To have two synonym is same set we need to have same 16 bits index for PA & VA.
Assume that physical pages are colored and each set should have pages of same color so that any synonyms are in same set.
Since page size = 8KB ⇒ 13bits
These 13bits are not translated during VA→PA. So 13bits are same out of 16 Index bits, 13 are same we need to make 3bits (16-13) same now.
3bits can produce, 23 = 8 combinations which can be mapped on the different sets, so we need 8 different colors to color our pages. >br> In physically indexed cache indexing is done via physical address bits, but in virtual indexed cache, cache is indexed from (offset + set) bits. In physical Index cache indexing is done one to one (1 index maps to one page in one block of cache). In VIPT we have more/ extra bits, so mapping is not one-one. Hence these extra bits have to be taken care, such that if two virtual address refers to same page in cache block of different sets then they have to be assumed same i.e., we say of same color and put same color page in one set to avoid write update problems.
 Question 6
Let the page fault service time to 10 ms in a computer with average memory access time being 20 ns. If one page fault is generated for every 106 memory accesses, what is the effective access time for the memory?
 A 21 ns B 30 ns C 23 ns D 35 ns
Operating-Systems       Memory-Management       Gate 2011
Question 6 Explanation:
P = page fault rate
EA = p × page fault service time + (1 – p) × Memory access time
=1/106×10×106+(1-1/106)×20 ≅29.9 ns
 Question 7
 A 0 3 5 7 16 55 B 0 3 5 7 9 16 55 C 0 5 7 9 16 55 D 3 5 7 9 16 55
Operating-Systems       Memory-Management       Gate 2005-IT
Question 7 Explanation:
The cache is 2-way associative, so in a set, there can be 2 block present at a time.
So, Since, each set has only 2 places, 3 will be thrown out as its the least recently used block. So final content of cache will be
0 5 7 9 16 55
Hence, answer is (C).
 Question 8
 A (i) 80 MB (ii) 2040 MB B (i) 2040 MB (ii) 80 MB C (i) 80 MB (ii) 360 MB D (i) 80 MB (ii) 360 MB
Operating-Systems       Memory-Management       Gate 2005-IT
Question 8 Explanation:
Constant linear velocity:
Diameter of inner track = d = 1 cm
Circumference of inner track
= 2 * 3.14 * d/2
= 3.14 cm
Storage capacity = 10 MB (given)
Circumference of all equidistant tracks
= 2 * 3.14 * (0.5+1+1.5+2+2.5+3+3.5+4)
= 113.14 cm
Here, 3.14 cm holds 10 MB
Therefore, 1 cm holds 3.18 MB.
So, 113.14 cm holds
113.14 * 3.18 = 360 MB
So, total amount of data that can be hold on the disk = 360 MB.
For constant angular velocity:
In case of CAV, the disk rotates at a constant angular speed. Same rotation time is taken by all the tracks.
Total amount of data that can be stored on the disk
= 8 * 10 = 80 MB
 Question 9
 A 13.5 ms B 10 ms C 9.5 ms D 20 ms
Operating-Systems       Memory-Management       Gate 2005-IT
Question 9 Explanation:
Radius of inner track is 0.5cm (where the head is standing) and the radius of outermost track is 4cm.
So, the header has to seek (4 - 0.5) = 3.5cm.
For 10m ------- 1s
1m ------- 1/10 s
100cm ------- 1/(10×100) s
3.5cm ------- 3.5/1000 s = 3.5ms
So, the header will take 3.5ms.
Now, angulur velocity is constant and header is now at end of 5th sector. To start from front of 4th sector it must rotate upto 18 sector.
6000 rotation in 60000ms.
1 rotation in 10ms (time to traverse 20 sectors).
So, to traverse 18 sectors, it takes 9ms.
In 10ms, 10MB data is read.
So, 1MB data can be read in 1ms.
∴ Total time = 1+9+3.5 = 13.5ms
 Question 10
In a paged segmented scheme of memory management, the segment table itself must have a page table because
 A the segment table is often too large to fit in one page B each segment is spread over a number of pages C segment tables point to page table and not to the physical locations of the segment D the processor’s description base register points to a page table E Both A and B
Operating-Systems       Memory-Management       Gate-1995
Question 10 Explanation:
The segment table is often too large to fit in one page. This is true and the segment table can be divided into pages. Thus page table for each segment table, pages are created.
Segment paging is different from paged segmentation.
 Question 11
The capacity of a memory unit is defined by the number of words multiplied by the number of bits/word. How many separate address and data lines are needed for a memory of 4K × 16?
 A 10 address, 16 data lines B 11 address, 8 data lines C 12 address, 16 data lines D 12 address, 12 data lines
Computer-Organization       Memory-Management       Gate-1995
Question 11 Explanation:
ROM memory size = 2m × n
m = no. of address lines
n = no. of data lines
Given, 4K × 16 = 212 × 16
Address lines = 12
Data lines = 16
 Question 12
 A P = 12.5, Q = 2.5×106
Operating-Systems       Memory-Management       Gate-1993
Question 12 Explanation:
RPM = 2400
So, in DOS, the disk rotates 2400 times.
Average latency is the time for half a rotation
= 0.5×60/2400 s
= 12.5 ms
In one full rotation, entire data in a track can be transferred. Track storage capacity = 62500 bits
So, disk transfer rate
= 62500 × 2400/60
= 2.5 × 106 bps
So,
P = 12.5, Q = 2.5×106
There are 12 questions to complete.