Transcript Multiprocessor Memory Allocation

Operating Systems
CMPSCI 377
Lecture 18: Storage Systems
Emery Berger
University of Massachusetts, Amherst
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
Last Time:
Improving I/O Performance

Approaches:



Caching – reduces data transfer
Large data transfers
DMA
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
2
Today

Improving performance of storage systems
(i.e., the disk)




Scheduling
Interleaving
Read-ahead
Tertiary storage
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
3
Disk Operations

Disks are SLOW!


Latency:



1 disk seek = 40,000,000 cycles
Seek – position head over track/cylinder
Rotational delay – time for sector to rotate
underneath head
Bandwidth:

Transfer time – move bytes from disk to
memory
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
4
Calculating Disk Operations Time

Read/write n bytes:
I/O time = seek + rotational delay + transfer(n)


Can’t reduce transfer time
Can we reduce latency?
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
5
Reducing Latency

Minimize seek time & rotational latency:





Smaller disks
Faster disks
Sector size tradeoff
Scheduling
Layout
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
6
Disk Head Scheduling

Change order of disk accesses


Reduce length & number of seeks
Algorithms




FCFS
SSTF
SCAN, C-SCAN
LOOK, C-LOOK
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
7
I/O time = seek + rotational delay + transfer(n)
FCFS


First-come, first-serve
Example: disk tracks – (65, 40, 18, 78)

18
assume head at track 50
40
65
78
disk

Total seek time?

15+25+22+60 = 122
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
8
I/O time = seek + rotational delay + transfer(n)
SSTF


Shortest-seek-time first (like SJF)
Example: disk tracks – (65, 40, 18, 78)

18
assume head at track 50
40
65
78
disk

Total seek time?


10+22+47+13 = 92
Is this optimal?
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
9
I/O time = seek + rotational delay + transfer(n)
SCAN


Always move back and forth across disk
a.k.a. elevator
Example: disk tracks – (65, 40, 18, 78)

18
assume head at track 50, moving forwards
40
65
78
disk

Total seek time?


15+13+22+60+22=132
When is this good?
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
10
C-SCAN


I/O time = seek + rotational delay + transfer(n)
Circular SCAN:
Go back to start of disk after reaching end
Example: disk tracks – (65, 40, 18, 78)

18
assume head at track 50, moving forwards
40
65
78
disk

Total seek time?


15+13+22+100+18+40=208
Can be expensive, but more fair
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
11
LOOK variants

Instead of going to end of disk,
go to last request in each direction

SCAN, C-SCAN ) LOOK, C-LOOK
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
12
Exercises

5000 cylinders, currently at 143, moving forwards
(86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130 )







FCFS
SSTF
SCAN
LOOK
C-SCAN
C-LOOK
Find sequence & total seek time

First = 0, last = 4999
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
13
Solutions

FCFS


SSTF


143, 913, 948, 1022, 1470, 1509, 1750, 1774, 130, 86 – distance: 3,319
C-SCAN


143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 130, 86 – distance: 9,769
LOOK


143, 130, 86, 913, 948, 1022, 1470, 1509, 1750, 1774 – distance: 1,745
SCAN


143, 86, 1470, 913, 1774, 948, 1509, 1022, 1750, 130 – distance: 7,081
143, 913, 948, 1022, 1470, 1509, 1750, 1774, 4999, 0, 86, 130 – distance:
9,813
C-LOOK

143, 913, 948, 1022, 1470, 1509, 1750, 1774, 86, 130 – distance: 3,363
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
14
Today

Improving performance of storage systems
(i.e., the disk)




Scheduling
Interleaving
Read-ahead
Tertiary storage
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
15
Disk Interleaving

Contiguous allocation


Requires OS response before disk spins past
next block
Instead of physically contiguous allocation,
interleave blocks

Relative to OS speed, rotational speed
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
16
Read-Ahead

Reduce seeks by reading blocks from disk
ahead of user’s request


Place in buffer on disk controller
Similar to pre-paging

Easier to predict future accesses on disk?
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
17
Tertiary Storage

a.k.a. long-term storage



Examples:



disks = secondary storage
used for archival data or backups
tape drives
CD-R, DVD-R/W
Performance still somewhat important

not much OS can do
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
18
Summary

OS can improve storage system performance



Scheduling
Interleaving
Read-ahead
UNIVERSITY OF MASSACHUSETTS, AMHERST • Department of Computer Science
19