1 important announcements midterm 3 is on wednesday, april 20 from 7pm to 8:30pm —practice...

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Important Announcements

Midterm 3 is on Wednesday, April 20 from 7pm to 8:30pm— Practice Midterm 1 released tonight— Please email me ASAP in case you need a conflict

Final Exam (cumulative) is on Monday, May 9 from 1:30pm to 4:30pm— Please email me ASAP in case you need a conflict

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Pipelined datapath

Readaddress

Instructioncache

Instruction[31-0]

Address

Writedata

Data cache

Readdata

MemWrite

MemRead

1

0

MemToReg

4

Shiftleft 2

Add

Signextend

ALUSrc

Result

ZeroALU

ALUOp

Instr [15 - 0] RegDst

Readregister 1

Readregister 2

Writeregister

Writedata

Readdata 2

Readdata 1

Registers

RegWrite

Add

Instr [15 - 11]

Instr [20 - 16]0

1

0

1

IF/ID ID/EX EX/MEM MEM/WB

1

0

PCSrc

PC

3

PC register stores virtual address

The PC registers stores PCv = virtual address of the next instruction

Readaddress

Instructioncache

Instruction[31-0]

4

Add

1

0

PCSrc

P Cv

4

PC register stores both virtual and physical address

The PC registers stores both PCv and PCp

Readaddress

Instructioncache

Instruction[31-0]

4

Add

1

0

PCSrc

P Cv

P Cp

5

Virtual Memory system

virtual address

data

physical address

TLB

page table

memory

cachedisk

page offset

page offsetvirtual page number (VPN)

PPN

tag index blockoffset

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Hard drives

The textbook shows the ugly guts of a hard disk—Data is stored on double-sided magnetic disks called platters—Each platter is arranged like a record, with many concentric

tracks—Tracks are further divided into individual sectors, which are the

basic unit of data transfer—Each surface has a read/write head like the arm on a record

player, but all the heads are connected and move together

A 75GB IBM Deskstar has roughly:—5 platters (10 surfaces),—27,000 tracks per surface,—512 bytes per sector,—~512 sectors per track…

…but this number increases goingfrom the center to the rim

Platter

Track

Platters

Sectors

Tracks

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There are two fundamental performance metrics for I/O systems:

1. Latency Time to initiate data-transfer (units = sec)

2. Bandwidth Rate of initiated data-transfer (units = bytes/sec)

Time = latency + transfer_size / bandwidth

sec bytes / (bytes/sec)

I/O Performance

Dominant term forsmall transfers

Dominant term forlarge transfers

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Accessing data on a hard disk Factors affecting latency:

—Seek time measures the delay for the disk head to reach the track—A rotational delay accounts for the time to get to the right sector

Factors affecting bandwidth:—Usually the disk can read/write as fast as it can spin—Bandwidth is determined by the rotational speed, which also

determines the rotational delay

We can compute average seek time/rotational delay/etc. but careful placement of data on the disk can speed things up considerably:—head is already on or near the desired track—data in contiguous sectors—in other words, locality!

Even so, loading a page from the hard-disk can take tens of milliseconds

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Parallel I/O Many hardware systems use parallelism for increased speed

A redundant array of inexpensive disks or RAID system allows access to several hard drives at once, for increased bandwidth—similar to interleaved memories from last week