Cheylon Woods
1. If you buy some 4.7GB write-once DVD-R disks that can be written by
the DVD+/-R/RW drive, how many disks would you need to buy to back up a
full hard drive once (assume no compression)? At 30 cents per DVD-R disk,
how much would a full backup cost? At 10 minutes per DVD-R disk, how
long would a full backup take?
If I had to back up a 350 GB hard drive on 4.7 GB discs it would take 75 discs
Work: 350 divided by 4.7= 74.46. Since you cannot have a partial disc, this
number rounds up to 75.
Since the discs cost $.30 each, it would cost $22.50 cents to back up the hard drive.
Work: 75 discs each cost $.30. So 75 multiplied by .30 = 22.50 which translates to
$22.50
It takes the system 10 minutes write each disc, so it would take approximately 12.41 hours or
744.6 minutes to perform the backup
Work: 74.46 multiplied by 10 = 744.6 min; 744.6 min divided by 60 min = 12.41
hours.
2. Now, let's see how much stuff that hard drive can hold. Assume you
have access to the following information stored for all 314 million
people in the United States
Name:
40 characters
Phone Number:
10 characters
Library Card Number: 9 characters
Unpaid Fines:
one 4-byte number
and that each character is stored in one byte.
Would all of this data fit on the hard drive of the computer described above? Yes
Work: 40 bytes(name) multiplied by 314 million people = 12,560,000,000
bytes
10 bytes(phone number) multiplied by 314 million people=3,140,000,000 bytes
9 bytes(library card number) multiplied by 314 million people = 2,626,000,000 bytes
4 bytes(library fines) multiplied by 314 million people = 1,256,000,000 bytes.
The sum of these products is 19,782,000,000 bytes
Divide 19,782,000,000 bytes by 1024 gives you 19318359.375 kilobytes
Divide 19318359.375 kilobytes by 1024 gives you 18865.58533 megabytes
Divide 18865.58533 megabytes by 1024 gives you 18.42342 gigabytes
*note: 1024^3 power= 1073741824*
If not, how big a hard drive would you need? If so, what fraction of the disk would this
fill?
Work: 18.42342 gigabytes divided by 350gigabytes = .05263 gigabytes
.05263 multiplied by 100 =5.263%
3. Now lets see how long it would take to read that much data off the disk. Assume that
you access the data in a random order, and that you start a new disk access for each
person. How long would it take to add up the library fines for all 314 million people?
Could this be done in a second? In a minute? In an hour? In a day? In a month?
In a year?
Work:
Access time is 10 milliseconds. So .001 multiplied by 10 =.o1 sec.
There are 314,000,000 people that have to have their library fines information accessed. So
314,000,000 multiplied by .01 sec = 3,140,000 seconds to access the information.
1 byte= 8 bits, and the computer runs at 3 gigabits per sec to determine how many bytes per
second it runs divide 3 into 8. 3 divided by 8 = .375
There is 1,256,000,000 bytes of information which converts to 1.16974GB of information
Work:
1,2556,000,000 divided by 1024^3 = 1.16974
It takes 3.11931 seconds to move the data from the hard disk to the CPU
Work:
1.16974 divided by .375= 3.11931
In total it takes 3,140,003.11931 seconds to access and move the data in the computer.
Work: 3,140,000 added to 3.11931 = 3,140,003.11931
Because the number is too manage in seconds, we divide the number by 60 to get the number of
minutes:
Work: 3,140,003.11931 divided by 60=52,333.38532 minutes.
This number is still hard to manage, so we divide the quotient by 60.
Work: 52,333.38532 divided by 60= 872.22309 hours.
This number is still too large to manage, so we divide this quotient by 24.
Work: 872.22309 divided by 24= 36.34263 days, which rounds up to 36.3
days.
So It would take 36.3 days to access and read all of the library fines for 314,000,000 people.
4. Assume for the sake of comparison that all of this data could fit in RAM (it may not; you
should convince yourself of that). How long would it take the processor to perform 314
million additions if it can perform one addition instruction for every two clock cycles (this
means that 3.6 GHz equates to 1,800 MIPS because 3.6 GHz is 3,600 million cycles)? Could
this be done in a second? In a minute? In an hour? In a day? In a month? In a year?
Note for this question, you are not required to calculate the time for transferring the data
from RAM to CPU. Also note that ''quad core'' means 4 CPUs.
For a computer with 4 cores running 3.6 GHz each, the total number of GHz that the system can
process is 14.4 billion processes per clock cycle.
Work:
3.6 GHz multiplied by 4= 14.4 GHz
3.6 GHz equates to 1,800 MIPS, then the 4 cores are running 7, 200 MIPS. Because this
particular computer runs one instruction per 2 clock cycles, so I would divide 3.6GHz by 2 to
determine how many Million Instructions Per Second the processor runs.
If the system can perform 1 addition for every 2 clock cycles, then it will take 628 million clock
cycles to perform all of the additions for the 314 million files. We know that the computer can
process 7,200 MIPS, so to determine how long it will take the processor to add the information,
we divide 7,200 MIPS into 314,000,000.
Work: 314,000,000 divided by 7,200,000,000= .04361(with the 1 repeating).
This system can perform 314 million additions in less than 1 second.