CPSC 310 HW 5, Summer 2008, Ronnie Ward
Assigned 7/7/08, DUE in class on 7/15/08
Turn in a hardcopy with the Assignment Cover page and post your solutions to CSNet
These exercises were selected from Chapter 11, and one from SimpleDB to help you
better understand disk technology, performance, reliability and implementation issues.
1. 11.3.1 (a, c, d, e, f, g), and 11.3.2 (graduate students) p. 524
a.) 777 capacity? 512 bytes/sector*1000 sectors/trk*10,000 trks/surface*10 surfaces =
51GB
c.) Max seek time? 1+0.001*10,000 = 1 + 10 = 11 milliseconds
d.) Max rotational latency = one rotation, which is 1 trk r = 1/10,000 r/min*60
seconds/min = 0.006 seconds = 6 milliseconds
e.) Blk transfer time? Blk=16,384 bytes=32 sectors, 0.006ms/trk=0.000006 sec/sector =
0.006ms, which includes the 20% overhead for the gap. So transfer time for one blk is
32*0.006 = 0.192ms. Or using the book’s methodology, the number of degrees occupied
by a blk is 72*(31/100)+288*(32/1000)=11.448. Dividing this by 360 degrees and
multiplying by 6ms, we get 0.191ms to read a blk.
f.) Avg seek time? 1+0.001*10,000/3 ms, or 4.3ms
e.) Avg rotational delay? One-half 6ms, or 3ms
1) 11.3.2 (graduate students) p. 524 using the 747 disk
The sum of the first N integers is N*(N+1)/2. Starting at 2048, moving backwards an
average of (1+2+…+2047)/2048 is 1024 with 1/8th of that as an expected, or 128. Moving
forward average is (1+2+…+14336)/14336 is 7168 with a 7/8th probability implies an
expected value of 6272. Adding 128 and 6272, we get an expected average seek distance
of 6400, which is 1+0.001*6400, or 7.4ms. Adding the average blk access time of
(4.17+0.25)ms gives a total of 11.82ms.
2. 11.4.1 p. 532 using the 777 disk
Time to sort a 10,000,000 tuple relation R with a unique sort key? Blk size=16,384, and
100 records per blk with a record length of 160 bytes plus overhead.
So R is 10,000,000/100 blks, or 100,000 = 1.64GB.
There is 100MB of RAM available for sorting, implying 6400 blks can fit in RAM. The
first phase of a TPMMS with the disk in 11.3.1 is 200,000 blk I/Os, or
4.3ms+3ms+0.192ms = 7.5ms giving a 1st phase I/O time of 7.5*200,000 = 1498 seconds
= 25 minutes. The second phase takes the same number of I/Os, so the total sort time is
50 minutes.
3. 11.5.1 p. 544, but instead of the 747 disk, use the 777 disk described in exercise 11.3.1
At what time is each request fully serviced using the 777 disk where the heads are at trk
8000 initially? Avg seek, rotational delay, and transfer time of 4.3ms, 3ms, and 0.192ms
respectively. Assume each blk accessed takes 3ms+0.192ms, or 3.192ms, plus the seek
time (1+0.001n). Requests are 2000 at 0, 12000 at 1, 1000 at 10 and 10000 at 20.
