Sistemas-Operativos-Modernos 3Ra Edicion Solicionario A. Tanenmaun

MODERN
OPERATING
SYSTEMS
THIRD EDITION

PROBLEM SOLUTIONS

ANDREW S. TANENBAUM
Vrije Universiteit
Amsterdam, The Netherlands

PRENTICE HALL
UPPER SADDLE RIVER, NJ 07458

Copyright Pearson Education, Inc. 2008

SOLUTIONS TO CHAPTER 1 PROBLEMS
1. Multiprogramming is the rapid switching of the CPU between multiple processes in memory. It is commonly used to keep the CPU busy whileone or
more processes are doing I/O.
2. Input spooling is the technique of reading in jobs, for example, from cards,
onto the disk, so that when the currently executing processes are finished,
there will be work waiting for the CPU. Output spooling consists of first
copying printable files to disk before printing them, rather than printing directly as the output is generated. Input spooling ona personal computer is not
very likely, but output spooling is.
3. The prime reason for multiprogramming is to give the CPU something to do
while waiting for I/O to complete. If there is no DMA, the CPU is fully occupied doing I/O, so there is nothing to be gained (at least in terms of CPU utilization) by multiprogramming. No matter how much I/O a program does, the
CPU will be 100% busy. Thisof course assumes the major delay is the wait
while data are copied. A CPU could do other work if the I/O were slow for
other reasons (arriving on a serial line, for instance).
4. It is still alive. For example, Intel makes Pentium I, II, and III, and 4 CPUs
with a variety of different properties including speed and power consumption.
All of these machines are architecturally compatible. Theydiffer only in
price and performance, which is the essence of the family idea.
5. A 25 × 80 character monochrome text screen requires a 2000-byte buffer. The
1024 × 768 pixel 24-bit color bitmap requires 2,359,296 bytes. In 1980 these
two options would have cost $10 and $11,520, respectively. For current
prices, check on how much RAM currently costs, probably less than $1/MB.
6. Considerfairness and real time. Fairness requires that each process be allocated its resources in a fair way, with no process getting more than its fair
share. On the other hand, real time requires that resources be allocated based
on the times when different processes must complete their execution. A realtime process may get a disproportionate share of the resources.
7. Choices (a), (c), and (d) shouldbe restricted to kernel mode.
8. It may take 20, 25 or 30 msec to complete the execution of these programs
depending on how the operating system schedules them. If P0 and P1 are
scheduled on the same CPU and P2 is scheduled on the other CPU, it will
take 20 mses. If P0 and P2 are scheduled on the same CPU and P1 is
scheduled on the other CPU, it will take 25 msec. If P1 and P2 are scheduledon the same CPU and P0 is scheduled on the other CPU, it will take 30 msec.
If all three are on the same CPU, it will take 35 msec.

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PROBLEM SOLUTIONS FOR CHAPTER 1

9. Every nanosecond one instruction emerges from the pipeline. This means the
machine is executing 1 billion instructions per second. It does not matter at
all how many stages the pipeline has. A 10-stage pipeline with 1nsec per
stage would also execute 1 billion instructions per second. All that matters is
how often a finished instruction pops out the end of the pipeline.
10. Average access time =
0.95 × 2 nsec (word is cache)
+ 0.05 × 0.99 × 10 nsec (word is in RAM, but not in cache)
+ 0.05 × 0.01 × 10,000,000 nsec (word on disk only)
= 5002.395 nsec
= 5.002395 μsec
11. The manuscript contains 80 × 50 ×700 = 2.8 million characters. This is, of
course, impossible to fit into the registers of any currently available CPU and
is too big for a 1-MB cache, but if such hardware were available, the
manuscript could be scanned in 2.8 msec from the registers or 5.8 msec from
the cache. There are approximately 2700 1024-byte blocks of data, so scanning from the disk would require about 27 seconds,...