2 PROBLEM SOLUTIONS FOR CHAPTER 1
stage would also execute 1 billion instructions per second. All that matters is
how often a finished instructions pops out the end of the pipeline.
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 scan-
ning from the disk would require about 27 seconds, and from tape 2 minutes 7
seconds. Of course, these times are just to read the data. Processing and
rewriting the data would increase the time.
12. Logically, it does not matter if the limit register uses a virtual address or a
physical address. However, the performance of the former is better. If virtual
addresses are used, the addition of the virtual address and the base register
can start simultaneously with the comparison and then can run in parallel. If
physical addresses are used, the comparison cannot start until the addition is
complete, increasing the access time.
13. Maybe. If the caller gets control back and immediately overwrites the data,
when the write finally occurs, the wrong data will be written. However, if the
driver first copies the data to a private buffer before returning, then the caller
can be allowed to continue immediately. Another possibility is to allow the
caller to continue and give it a signal when the buffer may be reused, but this
is tricky and error prone.
14. A trap is caused by the program and is synchronous with it. If the program is
run again and again, the trap will always occur at exactly the same position in
the instruction stream. An interrupt is caused by an external event and its
timing is not reproducible.
15. Base = 40,000 and limit = 10,000. An answer of limit = 50,000 is incorrect
for the way the system was described in this book. It could have been imple-
mented that way, but doing so would have required waiting until the address
+ base calculation was completed before starting the limit check, thus slow-
ing down the computer.
16. The process table is needed to store the state of a process that is currently
suspended, either ready or blocked. It is not needed in a single process sys-
tem because the single process is never suspended.
17. Mounting a file system makes any files already in the mount point directory
inaccessible, so mount points are normally empty. However, a system
administrator might want to copy some of the most important files normally
located in the mounted directory to the mount point so they could be found in
their normal path in an emergency when the mounted device was being
checked or repaired.
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ljb200625492012-08-08好像版本有点低
