Download Homework 1 Solutions

Homework 1 Solutions
May 12, 2008
1. A computer has a pipeline with four stages. Each stage takes the same time time to do its
work0.4 ns (nanoseconds). How many instructions can this machine execute per second?
The machine can execute 109 instructions per 0.8 seconds. 0.8 + 0.2 = 1 second. So the formula is 109 +
109 /4 = 109 + 109 /4 = 1,000,000,000 + 250,000,000 = 1,250,000,000 instructions per second.
2. What is a trap instruction? What is its use in operating systems? What is the key difference between a trap and an interrupt?
A trap instruction switches from user mode to kernel mode. An interrupt is used to save the state of a
process during a context switch.
3. Which of the following operations should be allowed only in kernel mode? Briefly explain
your decision for each one.
(a) Cause a trap: This can be done by user programs. It’s the only way that they can enter the kernel to
do a system call. It’s often useful to allow traps in the OS kernel as well.
(b) Disable all interrupts: This can only be done in kernel mode. Interrupts are used in context switches
or to notify the system when retrieving data from the hard disk.
(c) Write the memory map for the current process: This can only be done in kernel mode. If you
do it in user mode, a process could overwrite another process’ (or the kernel’s) memory.
(d) Read the time-of-day clock: User mode can do this. You have the clock displayed on your desktop,
right?
(e) Receive a packet of data from the network: This can only be done in kernel mode. The kernel
has to decide which process gets the packet, and needs to keep the contents secret from processes that aren’t
going to get the packet.
(f ) Shutdown the computer: This can only be done in kernel mode. You don’t want user processes to
be able to halt the computer and destroy everyone else’s work. (Imagine what this could do if someone
shutdown unix.ic.ucsc.edu while you were working on a program.)
4. What is the difference between the following two function calls? Assume that str is a
properly null-terminated string. Also, file descriptor 2 (in Unix) is the same as the stdio file
stderr. HINT: what does the operating system do for each function?
fprintf (stderr, %s, str);
write (2, str, strlen (str));
write() is a system call, while fprintf() is a library function. This fact can lead to major differences in
the process which is triggered on either of the above calls. When write() is called, a trap is always made
to kernel mode and the output is written without delay. On the other hand, fprintf() is not a system
call and may be writing to a buffer, which would be flushed via a write() call once it is full. Thus, many
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fprintf() calls could lead to only one write() call.
5. While there are many different processor architectures, most desktop computers and servers
use the Intel x86 architecture. From the point of view of operating systems, what are the advantages to this approach? Are there any disadvantages?
Advantages:
- Hardware development costs are spread over as many systems as possible.
- Byte ordering is the same on every system.
- Machine code may be somewhat portable between different computers.
- The OS and drivers are portable between computers (booting XP or Linux on a MacBook Pro).
Disadvantages:
- Problems in the x86 architecture will affect nearly all computers. This includes hardware issues as well as
(possibly) viruses.
- The architecture may not be well-optimized for different types of systems (low-power, high-performance,
etc.).
- The x86 architecture tends to sacrifice performance for backwards compatibility, and legacy features make
certain OS implementation details harder.
6. Why is the process table needed in a timesharing system? Is it also needed in personal
computer systems in which only one process exists, that process taking over the entire machine
until it is finished?
A process table is needed in a timesharing system because we need to know what information went along
with a process when we swap it in and out to run it and other processes. In a system where only run process
runs at a time, and it runs to completion, a process table is not needed.
7. For each of the following system calls, give a condition that causes it to fail:
(a) read(): Read can fail on a closed file handle, end of file already read, a bad string (buffer) pointer, or a
file not opened for reading. Problems reading the file itself can also cause failure (disk failure, for example).
(b) write(): Write can fail on a closed file handle, file not opened for writing, bad string pointer, lack of
space on the device, or I/O error.
(c) close(): You give close an incorrect file handle/descriptor.
(d) fork(): Fork can fail if there are too many processes (either for this user for the whole operating system)
or if there’s not enough memory.
(e) execvp(): Execvp can fail if the path name to the file you want to execute is incorrect, there are too
many symbolic links in translating the path or file, or your argument list can be too long.
(f ) waitpid(): Waitpid will fail if the process id given to it no longer exists.
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8. There are several different approaches to operating system design.
(a) Describe one advantage the monolithic modular approach has over the microkernel approach.
Monolithic kernels are typically higher performance, and may be easier to write.
(b) Describe one advantage the microkernel approach has over the monolithic modular approach.
When one portion of the microkernel has a critical error, it doesn’t necessarily bring down the rest of the
microkernel with it. This can make microkernels more reliable.
(c) If security is a bigger concern than speed (as it perhaps is today), which operating system
design approach would you choose?
Either answer is fine, as long as there is a good defense. Having said that, a microkernel makes it harder
for rogue kernel processes to mess up the system. Compromising a single server process wouldn’t necessarily
compromise others. On the other hand, it may be easier to debug a monolithic kernel...
9. Here are some questions for practicing unit conversions:
(a) How long is a microyear in seconds?
We know that micro = 10−6 . We also know that a year is 3.155815296 * 107 seconds (source: ask.com).
Then, a microyear is 10−6 * 3.155815296 * 107 ≈ 31.558 seconds. This is one of those cases where you don’t
need 8 digits of precision.
(b) If a car is traveling 25200 miles per fortnight, fast is it traveling in inches per second?
A fortnight is 2 weeks = 14 days = 336 hours = 20,160 minutes = 1,209,600 seconds. 25,200 miles =
133,056,000 feet = 1,596,672,000 inches. 1,596,672,000 inches / 1,209,600 seconds = 1,320 inches/second
(equivalent to 75 mph).
(c) Micrometers are called microns. How long is a gigamicron?
Again, micro = 10−6 . Giga = 109 . A gigamicron is 10−6 * 109 meters = 103 meters = 1 kilometer (km).
(d) How many bytes are there in 1 terabyte of memory?
Memory is measured in powers of 2, not 10, though a gigabyte in disk drive terms is typically a billion
bytes. We want to know how many bytes are in a terabyte of memory. A terabyte is 240 bytes, which is
1099511627776 bytes. You get half credit for saying that a terabyte is 1012 bytesit’s approximately right,
and the point of these questions is to get the answers right.
(e) The mass of the earth ia 6000 yottagrams. What is that in kilograms?
The prefix ’yotta’ means 1024 . 6000 yottagrams is 6 * 103 * 1024 = 6 * 1027 grams. Since 1 kg = 103 grams,
the earth weighs 6 * 1024 kg.
10. On all current computers, at least part of the interrupt handlers are written in assembly
language. Why?
There are actions (such as saving registers) that happen while an interrupt is being handled. These actions
need to be written in assembly because they cannot be expressed in higher level languages.
11. When an interrupt or a system call transfers control to the operating system, a kernel
stack area separate from the stack of the interrupted process is generally used. Why?
Having a separate area reduces in the need to check that user programs don’t overwrite data onto the kernel
stack and vice versa.
12. What is the biggest advantage of implementing threads in user space? What is the biggest
disadvantage?
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Implementing threads in user space can allow a program to run on top of any OS, even if it doesn’t support
threading. The biggest disadvantage is allowing threads to make system calls that normally cause the OS
to block to be made without halting the entire process since that would defeat the point of threading in the
first place.
13. Suppose that we have a message-passing system using mailboxes. When sending to a full
mailbox or trying to receive from an empty one, a process does not block. Instead, it gets an
error code back. The process responds to the error code by just trying again, over and over,
until it succeeds. Does this scheme lead to race conditions? Why or why not?
The message-passing system itself will not lead to race conditions, in the worst case it will lead the process
starvation. You can argue about what the system does with messages once they are placed in the mailbox,
but that is a different part of the problem.
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