453 csc/cpe Midterm Solutions
Winter 2003
Name: Section:
Key All
Rules: Do all your own work. Nothing says your neighbor has any better idea what the answer is. Do not discuss this exam outside of class until after 4:00pm. Suggestions(mostly the obvious): When in doubt, state any assumptions you make in solving a problem. If you think there is a misprint, ask me. Read the questions carefully. Be sure to answer all parts. Identify your answers clearly. Watch the time/point tradeoff: 95ts/50min works out to 31.6s/pt. Problems are not necessarily in order of difficulty. Be sure you have all pages. Pages other than this one are numbered "n of 7". Encouragement: Good Luck!
Problem 1 2 3 4 5 6 7 Total:
Possible 10 10 10 10 15 20 20 95
1
Score
csc/cpe 453
Midterm Solutions
Winter 2003
Answer the following questions clearly, concisely, and (where possible) correctly. 1. (10) Busywaiting is often denegrated in the concurrent programming community as a crude and inefficient practice. However, it has its place. (a) First, what is busywaiting? Solution: Busywaiting is the process of repeatedly checking a condition (in a tight loop, e.g.) until it becomes true.
(b) Under what circumstances would it be appropriate to choose a busywaiting approach? Solution: It would be appropriate to choose a busywaiting solution: (a) when the expected wait time is small, (b) when there is nothing else to do until the condition is satisfied, or (c) when there is no other mechanism available through which to synchronize. The first and third of these points are often true of synchronizing between processors of a multiprocessor.
2. (10) Semaphore waiting lists are often implemented as queues served in FIFO order. Could they be implemented as stacks? What problems might this cause? Solution: Certainly one could implement semaphore waiting lists as stacks, but doing so would make concurrent processes synchronizing via semaphores much more vulnerable to starvation.
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csc/cpe 453
Midterm Solutions
Winter 2003
3. (10) What will be the output of the following program? Explain. int main(int argc, char *argv[]){ printf("Hello.\n"); fork(); fork(); fork(); printf("Goodbye!\n"); return 0; } Solution: % a.out Hello. Goodbye! Goodbye! Goodbye! Goodbye! Goodbye! Goodbye! Goodbye! Goodbye! % This program prints "Hello" once in the parent, and "Goodbye!" in each of the 8 resulting processes. There are 7 children created, not 3, because each of the children executes the following fork()s, too.
parent c1 parent c2 c1 c3 parent "Goodbye!" c4 "Goodbye!" c2 "Goodbye!" c6 "Goodbye!" c1 "Goodbye!" c5 "Goodbye!" c3 "Goodbye!" c7 "Goodbye!" parent "Hello."
4. (10) A programmer dissatisfied with the behavior of a C library function can redefine it without limiting the capabilities of the program. (That is, there is nothing the program could have done before the redefinition that it could not do afterwards.) A system call, however cannot be replaced without limiting the program. Why is this? Solution: It all has to do with privilege: A system call is the means by which the kernel provides access to a particular operating system service, and the services available through system calls (e.g., reading and writing the disk, allocating memory, starting new processes, etc.) are reserved to the kernel; there is no other way of doing these things. Replacing a system call eliminates that particular entry to the kernel, limiting the program. A library call, in contrast, is code that runs entirely within the user's program requiring only the user's privileges. This code can be replaced by a do-ityourselfer (masochist?) without giving up anything irreplaceable.
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csc/cpe 453
Midterm Solutions
Winter 2003
5. (15) Given the following program: int main(int argc, char *argv[]){ int n; for(n = 1; n < argc; n++ ) execvp( argv[n], &(argv[n]) ); return 0; } Solution: Suppose that it is compiled in the current directory as myprog and that one gives the command: % myprog myprog ls myprog myprog myprog (a) How many times will execvp() be called during the run of this program?
execvp() will be called twice.
(b) What will be the expected output of the run?
myprog myprog myprog
(c) Explain.
The program is called with argv containing the list above. Each time around the loop (1...n-1) it will exec() argv[n] with arguments argv[n+1] . . . argv[argc-1]. Because the exec() replaces the current program, the loop never actually loops in any of them. So, the original instance of myprog calls execvp() with "myprog ls myprog myprog myprog" as its arguments. This new instance of myprog will call execvp() with "ls myprog myprog myprog" as its arguments. ls will proceed to list each of its arguments. Since we know myprog is in the current directory, it will be found, so the result of the listing will be "myprog myprog myprog".
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csc/cpe 453
Midterm Solutions
Winter 2003
6. (20) Assume you have hacked the kernel and are using a version of minix with a 1 second(!) quantum. Suppose there are five user-level jobs, labeled A through E , that require the following amounts of time to run (if they could get the whole CPU to themselves). Job Time A 2 sec. B 3 sec. C 4 sec. D 5 sec. E 2 sec. Supposing that these five jobs become runnable at very close to the same time and are initially added to the run queue in alphabetical order, answer the following questions. Recall that within a priority class, minix uses round-robin scheduling. You may assume that these are the only five jobs in the system and that the overhead required for context switches is negligible. (a) After how many seconds will each job terminate? Briefly explain your reasoning.
Solution Time Ending 0.0s 1.0s 2.0s 3.0s 4.0s 5.0s 6.0s 7.0s 8.0s 9.0s 10.0s 11.0s 12.0s 13.0s 14.0s 15.0s 16.0s A -- 1.0s -- -- -- -- 2.0s B -- -- 1.0s -- -- -- -- 2.0s -- -- -- 3.0s C -- -- -- 1.0s -- -- -- -- 2.0s -- -- -- 3.0s -- 4.0s D -- -- -- -- 1.0s -- -- -- -- 2.0s -- -- -- 3.0s -- 4.0s 5.0s E -- -- -- -- -- 1.0s -- -- -- -- 2.0s
Job Finish Time A 6 sec. B 11 sec. C 14 sec. D 16 sec. E 10 sec.
5
A
A
4
B
B
B
3
C
C
C
C
2
D
D
D
D
D
1
E
E
0
0
16
32
48
64
80
96
112
128
144
160
Time (x100.0ms)
(b) Now suppose that job D issues an I/O request after each 1.4s it is allowed run. to This I/O request takes 200ms (0.2s) to complete and causes D to block. Now how long will it take for each job to complete? Again, explain your reasoning.
Solution
5
A
A
Job A B C D E
Finish Time 6.0 sec. 10.4 sec. 13.4 sec. 16.4 sec. 9.4 sec. Solution:
D's first IO is masked by E's execution, but for the second two, there is nothing else to do.
Time End 0.0s 1.0s 2.0s 3.0s 4.0s 5.0s 6.0s 7.0s 8.0s 8.4s 9.4s 10.4s 11.4s 12.4s 13.4s 13.8s 14.0s 15.0s 15.4s 15.6s 16.4s
A -- 1.0s -- -- -- -- 2.0s
B -- -- 1.0s -- -- -- -- 2.0s -- -- -- 3.0s
C -- -- -- 1.0s -- -- -- -- 2.0s -- -- -- 3.0s -- 4.0s
D -- -- -- -- 1.0s -- -- -- -- 1.4s -- -- -- 2.4s -- 2.8s -- 3.8s 4.2s -- 5.0s
E -- -- -- -- -- 1.0s -- -- -- -- 2.0s
4
B
B
B
3
C
C
C
C
2
D
D
D
D
D
D
D
1
E
E
0
IO
IO
IO
0
16
32
49
65
82
98
114
131
147
163
Time (x100.0ms)
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csc/cpe 453
Midterm Solutions
Winter 2003
7. (20) Consider the following situation: A very narrow hurricane has washed out all but one lane of the Lake Pontchartrain Causeway1 . Given that it is a very large lake, going around is impractical, so it is necessary to come up with a system to keep the bridge open. The conditions:
Cars arrive at random intervals from either the north or south. The remains of the bridge are only one car wide and cars cannot back up. That is, a car that meets another car is stuck forever. Whenever a car wants to enter the bridge, it calls the function enter bridge(int direction) with a pre-defined integer constant indicating the direction. This will be either NORTH or SOUTH. When it wants to leave, it calls exit bridge(int dir).
Using semaphores and the C-like syntax used for semaphore examples in class and in Tanenbaum and Woodhull, develop a solution to the problem. Implement enter bridge() and exit bridge() and whatever auxilliary data and functions you may need. Be sure to explain briefly why your solution works.
For partial credit: produce a solution that allows cars to cross the bridge without risking meeting another car on the way (and getting stuck forever). For more partial credit: produce a solution that guarantees that no car will have to wait forever to cross. For full credit: produce a solution that does all of the above and allows multiple cars travelling in the same direction to be on the bridge at a time. (It is 24 miles long, after all.)
Write your code below (and on page 7 if necessary): Solution:
The first two partial credit levels above can be handled by a very simple solution that uses a single semaphore for exclusive access to the bridge. Exclusive access guarantees that there will be no accidents or deadlocks, and the semaphore's own queueing mechanisms guarantee that nobody will have to wait forever. See below. A better solution is presented on page 7.
semaphore mutex; /* initialized to 1 */
void enter bridge(direction d) { /* make sure nobody else is on the bridge */ DOWN(mutex); } void exit bridge(direction d) { /* indicate that the bridge is free */ UP(mutex); }
A semaphore-based solution to problem 7 that allows one car at a time.
1 Lake Pontchartrain is a lake 41 miles long and 24 miles wide north of New Orleans, La. The causeway is the bridge that spans the "short" direction and is one of the two roads out of the city.
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csc/cpe 453
Midterm Solutions
Winter 2003
Extra space for problem 7
typedef int direction; #define NORTH 0 #define SOUTH 1 semaphore mutex; semaphore waitsem[2]; /* initialized to 1 */ /* initialized to 0,0 */
The full solution is much trickier: To meet all three requirements, we must not allow traffic in one direction to hold up traffic in the other direction forever, but we can't just take turns, because a car may never arrive on the other side, and we still have to allow multiple cars to cross in the same direction. The technique is to check to s...