MATERIALS SCIENCE AND ENGINEERING: INTRO
10th Edition
ISBN: 9781119571308
Author: Callister
Publisher: WILEY
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Chapter 5, Problem 13QAP
To determine
The carburizing time.
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I need expert to solve this,if you don't know don't attempt,also show the signal flow diagram
Suppose we alter do stuff to look as follows:
void *do_stuff(void *arg attribute ((unused))) {
int a;
for (a = 0; a < 1000; a++)
i++;
return NULL;
}
Because the code is not optimized, there will be one load-increment-store sequence per iteration of the
loop.
B.
For each number, tell whether or not our program could output it, and briefly explain why
or why not.
• 2000
• 1500
.
2
1
The following problem concerns the way virtual addresses are translated into physical addresses.
Thememory is byte addressable.
• Memory accesses are to 1-byte words (not 4-byte words).
• Virtual addresses are 16 bits wide.
• Physical addresses are 13 bits wide.
•The page size is 512 bytes.
• The TLB is 8-way set associative with 16 total entries.
• The cache is 2-way set associative, with a 4 byte line size and 16 total lines.
In the following tables, all numbers are given in hexadecimal. The contents of the TLB, the page table for the first 32 pages, and the cache
are as follows:
TLB
Page Table
2-way Set Associative Cache
Index
Tag
PPN Valid
VPN PPN Valid VPN PPN Valid
Index Tag Valid Byte 0 Byte 1 Byte 2 Byte 3 Tag Valid Byte 0 Byte 1 Byte 2 Byte 3
0
09
4
1
00
6
1
10 0
1
0
19
1
99
11
23
11
00 0 99
11 23 11
12
2
1
01
5
0
11
5
0
1
15
0
4F
22
EC
11
2F 1
55
59
OB
41
10
0
1
02
3
1
12
2
1
2
1B
1
00
02
04
08
OB
1
01
03
05
07
08
5
03
4
1
13
4
0
3
06
0
84
06
B2
9C
12
0
84
06 B2 9C
05
7
1
04…
Chapter 5 Solutions
MATERIALS SCIENCE AND ENGINEERING: INTRO
Ch. 5 - Prob. 1QAPCh. 5 - Prob. 2QAPCh. 5 - Prob. 3QAPCh. 5 - Prob. 4QAPCh. 5 - Prob. 5QAPCh. 5 - Prob. 6QAPCh. 5 - Prob. 7QAPCh. 5 - Prob. 8QAPCh. 5 - Prob. 10QAPCh. 5 - Prob. 12QAP
Ch. 5 - Prob. 13QAPCh. 5 - Prob. 14QAPCh. 5 - Prob. 15QAPCh. 5 - Prob. 17QAPCh. 5 - Prob. 18QAPCh. 5 - Prob. 19QAPCh. 5 - Prob. 20QAPCh. 5 - Prob. 21QAPCh. 5 - Prob. 24QAPCh. 5 - Prob. 26QAPCh. 5 - Prob. 27QAPCh. 5 - Prob. 31QAPCh. 5 - Prob. 32QAPCh. 5 - Prob. 34QAPCh. 5 - Prob. 35QAPCh. 5 - Prob. 1DPCh. 5 - Prob. 2DPCh. 5 - Prob. 3DPCh. 5 - Prob. 4DPCh. 5 - Prob. 1SSPCh. 5 - Prob. 2SSPCh. 5 - Prob. 3SSPCh. 5 - Prob. 4SSPCh. 5 - Prob. 1FEQPCh. 5 - Prob. 2FEQP
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- Part II (8 points) int counter = 0; void handler(int sig) { counter++; } int main() { int i; signal(SIGCHLD, handler); for (i = 0; i < 5; i ++){ if (fork() == 0){ } } exit(0); /* wait for all children to die */ while (wait(NULL) != -1); printf("counter = %d\n", counter); return 0; } A. Does the program output the same value of counter every time we run it? Yes No B. If the answer to A is Yes, indicate the value of the counter variable. Otherwise, list all possible values of the counter variable. Answer: counter =arrow_forwardA useful helper function for opening a socket connection to a server was introduced in the course: int open_clientfd(char *hostname, int port) { int clientfd; struct hostent *hp; struct sockaddr_in serveraddr; if ((clientfd = socket(AF_INET, SOCK_STREAM, 0)) h_addr_list[0], (char *)&serveraddr.sin_addr.s_addr, hp->h_length); serveraddr.sin_port = htons(port); /* Establish a connection with the server */ if (connect(clientfd, (SA *) &serveraddr, sizeof(serveraddr)) < 0) return -1; return clientfd; } The problem with this function is that it is not thread safe: it uses function gethostbyname, which returns a pointer to a static variable. On the following page, reimplement open_clientfd in a thread safe way using the lock-and-copy technique. (Note 1: that the new function will take an additional input: a semaphore that will control access to the critical section so only one thread at a time can call gethostbyname and read the result stored in a static variable. Note 2: a call to malloc is…arrow_forwardA. The box below shows the format of a virtual address. Indicate (by labeling the diagram) the fields (if they exist) that would be used to determine the following: (If a field doesn't exist, don't draw it on the diagram.) VPO The virtual page offset VPN The virtual page number TLBI The TLB index TLBT The TLB tag 15 14 13 12 11 10 98 76543210 B. The box below shows the format of a physical address. Indicate (by labeling the diagram) the fields that would be used to determine the following: PPO The physical page offset PPN The physical page number CO The block offset within the cache line The cache index Cl ст The cache tag 12 11109876543210arrow_forward
- Suppose we have a program as follows: #include #include int i = 0; void *do_stuff(void *arg) { } i++; return NULL; int main() { } pthread_t tid1, tid2; pthread_create(&tid1, NULL, do_stuff, NULL); pthread_create(&tid2, NULL, do_stuff, NULL); pthread_join(tid1, NULL); pthread_join(tid2, NULL); printf("%d\n", i); return 0; Recall that because i is a global variable, i++; will compile to something like this: 400728: 8b 04 25 40 10 60 00 mov 0x601040,%eax 40072f: 83 c0 01 add $0x1,%eax 400732: 89 04 25 40 10 60 00 mov %eax,0x601040 A. (8 points) What are all possible outputs of this program? For each output, explain how the kernel could interleave execution of the two child threads to produce it.arrow_forwardFor the given virtual address, indicate the TLB entry accessed, the physical address, and the cache byte value returned in hex. Indicate whether the TLB misses, whether a page fault occurs, and whether a cache miss occurs. If there is a cache miss, enter "-" for "Cache Byte returned". If there is a page fault, enter "-" for "PPN" and leave parts C and D blank. Virtual address: 1DDE A. Virtual address format (one bit per box). 15 14 13 12 11 10 9 8 7 6 4 3 2 1 0 B. Address translation Parameter Value VPN Ox TLB Index Ox TLB Tag Ox TLB Hit? (Y/N) Page Fault? (Y/N) PPN Ox C. Physical address format (one bit per box) 12 11 10 9 8 765 D. Physical memory reference Parameter Value Byte offset Ox Cache Index Ox Cache Tag Ox Cache Hit? (Y/N) Cache Byte returned Ox 4 3 2 1 0arrow_forwardConsider the C program below. (For space reasons, we are not checking error return codes, so assume that all functions return normally.) int main() { if (fork() == 0) { if (fork() == 0) { printf("3"); } pid_t pid; int status; wait(&status)) > 0) { printf("4"); } else { if ((pid } } } else { } printf("2"); exit(0); printf("0"); return 0; For each of the following strings, circle whether (Y) or not (N) this string is a possible output of the program. A. 32040 Y N B. 34002 Y N C. 30402 Y D. 23040 E. 40302 Y > > Y N Z Z Z N Narrow_forward
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