Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms
4th Edition
ISBN: 9780534423551
Author: Wayne L. Winston
Publisher: Cengage Learning
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Expert Solution & Answer
Chapter 3, Problem 24RP
Explanation of Solution
Formulation of a Linear
Let “i = 1” is diagnostic services, 2 is bed day service, 3 is nursing care use and 4 is drugs supplied.
Let “xij” be the service “i” and “j” diagnostic-related groups (DRGs), for (i = 1, 2, 3, 4 and j = 1, 2, 3, 4).
The objective is to find the optimal mix of DRGs.
Therefore, the objective function to maximize the profit is,
Constraint 1:
At most, 570 hours of diagnostic service are available.
Constraint 2:
At most, 1000 bed-days are available.
Constraint 3:
At most, 50000 hours of nursing are available.
Constraint 4:
At most, 50000 dollars of drugs are available
Expert Solution & Answer
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Students have asked these similar questions
The next problem concerns the following C code:
/copy input string x to buf */
void foo (char *x) {
char buf [8];
strcpy((char *) buf, x);
}
void callfoo() {
}
foo("ZYXWVUTSRQPONMLKJIHGFEDCBA");
Here is the corresponding machine code on a Linux/x86 machine:
0000000000400530 :
400530:
48 83 ec 18
sub
$0x18,%rsp
400534:
48 89 fe
mov
%rdi, %rsi
400537:
48 89 e7
mov
%rsp,%rdi
40053a:
e8 di fe ff ff
callq
400410
40053f:
48 83 c4 18
add
$0x18,%rsp
400543:
c3
retq
400544:
0000000000400544 :
48 83 ec 08
sub
$0x8,%rsp
400548:
bf 00 06 40 00
mov
$0x400600,%edi
40054d:
e8 de ff ff ff
callq 400530
400552:
48 83 c4 08
add
$0x8,%rsp
400556:
c3
This problem tests your understanding of the program stack. Here are some notes to
help you work the problem:
⚫ strcpy(char *dst, char *src) copies the string at address src (including
the terminating '\0' character) to address dst. It does not check the size of
the destination buffer.
• You will need to know the hex values of the following characters:
1234
3. Which line prevents compiler optimization? Circle one: 1234
Suggested solution:
Store strlen(str) in a variable before the if statement.
⚫ Remove the if statement.
Replace index 0 && index < strlen(str)) {
5 }
}
=
str [index] = val;
Character Hex value | Character Hex value Character Hex value
'A'
0x41
'J'
Ox4a
'S'
0x53
'B'
0x42
'K'
0x4b
"T"
0x54
0x43
'L'
Ox4c
'U'
0x55
0x44
'M'
0x4d
'V'
0x56
0x45
'N'
Ox4e
'W'
0x57
0x46
'0'
Ox4f
'X'
0x58
0x47
'P'
0x50
'Y'
0x59
0x48
'Q'
0x51
'Z'
Ox5a
'T'
0x49
'R'
0x52
'\0'
0x00
Now consider what happens on a Linux/x86 machine when callfoo calls foo with
the input string "ZYXWVUTSRQPONMLKJIHGFEDCBA".
A. On the left draw the state of the stack just before the execution of the instruction
at address Ox40053a; make sure to show the frames for callfoo and foo and
the exact return address, in Hex at the bottom of the callfoo frame.
Then, on the right, draw the state of the stack just after the instruction got
executed; make sure to show where the string "ZYXWVUTSRQPONMLKJIHGFEDCBA"
is placed and what part, if any, of the above return address has been overwritten.
B. Immediately after the ret instruction at address 0x400543 executes, what is
the value of the program counter register %rip?…
Chapter 3 Solutions
Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms
Ch. 3.1 - Prob. 1PCh. 3.1 - Prob. 2PCh. 3.1 - Prob. 3PCh. 3.1 - Prob. 4PCh. 3.1 - Prob. 5PCh. 3.2 - Prob. 1PCh. 3.2 - Prob. 2PCh. 3.2 - Prob. 3PCh. 3.2 - Prob. 4PCh. 3.2 - Prob. 5P
Ch. 3.2 - Prob. 6PCh. 3.3 - Prob. 1PCh. 3.3 - Prob. 2PCh. 3.3 - Prob. 3PCh. 3.3 - Prob. 4PCh. 3.3 - Prob. 5PCh. 3.3 - Prob. 6PCh. 3.3 - Prob. 7PCh. 3.3 - Prob. 8PCh. 3.3 - Prob. 9PCh. 3.3 - Prob. 10PCh. 3.4 - Prob. 1PCh. 3.4 - Prob. 2PCh. 3.4 - Prob. 3PCh. 3.4 - Prob. 4PCh. 3.5 - Prob. 1PCh. 3.5 - Prob. 2PCh. 3.5 - Prob. 3PCh. 3.5 - Prob. 4PCh. 3.5 - Prob. 5PCh. 3.5 - Prob. 6PCh. 3.5 - Prob. 7PCh. 3.6 - Prob. 1PCh. 3.6 - Prob. 2PCh. 3.6 - Prob. 3PCh. 3.6 - Prob. 4PCh. 3.6 - Prob. 5PCh. 3.7 - Prob. 1PCh. 3.8 - Prob. 1PCh. 3.8 - Prob. 2PCh. 3.8 - Prob. 3PCh. 3.8 - Prob. 4PCh. 3.8 - Prob. 5PCh. 3.8 - Prob. 6PCh. 3.8 - Prob. 7PCh. 3.8 - Prob. 8PCh. 3.8 - Prob. 9PCh. 3.8 - Prob. 10PCh. 3.8 - Prob. 11PCh. 3.8 - Prob. 12PCh. 3.8 - Prob. 13PCh. 3.8 - Prob. 14PCh. 3.9 - Prob. 1PCh. 3.9 - Prob. 2PCh. 3.9 - Prob. 3PCh. 3.9 - Prob. 4PCh. 3.9 - Prob. 5PCh. 3.9 - Prob. 6PCh. 3.9 - Prob. 7PCh. 3.9 - Prob. 8PCh. 3.9 - Prob. 9PCh. 3.9 - Prob. 10PCh. 3.9 - Prob. 11PCh. 3.9 - Prob. 12PCh. 3.9 - Prob. 13PCh. 3.9 - Prob. 14PCh. 3.10 - Prob. 1PCh. 3.10 - Prob. 2PCh. 3.10 - Prob. 3PCh. 3.10 - Prob. 4PCh. 3.10 - Prob. 5PCh. 3.10 - Prob. 6PCh. 3.10 - Prob. 7PCh. 3.10 - Prob. 8PCh. 3.10 - Prob. 9PCh. 3.11 - Prob. 1PCh. 3.11 - Show that Finco’s objective function may also be...Ch. 3.11 - Prob. 3PCh. 3.11 - Prob. 4PCh. 3.11 - Prob. 7PCh. 3.11 - Prob. 8PCh. 3.11 - Prob. 9PCh. 3.12 - Prob. 2PCh. 3.12 - Prob. 3PCh. 3.12 - Prob. 4PCh. 3 - Prob. 1RPCh. 3 - Prob. 2RPCh. 3 - Prob. 3RPCh. 3 - Prob. 4RPCh. 3 - Prob. 5RPCh. 3 - Prob. 6RPCh. 3 - Prob. 7RPCh. 3 - Prob. 8RPCh. 3 - Prob. 9RPCh. 3 - Prob. 10RPCh. 3 - Prob. 11RPCh. 3 - Prob. 12RPCh. 3 - Prob. 13RPCh. 3 - Prob. 14RPCh. 3 - Prob. 15RPCh. 3 - Prob. 16RPCh. 3 - Prob. 17RPCh. 3 - Prob. 18RPCh. 3 - Prob. 19RPCh. 3 - Prob. 20RPCh. 3 - Prob. 21RPCh. 3 - Prob. 22RPCh. 3 - Prob. 23RPCh. 3 - Prob. 24RPCh. 3 - Prob. 25RPCh. 3 - Prob. 26RPCh. 3 - Prob. 27RPCh. 3 - Prob. 28RPCh. 3 - Prob. 29RPCh. 3 - Prob. 30RPCh. 3 - Graphically find all solutions to the following...Ch. 3 - Prob. 32RPCh. 3 - Prob. 33RPCh. 3 - Prob. 34RPCh. 3 - Prob. 35RPCh. 3 - Prob. 36RPCh. 3 - Prob. 37RPCh. 3 - Prob. 38RPCh. 3 - Prob. 39RPCh. 3 - Prob. 40RPCh. 3 - Prob. 41RPCh. 3 - Prob. 42RPCh. 3 - Prob. 43RPCh. 3 - Prob. 44RPCh. 3 - Prob. 45RPCh. 3 - Prob. 46RPCh. 3 - Prob. 47RPCh. 3 - Prob. 48RPCh. 3 - Prob. 49RPCh. 3 - Prob. 50RPCh. 3 - Prob. 51RPCh. 3 - Prob. 52RPCh. 3 - Prob. 53RPCh. 3 - Prob. 54RPCh. 3 - Prob. 56RPCh. 3 - Prob. 57RPCh. 3 - Prob. 58RPCh. 3 - Prob. 59RPCh. 3 - Prob. 60RPCh. 3 - Prob. 61RPCh. 3 - Prob. 62RPCh. 3 - Prob. 63RP
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