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 6.5, Problem 3P
Explanation of Solution
Duals of LP:
- Consider the linear
programming problem given below: - max z=
- Subject to constraints
- Dual of the above LPP can be found with the use of three steps:
- Step1:
- Construct the following table according to the given LP,
| Max Z | |||||
| Min W | |||||
| 1 | 1 | 0 | |||
| 2 | 1 | 0 | |||
| 0 | 2 | 1 | |||
| 1 | 0 | 1 | |||
- Step 2:
- From the above given table, it can be said that the third primal constraint is
- From the above given table, it can be said that the third primal constraint is
Expert Solution & Answer
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Students have asked these similar questions
Match each of the assembler routines on the left with the equivalent C function on
the right. Write the name of the label (e.g., foo) to the right of the corresponding
function. Note: shrq is the logical right shift instruction, and sarq is the arithmetic
right shift instruction.
foo1:
leaq
0(,%rdi, 8), %rax
long choice1 (long x)
{
ret
return x -
8 >8;
foo3:
}
movq
sarq
%rdi, %rax
$8, %rax
long choice4 (long x)
ret
{
return x*256;
}
foo4:
long choice5 (long x)
leaq
-8 (%rdi), %rax
{
ret
return x-8;
}
long choice6 (long x)
foo5:
{
leaq
-8 (%rdi), %rax
return x+8;
shrq
$63, %rax
}
ret
Given the variables and code in the text below, identify where in memory they will
live once the code is compiled.
1 char
big_array [1L<<24]; /* 16 MB */
2 GB *
:/
2 char huge_array [1L<<31]; /*
3
4 int global = 0;
5
6 int useless () { return 0; }
7
8 int main()
9 {
10
void *p1, p2, *p3, *p4;
int local =
0;
malloc (1L << 28); /* 256 MB *,
11
12
p1
13
p2
=
malloc (1L << 8);
/* 256 B *
14
p3
15
p4
=
malloc (1L << 32);
malloc (1L << 8);
/* 4
GB *
*/
/* 256
B */
16 }
Note: *pN is the thing at which pN points.
1. big_array
2. huge_array
3. global
4. useless
5. void* p1
6. *p1
7. void* p2
8. *p2
9. void* p3
10. *p3
11. void* p4
12. *p4
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:
Chapter 6 Solutions
Student Suite Cd-rom For Winston's Operations Research: Applications And Algorithms
Ch. 6.1 - Prob. 1PCh. 6.1 - Prob. 2PCh. 6.1 - Prob. 3PCh. 6.1 - Prob. 4PCh. 6.1 - Prob. 5PCh. 6.2 - Prob. 1PCh. 6.2 - Prob. 2PCh. 6.3 - Prob. 1PCh. 6.3 - Prob. 2PCh. 6.3 - Prob. 3P
Ch. 6.3 - Prob. 4PCh. 6.3 - Prob. 5PCh. 6.3 - Prob. 6PCh. 6.3 - Prob. 7PCh. 6.3 - Prob. 8PCh. 6.3 - Prob. 9PCh. 6.4 - Prob. 1PCh. 6.4 - Prob. 2PCh. 6.4 - Prob. 3PCh. 6.4 - Prob. 4PCh. 6.4 - Prob. 5PCh. 6.4 - Prob. 6PCh. 6.4 - Prob. 7PCh. 6.4 - Prob. 8PCh. 6.4 - Prob. 9PCh. 6.4 - Prob. 10PCh. 6.4 - Prob. 11PCh. 6.4 - Prob. 12PCh. 6.4 - Prob. 13PCh. 6.5 - Prob. 1PCh. 6.5 -
Find the duals of the following LPs:
Ch. 6.5 - Prob. 3PCh. 6.5 - Prob. 4PCh. 6.5 - Prob. 5PCh. 6.5 - Prob. 6PCh. 6.6 - Prob. 1PCh. 6.6 - Prob. 2PCh. 6.7 - Prob. 1PCh. 6.7 - Prob. 2PCh. 6.7 - Prob. 3PCh. 6.7 - Prob. 4PCh. 6.7 - Prob. 5PCh. 6.7 - Prob. 6PCh. 6.7 - Prob. 7PCh. 6.7 - Prob. 8PCh. 6.7 - Prob. 9PCh. 6.8 - Prob. 1PCh. 6.8 - Prob. 2PCh. 6.8 - Prob. 3PCh. 6.8 - Prob. 4PCh. 6.8 - Prob. 5PCh. 6.8 - Prob. 6PCh. 6.8 - Prob. 8PCh. 6.8 - Prob. 9PCh. 6.8 - Prob. 10PCh. 6.8 - Prob. 11PCh. 6.9 - Prob. 1PCh. 6.9 - Prob. 2PCh. 6.9 - Prob. 3PCh. 6.10 - Prob. 1PCh. 6.10 - Prob. 2PCh. 6.10 - Prob. 3PCh. 6.11 - Prob. 1PCh. 6.11 - Prob. 3PCh. 6.11 - Prob. 4PCh. 6.12 - Prob. 5PCh. 6.12 - Prob. 6PCh. 6.12 - Prob. 7PCh. 6 - Prob. 1RPCh. 6 - Prob. 2RPCh. 6 - Prob. 3RPCh. 6 - Prob. 4RPCh. 6 - Prob. 5RPCh. 6 - Prob. 6RPCh. 6 - Prob. 7RPCh. 6 - Prob. 8RPCh. 6 - Prob. 9RPCh. 6 - Prob. 10RPCh. 6 - Prob. 11RPCh. 6 - Prob. 13RPCh. 6 - Prob. 14RPCh. 6 - Prob. 15RPCh. 6 - Prob. 17RPCh. 6 - Prob. 18RPCh. 6 - Prob. 19RPCh. 6 - Prob. 20RPCh. 6 - Prob. 21RPCh. 6 - Prob. 22RPCh. 6 - Prob. 25RPCh. 6 - Prob. 29RPCh. 6 - Prob. 33RPCh. 6 - Prob. 34RPCh. 6 - Prob. 35RPCh. 6 - Prob. 36RPCh. 6 - Prob. 37RP
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