EBK BUSINESS DRIVEN TECHNOLOGY
7th Edition
ISBN: 8220103675451
Author: BALTZAN
Publisher: YUZU
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Question
Chapter 2, Problem 1OC
Program Plan Intro
Porter’s Five Forces Model:
To assess the potential for profitability in an industry, the competitive forces within the environment is analyzed by the Porter’s Five Forces.
- Buyer Power – The ability of the buyers to affect the price that they need to pay for an item.
- Supplier Power – The ability to influence the prices that are charged for the suppliers.
- Threat of substitute products or services – It will be high when there are many alternatives to a product or service and it will remain low when there are very few alternatives from which to choose.
- Threat of new entrants – It will be high when it becomes easy for new competitors to enter a market and it will remain low when there are some entry barriers to enter a market.
- Rivalry among existing competitors – It will be high when the competition is fierce in the market and it will remain low when competition becomes more complacent.
Expert Solution & Answer
Explanation of Solution
Analyzing Buyer power and Supplier power for Costco:
- Costco is having many choices for competing stores such as Sam’s club, Wal-Mart, or generic brands at grocery stores so that the buyer power for Costco is also high.
- Whereas the supplier power for Costco is low since the company competes on prices if the supplier chooses to increase prices then Costco will just drop the product and they have reduced supplier power since their warehouse is changing products without any guaranty.
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1 typedef struct node* {
2
struct node* next;
3
char* key;
4
char* val;
5} node_t;
6
7 char* find_node (node_t* node, char* key_to_find) {
while(strcmp (node->key, key_to_find ) != 0 ) {
node = node->next;
8
9
10
}
11
return node->val;
12 }
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
Chapter 2 Solutions
EBK BUSINESS DRIVEN TECHNOLOGY
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