EBK ESSENTIALS OF MIS,
13th Edition
ISBN: 8220106778494
Author: LAUDON
Publisher: PEARSON
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Chapter 12, Problem 2RQ
Explanation of Solution
Alternative methods of building
An information system is a system which provides the necessary information to an organization to achieve the goals. The following alternative methods are using to building an information system...
Explanation of Solution
Traditional systems life cycle:
It is a formal methodology for managing the development of systems in development life cycle. It has the six step process to develop the system. They are:
- Analyzing the
system - Design the system
Programming - Test the system
- Implementation of system
- Maintenance of system.
Advantages for systems building:
The advantages of traditional system life cycle for system building is as follows:
- It is simple to implement and easy to understand...
Explanation of Solution
Information system prototyping:
It is an alternative model for building information system. It has a prototype which represents the requirements of the user.
Benefits of information system prototyping:
- The benefit of information system prototyping is exactly matches to the customer requirements and these model is mainly used to enhance the bigger projects.
- It is useful for routine problems and this information system is very structured and well-understood.
- It reduces the implementation cost of the project.
Limitations of information system prototyping:
- It is not appropriate for massive amounts of data...
Explanation of Solution
End-user development:
End-user development of information system is refers the programmers and system analysts. It gives the direct control to the project. By using the end-user development model, the project is rapidly developed with standard programming tools.
Advantages of end-user development:
- It takes minimal time to finish the software and the customer requirements in project.
- The tools are used by professional programmers in productivity levels...
Explanation of Solution
Application software packages and cloud software services (SaaS):
- The software packages is common tool for all business organizations for its daily process like payroll, inventory control, etc. Many organizations use cloud services for its higher security.
- SaaS is one of the cloud software services in cloud computing. In SaaS, the software is deployed from a cloud provider; internet is used to deliver the service and the subscribers access the service through the browser.
Advantages:
- Using application software packages, the organization got a solution in quick and easy way...
Explanation of Solution
Outsourcing:
Outsourcing means taking a company for contract to complete the specific task of some other company project using the in-house employees.
Advantages of outsourcing:
- Company takes specialized skills.
- Provides labor flexibility.
- Cost effective.
Drawbacks of outsourcing:
- Increases the confidence on third parties.
- Lacking the knowledge of in-house for complex business operations.
Favorable circumstances for the outsourcing strategy are as given below:
Lack in experience:
- Whenever there are less experienced professionals in any organization, then for a new system design, the organization always prefer the outsourcing strategy.
- Hence, the product is designed by the other firm.
Inadequate resources:
- If an organization does not have adequate resources, then outsourcing of the project must be done.
- Organization can hire different vendors, developers, or suppliers to meet their needs...
Explanation of Solution
Explanation of developing e-business applications:
- Developing e-business application is a needed one for every business for its development. RAD is the technology of process for developing systems in short period of time...
Explanation of Solution
Development of the mobile application:
- Smart phone tend to become more important part in every person’s day to day life routine.
- Every process required to be made in a smart manner and it is considered to be more important in case of the business process.
- Every customer requires process that is made should be effective a...
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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 d1 fe ff ff
40053f:
48 83 c4 18
add
callq 400410
$0x18,%rsp
400543:
c3
retq
0000000000400544 :
400544:
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:
A ROP (Return-Oriented Programming) attack can be used to execute arbitrary
instructions by chaining together small pieces of code called "gadgets." Your goal is
to create a stack layout for a ROP attack that calls a function located at '0x4018bd3'.
Below is the assembly code for the function 'getbuf', which allocates 8 bytes of stack
space for a 'char' array. This array is then passed to the 'gets' function. Additionally,
you are provided with five useful gadgets and their addresses. Use these gadgets to
construct the stack layout.
Assembly for getbuf
1 getbuf:
2
sub
$8, %rsp
3
mov
%rsp, %rdi
4
call
gets
56
add
$8, %rsp
ret
#Allocate 8 bytes for buffer
#Load buffer address into %rdi
#Call gets with buffer
#Restore the stack pointer
#Return to caller.
Stack
Layout
(fill
in
Gadgets
each
8-byte
section)
Address
Gadget
Address
Value (8 bytes)
0x4006a7 pop %rdi; ret
0x7fffffffdfc0
Ox4006a9 pop %rsi; ret
0x7fffffffdfb8
0x4006ab pop %rax; ret
0x7fffffffdfb0
0x7fffffffdfa8
Ox4006ad mov %rax,…
In each of the following C code snippets, there are issues that can prevent the compilerfrom applying certain optimizations. For each snippet:• Circle the line number that contains compiler optimization blocker.• Select the best modification to improve optimization.1. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Remove printf or move it outside the loop.• Remove the loop.• Replace arr[i] with a constant value.1 int sum( int ∗ ar r , int n) {2 int s = 0 ;3 for ( int i = 0 ; i < n ; i++) {4 s += a r r [ i ] ;5 p r i n t f ( ”%d\n” , s ) ;6 }7 return s ;8 }2. Which line prevents compiler optimization? Circle one: 2 3 4 5 6Suggested solution:• Move or eliminate do extra work() if it’s not necessary inside the loop.• Remove the loop (but what about scaling?).• Replace arr[i] *= factor; with arr[i] = 0; (why would that help?).1 void s c a l e ( int ∗ ar r , int n , int f a c t o r ) {2 for ( int i = 0 ; i < n ; i++) {3 a r r [ i ] ∗= f a c t o r…
Chapter 12 Solutions
EBK ESSENTIALS OF MIS,
Ch. 12.2 - Prob. 1CQ1Ch. 12.2 - Prob. 2CQ1Ch. 12.2 - Prob. 3CQ1Ch. 12.4 - Prob. 1CQ2Ch. 12.4 - Classify and describe the people, organization,...Ch. 12.4 - Prob. 3CQ2Ch. 12.4 - Why do you think the implementation of Snohomish...Ch. 12 - Prob. 1IQCh. 12 - Prob. 2IQCh. 12 - Prob. 3IQ
Ch. 12 - Prob. 4IQCh. 12 - Prob. 5IQCh. 12 - What are the core problem-solving steps for...Ch. 12 - Prob. 2RQCh. 12 - Prob. 3RQCh. 12 - Prob. 5DQCh. 12 - Prob. 6DQCh. 12 - Prob. 7DQCh. 12 - The Warm and Toasty Heating Oil Company used to...Ch. 12 - Prob. 9HMPCh. 12 - Prob. 11HMPCh. 12 - Prob. 12CSQCh. 12 - What problems was the Philly311 project designed...Ch. 12 - Prob. 14CSQCh. 12 - Prob. 15CSQCh. 12 - Prob. 18MLMCh. 12 - Prob. 19MLM
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Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, computer-science and related others by exploring similar questions and additional content below.Similar questions
- 123456 A ROP (Return-Oriented Programming) attack can be used to execute arbitrary instructions by chaining together small pieces of code called "gadgets." Your goal is to create a stack layout for a ROP attack that calls a function located at 'Ox4018bd3'. Below is the assembly code for the function 'getbuf, which allocates 8 bytes of stack space for a 'char' array. This array is then passed to the 'gets' function. Additionally, you are provided with five useful gadgets and their addresses. Use these gadgets to construct the stack layout. Assembly for getbuf 1 getbuf: sub mov $8, %rsp %rsp, %rdi call gets add $8, %rsp 6 ret #Allocate 8 bytes for buffer #Load buffer address into %rdi #Call gets with buffer #Restore the stack pointer #Return to caller Stack each Layout (fill in Gadgets 8-byte section) Address Gadget Address Value (8 bytes) 0x7fffffffdfc0 0x7fffffffdfb8 0x7fffffffdfb0 0x7fffffffdfa8 0x7fffffffdfa0 0x7fffffffdf98 0x7fffffffdf90 0x7fffffffdf88 Original 0x4006a7 pop %rdi;…arrow_forwardCharacter Hex value || Character Hex value | Character Hex value 'A' 0x41 יני Ox4a 'S' 0x53 0x42 'K' 0x4b 'T" 0x54 0x43 'L' Ox4c 0x55 0x44 'M' Ox4d 0x56 0x45 'N' Ox4e 'W' 0x57 0x46 Ox4f 'X' 0x58 0x47 'P' 0x50 'Y' 0x59 'H' 0x48 'Q' 0x51 'Z' Охба '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? (That is…arrow_forwardDraw out the way each of these structs looks in memory, including padding! Number the offsets in memory. 1 struct okay Name 2 { short a; 3 4 long number; 5 int also_a_number; 6 7 }; char* text; 1 struct badName 2 { 3 4 5 }; short s; struct okay Name n;arrow_forward
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