Answer:
- The problem as they see it has been addressed
- The solution proposed will meet their needs
- The needs of their colleagues are being met
- All issues and constraints have been identified
Explanation:
The stakeholders in a project or business involves every human resource of the business/project that is essential to the successful completion of the business/project at hand.
The factors to be addressed by stakeholders when reviewing a problem statement is very important for the success of the project hence : the problem as seen in the problem statement has to be addressed, the solution been proposed to solve every problem identified should meet the needs of every stakeholder, also every stakeholder in a project has to be carried along hence the need of every stakeholder has to be considered as well.
Answer:
backup() {
read dirname;
if [[ whereis . /`$dirname` 2> sterr.exe]]
then
mkdir $dirname
for f in . / *.cpp
do
cp f "path_to_dirname"
echo "file backup complete"
}
backup( )
Explanation:
The bash script above is used to backup C++ source files in a directory to a backup directory which is created if it does not exist, and copy's each .cpp file to backup, then sends a message to declare its completion.
Answer:
For 32 bits Instruction Format:
OPCODE DR SR1 SR2 Unused bits
a) Minimum number of bits required to represent the OPCODE = 3 bits
There are 8 opcodes. Patterns required for these opcodes must be unique. For this purpose, take log base 2 of 8 and then ceil the result.
Ceil (log2 (8)) = 3
b) Minimum number of bits For Destination Register(DR) = 4 bits
There are 10 registers. For unique register values take log base 2 of 10 and then ceil the value. 4 bits are required for each register. Hence, DR, SR1 and SR2 all require 12 bits in all.
Ceil (log2 (10)) = 4
c) Maximum number of UNUSED bits in Instruction encoding = 17 bits
Total number of bits used = bits used for registers + bits used for OPCODE
= 12 + 3 = 15
Total number of bits for instruction format = 32
Maximum No. of Unused bits = 32 – 15 = 17 bits
OPCODE DR SR1 SR2 Unused bits
3 bits 4 bits 4 bits 4 bits 17 bits
Answer:
O(N!), O(2N), O(N2), O(N), O(logN)
Explanation:
N! grows faster than any exponential functions, leave alone polynomials and logarithm. so O( N! ) would be slowest.
2^N would be bigger than N². Any exponential functions are slower than polynomial. So O( 2^N ) is next slowest.
Rest of them should be easier.
N² is slower than N and N is slower than logN as you can check in a graphing calculator.
NOTE: It is just nitpick but big-Oh is not necessary about speed / running time ( many programmers treat it like that anyway ) but rather how the time taken for an algorithm increase as the size of the input increases. Subtle difference.
