Answer:
The basic comaprism of OSSTMN and PTES includes the following: OSSTMN is more theoretical, security assessment methodology, and Metrics based why PTES is technology oriented, penetration testing methodology
, extended analysis of all stages
Explanation:
Solution
Penetration testing has several methodologies which include :OSSTMM and PTES
The comparison between OSSTMM and PTES is stated as follows:
OSSTMM:
Security assessment methodology
More Theoretical
Metrics based
PTES
:
Technology oriented
Penetration testing methodology
Extended analysis of all stages
Now,
There are 7 stages which is used to define PTES for penetration testing.(Penetration Testing Execution Standard)
- Pre-engagement Interactions
Now,
The OSSTMM is used to obtain security metrics and performing penetration testing .The OSSTMM provides transparency to those who have inadequate security policies and configurations.
The OSSTMM includes the entire risk assessment process starting from requirement analysis to report creation.
Six areas are covered by OSSTMM which are:
-
Internet technology security
A burn-in test is a test which is usually performed on a system or component by running it for a long time in order to bring out any errors or system failures etc.
While doing it on CPU the data must be backed up as any kind of error or failure may result in the loss of data, at time systems can be repaired to retrieve data but still there is no guarantee, backing up is the best option.
Answer:
The code is given below
Explanation:
The correct syntax would be to place appropriate parenthesis.
(month==1?"jan":(month==2?"feb":(month==3?"mar":(month==4?"apr":(month==5?"may":(month==6?"jun":(month==7?"jul":(month==8?"aug":(month==9?"sep":(month==10?"oct":(month==11?"nov":"dec")))))))))));
Similarly, you can also use the following code:
String[] months = { "jan", "feb", "mar", "apr", "may", "jun", "jul", "aug", "sep", "oct", "nov", "dec" };
int month = 1;
String monthDescription = months[month - 1];
Answer:
See explaination
Explanation:
StackExample.java
public class StackExample<T> {
private final static int DEFAULT_CAPACITY = 100;
private int top;
private T[] stack = (T[])(new Object[DEFAULT_CAPACITY]);
/**
* Returns a reference to the element at the top of this stack.
* The element is not removed from the stack.
* atreturn element on top of stack
* atthrows EmptyCollectionException if stack is empty
*/
public T peek() throws EmptyCollectionException
{
if (isEmpty())
throw new EmptyCollectionException("stack");
return stack[top-1];
}
/**
* Returns true if this stack is empty and false otherwise.
* atreturn true if this stack is empty
*/
public boolean isEmpty()
{
return top < 0;
}
}
//please replace "at" with the at symbol
Note:
peek() method will always pick the first element from stack. While calling peek() method when stack is empty then it will throw stack underflow error. Since peek() method will always look for first element ffrom stack there is no chance for overflow of stack. So overflow error checking is not required. In above program we handled underflow error in peek() method by checking whether stack is an empty or not.
Answer:
CPU need 50% much faster
disk need 100% much faster
Explanation:
given data
workload spend time CPU = 60%
workload spend time I/O = 40%
achieve overall system speedup = 25%
to find out
How much faster does CPU need and How much faster does the disk need
solution
we apply here Amdahl’s law for the overall speed of a computer that is express as
S = 
.............................1
here f is fraction of work i.e 0.6 and S is overall speed i.e 100% + 25% = 125 % and k is speed up of component
so put all value in equation 1 we get
S =
1.25 = 
solve we get
k = 1.5
so we can say CPU need 50% much faster
and
when f = 0.4 and S = 125 %
put the value in equation 1
S =
1.25 = 
solve we get
k = 2
so here disk need 100% much faster
