1. Proper planning
2. Honesty
3. Being reasonable in pricing
4. Knowledge of what you’re offering
5. Being kind/good to your clients
Answer:
its A. the answer to your question is preserved. :)
Answer:
a) Speedup gain is 1.428 times.
b) Speedup gain is 1.81 times.
Explanation:
in order to calculate the speedup again of an application that has a 60 percent parallel component using Anklahls Law is speedup which state that:
Where S is the portion of the application that must be performed serially, and N is the number of processing cores.
(a) For N = 2 processing cores, and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.428 times.
(b) For N = 4 processing cores and a 60%, then S = 40% or 0.4
Thus, the speedup is:
Speedup gain is 1.81 times.
Answer:
See explaination for the program code
Explanation:
The code below
Pseudo-code:
//each item ai is used at most once
isSubsetSum(A[],n,t)//takes array of items of size n, and sum t
{
boolean subset[n+1][t+1];//creating a boolean mtraix
for i=1 to n+1
subset[i][1] = true; //initially setting all first column values as true
for i = 2 to t+1
subset[1][i] = false; //initialy setting all first row values as false
for i=2 to n
{
for j=2 to t
{
if(j<A[i-1])
subset[i][j] = subset[i-1][j];
if (j >= A[i-1])
subset[i][j] = subset[i-1][j] ||
subset[i - 1][j-set[i-1]];
}
}
//returns true if there is a subset with given sum t
//other wise returns false
return subset[n][t];
}
Recurrence relation:
T(n) =T(n-1)+ t//here t is runtime of inner loop, and innner loop will run n times
T(1)=1
solving recurrence:
T(n)=T(n-1)+t
T(n)=T(n-2)+t+t
T(n)=T(n-2)+2t
T(n)=T(n-3)+3t
,,
,
T(n)=T(n-n-1)+(n-1)t
T(n)=T(1)+(n-1)t
T(n)=1+(n-1)t = O(nt)
//so complexity is :O(nt)//where n is number of element, t is given sum
you can install a 64 bit operating system on a 32 bit machine.
hope this helps!
