The linear programming ingredient or blending problem model allows one to include not only the cost of the resource, but also th
1 answer:
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Solution :
Initial array =
(length of the array)
Iteration:
i = 1
j = 1
= C
a[j] = Q
since < a[j] , break from inner loop
Number of comparisons in 1st Iteration = 1
After 1st Iteration:
Array : C,Q,S,A,X,B,T
2nd Iteration:
i = 2
j = 2
a[j-1] = Q
a[j] = S
since a[j-1] < a[j], break from inner loop
Number of comparisons in 2nd Iteration = 1
After 2nd Iteration:
Array : C,Q,S,A,X,B,T
3rd Iteration:
i = 3
j = 3
a[j-1] = S
a[j] = A
since a[j-1] > a[j], exchange a[2] with a[3]
Array : C,Q,A,S,X,B,T
j = 2
a[j-1] = Q
a[j] = A
since a[j-1] > a[j], exchange a[1] with a[2]
Array : C,A,Q,S,X,B,T
j = 1
a[j-1] = C
a[j] = A
since a[j-1] > a[j], exchange a[0] with a[1]
Array : A,C,Q,S,X,B,T
j = 0, break from inner loop
Number of comparisons in 3rd Iteration = 3
After 3rd Iteration:
Array : A,C,Q,S,X,B,T
4th Iteration:
i = 4
j = 4
a[j-1] = S
a[j] = X
since a[j-1] < a[j], break from inner loop
Number of comparisons in 4th Iteration = 1
After 4th Iteration:
Array : A,C,Q,S,X,B,T
5th Iteration:
i = 5
j = 5
a[j-1] = X
a[j] = B
since a[j-1] > a[j], exchange a[4] with a[5]
Array : A,C,Q,S,B,X,T
j = 4
a[j-1] = S
a[j] = B
since a[j-1] > a[j], exchange a[3] with a[4]
Array : A,C,Q,B,S,X,T
j = 3
a[j-1] = Q
a[j] = B
since a[j-1] > a[j], exchange a[2] with a[3]
Array : A,C,B,Q,S,X,T
j = 2
a[j-1] = C
a[j] = B
since a[j-1] > a[j], exchange a[1] with a[2]
Array : A,B,C,Q,S,X,T
j = 1
a[j-1] = A
a[j] = B
since a[j-1] < a[j], break from inner loop
Number of comparisons in 5th Iteration = 5
After 5th Iteration:
Array : A,B,C,Q,S,X,T
6th Iteration:
i = 6
j = 6
a[j-1] = X
a[j] = T
since a[j-1] > a[j], exchange a[5] with a[6]
Array : A,B,C,Q,S,T,X
j = 5
a[j-1] = S
a[j] = T
since a[j-1] < a[j], break from inner loop
Number of comparisons in 6th Iteration = 2
After 6th Iteration:
Array : A,B,C,Q,S,T,X
Sorted Array : A B C Q S T X
Using the knowledge in computational language in mathlab it is possible to write a code that while loop that continues to increment uservalue by 5 as long as uservalue is less than 0.
<h3>Writting the code:</h3><em>function </em><em>userValue </em><em>= IncreaseValue(userValue)</em>
<em>while(userValue<0)</em>
<em>userValue = </em><em>userValue</em><em>+5;</em>
<em>end</em>
<em>end</em>
<h3>How to run a code in MATLAB?</h3>To run: If inside the Editor window: on the toolbar choose the Run icon; or in the Debug menu choose the Run option; or press F5. If the file is saved in memory, just type its name in the command window.
See more about mathlab at brainly.com/question/12975450
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