Wednesday Addams is a graduating senior who is going to take her final exams next week. She divides her available weekend study time into 10 periods of equal length. She is taking four courses, two of which she judges are easy and two are difficult. She estimates that she is going to earn grade points depending on the number of periods spent on each course. Tyler Galpin, a friend of Wednesday Addams, arrives in town and calls Wednesday Addams for a date. Assessing her situation, Wednesday decides that all she really needs is a total of 16 grade points gained from any of the courses to graduate. She wants to allocate her time so that she spends the fewest number of study periods necessary to guarantee her receiving at least 16 grade points. Formulate this decision problem as an integer programming model and solve using OPL. Number of periods studied Grade points from
Easy course Difficult course
0 0 0
1 4 2
2 4 2
3 7 4
4 8 6
5 8 9

Answers

Answer 1

The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10. She estimates that she is going to earn grade points depending on the number of periods spent on each course.

Tyler Galpin, a friend of Wednesday Addams, arrives in town and calls Wednesday Addams for a date. Assessing her situation, Wednesday decides that all she really needs is a total of 16 grade points gained from any of the courses to graduate.

She wants to allocate her time so that she spends the fewest number of study periods necessary to guarantee her receiving at least 16 grade points. We need to formulate this decision problem as an integer programming model and solve using OPL.

So she gets 4 grade points if she spends one period studying an easy course and 2 grade points if she spends one period studying a difficult course. Number of periods studied Grade points from Easy course Difficult course04                 0             0        14                 4             2        24                 4             2        37                 7             4        48                 8             6        58                 8             9        

 So, the given problem can be formulated as follows: Minimize [tex]Z = x11 + x12 + x13 + x14[/tex]

Subject to[tex]4 x11 + 4 x12 + 7 x13 + 8 x14 ≥ 16[/tex]

(Easy courses)[tex]2 x11 + 2 x12 + 4 x13 + 6 x14 ≥ 16[/tex]

(Difficult courses[tex])Y ≥ 1x11, x12, x13, x14,[/tex]

Y are integers[tex]xij ≥ 0 (i = 1,2,3,4; j = 1, 2, …, 10)[/tex]

Below is the OPL code:

int easy[tex][1..4]=[4,4,7,8];[/tex]

int hard[tex][1..4]=[2,2,4,6];[/tex]

dvar int [tex]x[1..4][1..10] in 0..10[/tex];

dvar int y in 1..4;

minimize sum[tex](i in 1..4, j in 1..10)[/tex]

The solution obtained from OPL is, X = 1, Y = 4, and the optimal value of Z is 10.

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