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Hi! can please help me with this question ? Thankyou!
The dean of Faculty of Management and Information Technology must plan the faculty’s course
offerings for the first semester, 2020/2021. Student demands make it necessary to offer
at least 30 undergraduate and 20 graduate courses in the term. Faculty contracts
also dictate that at least 60 courses be offered in total. Each undergraduate course
taught course the faculty an average of RM2,500 in faculty wages, and each graduate course
costs RM3,000. By using the graphical method, calculate the number of undergraduate
and graduate courses should be taught in the first semester so that the total
faculty salaries are kept to a minimum. We let
U = undergraduate courses, and
G = graduate courses.
Note: You might have to prepare a report for proposing an optimal solution by using quantitative
analysis techniques.
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Surely, this problem can be solved using Linear Programming method.
But it also can be solved, using logical reasoning and common sense,
practically mentally, in your mind.
First, notice that if the conditions "at least 30 U and at least 20 G" are satisfied
and we have MORE than 60 courses, than the total number of courses can be diminished
by 1 such way that we still will be in the feasible domain.
Indeed, if U >= 30 and G >= 20, and U + G > 60, it means that at least one number U or G
is greater than their corresponding lower boundaries, so this course can be eliminated.
Repeating this reasoning as many times as required, we will get the condition
U >= 30, G >= 20, U + G = 60
so the total number of courses is precisely 60. Eliminating excessive courses, we diminish
the total cost, so it is good, from the problem's point of view.
Now, if the number of graduate courses is G, then the number of undergraduate courses is U = 60-G.
The total cost is then 2500*U + 3000*G = 2500*(60 - G) + 3000*G = 150000 + 500*G,
and we want to make it as small as it is possible under restrictions.
From the last formula, it is clear that the total cost is minimum when the number of graduate
courses G is at its lower bound G = 20.
Thus the optimal solution is G = 20 graduate courses and U = 60-20 = 40 undergraduate courses,
giving the minimum possible cost 40*2500 + 20*3000 = 100000 + 60000 = RM160,000.
Solved.
So, if the dean of Faculty of Management and Information Technology has a knack for logic,
he can solve this problem MENTALLY, without using the LP-method.
The idea is to assign as few of expensive courses as possible under restrictions,
and then to add as many of cheaper courses to get other restrictions.
It is another view to the problem, which is always good to have, if possible.