This is the follow-up of yesterday’s post on “Maximizing Savings by Linear Programming- Part 1“.
The problem at hand:
Suppose I am a student earning Rs 20,000 per month from three different part-time jobs and I save a fixed amount per month for a holiday I am planning. Suppose I want to increase my savings so that I can have a bit left off as an emergency fund, without having to make a major reduction in expenditure. So my problem is: How to maximize this saving?
Applying the concepts of Linear Programming,
I can segregate my expenses as “higher range” and the “lower range”.
Let X1 and X2 be the basic expenditure on food and miscellany respectively, both of which falls under “low range of expenses”.
The expenditure on rent is 2 times that of food, i.e., 2X1 and the medical/ educational/ transport being grouped under one class is 3 times that of miscellany, i.e., 3X2.
The following depicts my spending trend for the past 2-3 months:
2X1 + 3X2 ≤ (4000+9000) IN THE HIGHER RANGE.
X1 + X2 ≤(2000+ 3000) IN THE LOWER RANGE.
My total expenses being about Rs 18,000 per month, I am able to save at least 2000 every month. Now this amount is enough for my holidays at the end of the year. But to have something left off for an emergency fund, I need to maximize my savings.
How to minimize ‘Z’ (expenditure) = X1 +2X1+ X2+3X2 or 3X1 +4X2,
subject to constraints: 2X1 + 3X2 ≤ (4000+9000) and X1 + X2 ≤(2000+ 3000). And both X1 and X2 ≥ 0.
Solving: Z when X1 (HORIZONTAL) and X2 (VERTICAL) is (0, 4.5) and 1 is equivalent to Rs 1000 on the scale.
Z= 3X1+ 4X2 = 0 + 4×4.5 = 18,000 (SAME AS MY CURRENT EXPENDITURE)
AND Z when X1 AND X2 is (5, 0)
Z= 3X1+ 4X2 = 3×5 + 0 = 15,000
and Z at (0, 0) is 0
Conclusion: After solving for Z from the vertices of the feasible zone, if I can keep my expenses to 15,000 per month, I will be saving the maximum under the given set-up. and can have an emergency fund without disturbing my current savings of Rs 2,000 per month for a holiday at the end of the year.
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©The Idea Bucket, 2013