Minimize cost function with constraint

Question

I have an optimization question that I'm not sure if I interpreted correctly.

A hospital wants to determine a drug amount in a patient's urine using an instrument/method. The maximum total inaccuracy must be less than 0.25 μg for at least 95% of the individual measurements. The current instrument method has a bias of 0.2 μg and a standard deviation of 0.1 μg. The analytical chemist estimates that the bias can never be less than 0.1 μg, and the lab will have to spend \$1,000 per 0.01 μg of bias reduced. The chemist also estimates that the standard deviation can never be less than 0.03 μg, and the lab will have to spend \$3,500 per 0.01 μg of standard deviation reduced. What is the least expensive way to validate this method/instrument, and what will be the cost?

Here is my current solution:

Define $x$ as the amount of μg the lab wants to reduce for bias and $y$ as the amount of μg for reduction in standard deviation.

In a normal distribution, 95% of values are within 1.96 standard deviations of the mean.

Our goal is to minimize $f(x,y) = 1000x + 3500y$ with the following three constraints:

$0.2 - 0.01x + 1.96*(0.1-0.01y) < 0.25$

$0.2-0.01x > 0.1$

$0.1-0.01y > 0.03$

We can rewrite the three constraints as:

$x + 1.96y > 14.6$

$x < 10$

$y < 7$

I plotted the three inequalities to get the following: plot

The inequality plot shows three vertices, one of which is the solution. We are left with three possibilities of $(x,y)$:

If $y = 7, x = 0.88$ as given by our first constraint. If $x = 10, y = 2.34$, and we can also have $x=10, y=7$

So our three x,y pairs are: (0.88, 7), (10, 2.34), (10,7). Since we are trying to minimize, it can’t be the last pair.

We compare the first two:

$0.88*1000 + 7 * 3500 = \$25,380$

$10*1000 + 2.34 * 3500 = \$18,190$

The cheapest way is thus the 2nd way, by reducing bias by 10 increments of 0.01 μg to 0.1 μg, and reducing the standard deviation by 2.34 increments of 0.01 μg to 0.0766 μg

Am I on the right track?

Answer 1

With two dimensional problems as that you can use the graphical representation to verify the solutions quality. In the attached figure we can observe the feasible region in light blue and the trace for the objective function

$$ f = 1000 x+3500 y $$

so choosing suitable values of $f$ we have in red, orange and blue the trace passing by the three vertex. Clearly the blue colored is the solution.

Answer 2

As far as I can see, you are on the right track. Goal is linear, so we should only check the boundaries. You then just check the points and correctly select the minimal one.