# How to transform a word problem into its primal and solve with graphical method…

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ENG 401.1: Mathematical Modelling and Operations Research (uniport)

## 2021 Regular exam Question 2

I really do not have much time, but let me squeeze out time to attend to this question so that we would see clearly how to transform a word problem into its objective function and constraints in form of:

Maximize or minimize…….

Subject to…..

## Now, to the question:

An electronics firm manufactures printed circuit boards and specialized electronic devices. Final assembly operations are completed by a small group of trained workers who labor simultaneously on the products. Because of limited space available in the plant, no more than ten assemblers can be employed. The standard operating budget in this functional department allows a maximum of N9000 per month as salaries for the workers. The existing wage structure in the community requires that workers with two or more years of experience receive N1000 per month, while recent trade-school graduates will work for only N800. Previous studies have shown that experienced assemblers produce N2000 in “value added” per month while new-hires add only N1800. To maximize the value added by the group, how many persons from each group should be employed? You are required to graphically solve the problem. (20 marks). Hint: Linear programming.

## SOLUTION:

To solve problems like this, follow these steps:

1.. Identify the variables required to be maximized or minimized. In this case, we are required to maximize the ‘value added’ by the group made up of experience and inexperienced workers.

Now let the number of experienced workers be represented by x,

and the number of inexperienced workers be represented by y.

from the question, the value added by each experienced worker is 2000, hence the total value offered by all the experienced workers will be 2000x.

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Similarly, the total value offered by the inexperienced workers will be 1800y

total value offered by all the workers will be 2000x + 1800y

hence, we now have the OBJECTIVE FUNCTION as:

Maximize V = 2000x + 1800y

Subject to……

We are now done with the first part which is to formulate the objective function, we will now move to the next step which is to find the constraints.

2.. Formulate the constraints by transforming the limitations stated in words into mathematical expressions.

• The first statement to be transformed is this: “no more than ten assemblers can be employed”. Transforming this will give us (x + y) is less or equal to 10. This is our first constraint.
• The second statement to be transformed is this: “a maximum of N9000 per month as salaries for the workers”. Transforming this will give us (1000x + 800y) is less or equal to 9000 since each experienced worker receives 1000 and the inexperienced receives 800. This is our second constraint.

3.. Assembling our objective function and constraints will give:

4.. Draw a combined graph of all the constraints and making sure you shade each line properly according to the inequality signs like this:

From this graph, the point of intersection between constraints 1 and 2, c1 and c2 becomes the solution.

Now at the point of intersection shown above, x = 5, and y = 5.

This means 5 experienced workers and 5 inexperienced workers should be employed in order to maximize value.

And this maximized value will be V = 2000x + 1800y

V = 2000(5) + 1800(5) = 10,000 + 9,000 = N19,000

But we were told that each experienced worker receives N1000, while each inexperienced worker receives N800.

Hence, after salary payments, maximum profit will be 19,000 – 5(1000) – 5(800) = 19,000 – 9000 = N10,000