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A manufacturer of printed circuits produces two products. Circuit A and Circuit B.

They have a daily stock of: 200 Resistors, 120 Transistors and 150 Capacitors.

**Circuit A** requires 20 Resistors, 10 Transistors and 10 Capacitors.

**Circuit B** requires 10 Resistors, 20 Transistors and 30 Capacitors.

The profit on circuit A is £5 at a selling price of £14

The profit on circuit B is £12 at a selling price of £20

**What output combination should the firm choose in order to maximize profits?**

Required:

Undertake a review of the case and in a report provide a fully developed analysis of the use of the factors of production. Your analysis should include full workings in an appendix, also clear tables and diagrams and a demonstration of how the initial solution might change should the cost of producing Circuit A and B rise.

**Secondary Research Level HE4** – It is expected that the Reference List will contain between **five and ten sources**. As a MINIMUM the Reference List should include **one** refereed academic journal and **two** academic books

**Specific Assessment Criteria:**

The following is a method statement of how the problem will be addressed:

1) Arrange the given information in tabular form. (5 marks)

2) Translate the problem into a linear programming one, identifying and writing down the objective function and the constraints. (15 marks)

3) Plot the inequalities on a graph and identify the feasible region. (20 marks)

4) Find the optimum solution that satisfies the objective function. (15 marks)

5) Calculate and interpret the shadow prices per Transistor and Capacitor. (15 marks)

6) Determine the new objective function, optimum solution and shadow prices, if the cost of producing circuit A and B were to rise and then compare and interpret the results. (30 marks)

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There are three stages to be followed for formulating the problem into linear programming one (Gärtner, Matoušek, 2006). The first step is to define unknowns. The second one is to identify the constraints. The third one is to find the profit maximization (Schrijver, 2003).

- Define the unknowns

Given, the two products have been produced by the manufacturer of printed circuits i.e. circuit A and circuit B. So, we define two unknown variables:

M=numbers of circuit A produced

N=numbers of circuit B produced

- Identify the constraints

The constraints that are available are total numbers of transistors, capacitors and resistors are:

Resistors: if circuit A and circuit B requires 20 and 10 resistors respectively, then

20M+10N≤200

as total 200 resistors are available.

Similarly, constraints for both transistors and capacitors are generated.

Transistors: 10M+20N≤120

Capacitors: 10M+30N≤150

Finally, inequalities stated will be M≥0 and N≥0

- Finding the profit maximization

If circuit A returns profit £ 5 and circuit B returns profit £12, then total profit returns are £G

G= 5M+12N

The problem can now be summarized as:

maximize G= 5M+12N

subject to 20M+10N≤200

10M+20N≤120

10M+30N≤150

where M≥0 and N≥0

The above statement is known as linear programming problem as both the constraints and function G are all linear in M and N.

In order to find a feasible solution for meeting the optimum solution that could satisfy the objective function. The optimum solution can be found out through the graphical method (Garey, Johnson, 1979). In order to plot a graph, we need to find all possible points of the feasible region, profit function at each point and select a feasible region which can provide profit function a maximum value (Kreveld, Overmars, Schwarzkopf, 2000).

- The vertices are calculated from the line 20M+10N=200 i.e. N=(200-20M)/10 are 0(0,20),A(1,18),B(2,16),C(3, 14) and D(4,12).

- The vertices are calculated from the line 10M+20N=120 i.e. N=(120-10M)/20 are 0(0,6),A(1,5.5),B(2,5),C(3,4.5) and D(4, 4).

- The vertices are calculated from the line 10M+30N=150 i.e. N=(150-10M)/30 are 0(0,5),A(1,4.67),B(2,4.33),C(3,4) and D(4, 3.67).

The profit lines given are G= 5M+12N. The maximum profit will be generated in between the line 20M+10N=120, 10M+20N=120, and 10M+30N=150.

The different values of M and N based on which the feasible solution is plotted are shown in appendix 1.

The feasible region at which all the three lines intersect is depicted in Appendix 2.

The family of straight lines as shown in appendix 3 can be plotted as:

C= 5x+12y where C can assume multiple values.