Answer to Question #255932 in Operations Research for aquero

Question

An electronics manufacturing company has three production plants, each of which produces three different models of a particular MP3 player. The daily capacities (in thousands of units) of the three plants are shown in the table. Basic model Gold model Platinum model Plant 1 8 4 8 Plant 2 6 6 3 TUTORIAL LETTER SEMESTER 2/2021 MATHEMATICS FOR ECONOMISTS 1B MFE512S 12 Plant 3 12 4 8 The total demands are 300,000 units of the Basic model, 172,000 units of the Gold model, and 249,500 units of the Platinum model. The daily operating costs are $50,000 for plant 1, $60,000 for plant 2, and $60,000 for plant 3. How many days should each plant be operated in order to fill the total demand while keeping the operating cost at a minimum? What is the minimum cost? Use the method of the dual

Answer 1

Let plant 1 be operated for $x$ days, plant 2 for $y$ days and plant 3 for $z$ days

The operating cost C will then be: C=$50,000x+60,000y+60,000z$

The objective is to find the minimum operating cost.

The constraints can be represented as:

$8x+6y+12z\ge300,000$.................(I) ​

$4x+6y+4z≥172,000 ​$ ​....................(II)

$8x+3y+8z≥249,500$......................(III)

(I)-(II) gives:

$8x+6y+12z-(8x+3y+8z)\ge300,000-249,500$

$3y+4z\ge50,500$

(II)-(III) gives;

$2(4x+6y+4z)-(8x+3y+8z)\ge2(172,000)-(249,500)$

$9y\ge94,500$

$\therefore y\ge10,500$

Substitute in $3y+4z\ge50,500$

$3(10,500)+4z\ge50,500$

$4z\ge40,000 \therefore z\ge10,000$

Substituting $y$ and $z$ in (I) gives;

$8x+6y+12z\ge300,000$

$8x+6(10,500)+12≥300,000$

$8x\ge117,000$

$x\ge14,625$

Thus $x=14,625$ days

$y=10,500$ days

$z=10,000$ days

These values will meet the full demand and at the same time keep the operating costs at a minimum as plants will be fully utilized.

Minimum cost C=50,000(14,625)+10,500(60,000)+10,000(60,000)=$\$$ 1,961,250,000