Answer to Question #190314 in Operations Research for Litgebew

Question

1.  The Eastern Iron and steel company makes nails, bolts, and washers from steel and coats them with zinc. The company has 24 tons of steel and 30 tons of zinc. The following table gives detailed information on the objective of the company and items production. (6 Points)

Products

Steel used in ton per unit

 Zinc used in tons/unit

Profit per unit

Nails (X1)

4

2

6

Bolts (X2)

1

6

2

Washer (X3)

3

3

12

Total Resource available

24

30

 

a)     Formulate the model for LP

b)     Solve the LPM using the simplex algorithm.

c)     If the company plans to make only nails and bolts with the existed steel and zinc, Solve the LPM using a graphical method.

d)    Interpret the optimal solution. 

Answer 1

(a) We have the given conditions,

Maximize $Z = 6x_1+2x_2+12x_3$

Subject to the given conditions,

$2x_1+6x_2+3x_3\le 30$ (zinc, tons)

$4x_1+x_2+3x_3\le 24$ (steel, tons)

$x_1,x_2,x_3\ge0$

(b)

Now we have to solve this model using the simple method.

We can clearly see that the last column have some negative numbers.

Hence we can apply some row transformations in the last row.

$R_2\rightarrow \dfrac{1}{3}R_2$

$R_1\rightarrow R_1-3R_2$

$R_3 \rightarrow R_3+12R_2$

Hence after applying these transformations we get a new table which is given below,

From this table we can clearly obtain the values of $x_1,x_2$ and $x_3$

$x_1 = 0$

$x_2= 0$

$x_3 = 8$

$Z = 96$

Hence, the maximum profit we can obtain is 96.

(c) Now according to conditions:

Maximize $Z = 6x_1+2x_2$

Subject to the given conditions,

$2x_1+6x_2\leq 30$ (zinc, tons)

$4x_1+x_2\le 24$ (steel, tons)

$x_1,x_2,\ge0$

So, Using Graphical Method:

 To draw constraint$2x_1+6x_2≤30→(1)$

Treat it as$2x_1+6x_2=30$

When $x_1=0\ then\ x_2=?$

$⇒2(0)+6x_2=30\\ ⇒6x_2=30\\ ⇒x_2=30/6=5$

When$x_2=0\ then\ x_1=?$

$⇒2x_1+6(0)=30\\ ⇒2x_1=30 \\ ⇒x_1=30/2=15$

To draw constraint $4x_1+x_2≤24→(2)$

Treat it as$4x_1+x_2=24$

When x1=0 then x2=?

$⇒4(0)+x_2=24\\ ⇒x_2=24$

When x2=0 then x1=?

$⇒4x_1+(0)=24\\ ⇒4x_1=24\\ ⇒x_1=24/4=6$

The maximum value of the objective function $Z=37.64$  occurs at the extreme point $(5.18,3.27).$

Hence, the optimal solution to the given LP problem is :$x_1=5.18,\ x _2=3.27\ and\ max Z=37.64.$

(d) If the company plans to make only nails and bolts with the existed steel and zinc, then its profit is reduced.