Question

A company blends three ingredients A, B, and C to form two types of fertilizers, standard (S) and premium (P). The company has to produce at least 500 pounds of each type of fertilizer. Standard fertilizer should have at least 35% of nitrogen. Premium fertilizer should have at least 35% chlorine. Composition, availability and cost of ingredients are as indicated below:

Ingredient	Nitrogen Content	Chlorine Content	Cost/lb. ($)
A	30%	40%	10
B	30%	30%	12
C	40%	30%	18

Formulate a linear programming model that will minimize the cost of production.

(a) What are the Decision Variables for the above problem?

ANSWER:

(b) What is the Objective Function for the above problem?

ANSWER:

(c) What are the constraints for the above problem?

ANSWER:

Answer 1

Q1) Answer:

A, B and C are the three ingradients used to blend Fertilizers P and S.

Let 'Xij' be the quantity of ingredient

Where,

i = ingradient used to produce product k that is j=a, b, c,

j= types of the fertilizer that is k=s, p

Thus we can say below are the decision variables for the given lpp problem:

Xas = Quantity of A to produce S

Xbs = Quantity of B to produce S

Xcs = Quantity of C to produce S

Xap = Quantity of A to produce P

Xbp = Quantity of B to produce P

Xcp = Quantity of C to produce P

Q2) Answer:

Cost for ingradients A, B and C are given as 10, 12, 18 respectively. Which means it takes $10 for ingradient A to produce combinely fertilizers P and S and similarly for B and C as well.

Thus objective function becomes,

Minimize Z = Total cost = 10*(Xas + Xap) + 12*(Xbs + Xbp) + 18*(Xcs + Xcp)

Q3) Answer:

Constraints will be as follows,

Xas + Xbs + Xcs >= 500..... { Total of fretilizer S must be atleast 500 lbs}

Xap + Xbp + Xcp >= 500..... { Total of fretilizer P must be atleast 500 lbs}

0.30*Xas + 0.30*Xbs + 0.40*Xcs >= 0.35*500 >=175 ...{ Standard fertilizer should have at least 35% of nitrogen.}

0.40*Xap + 0.30*Xbp + 0.30*Xcp >=0.35* 500 >=175 ..... {Premium fertilizer should have at least 35% chlorine}

Xas, Xbs, Xcs, Xap, Xbp, Xcp >= 0

Thus, complete LPP problem will be,

Minimize Z = Total cost = 10*(Xas + Xap) + 12*(Xbs + Xbp) + 18*(Xcs + Xcp)

Subject to,

Xas + Xbs + Xcs >= 500

Xap + Xbp + Xcp >= 500

0.30*Xas + 0.30*Xbs + 0.40*Xcs >= 0.35*500 >=175

0.40*Xap + 0.30*Xbp + 0.30*Xcp >=0.35* 500 >=175

Xas, Xbs, Xcs, Xap, Xbp, Xcp >= 0