8/17/2019 Blending Problem

1/3

IE 141: Operations Research 1

Sample Problem

BLENDING PROBLEM

The Philippine Petron Oil Corporation produces unleaded and premium-grade gasoline that is sold to

independent gasoline stations all over the Philippines. The company has a refinery in Bataan that manufactures

the products by blending three petroleum components. Each type of gasoline has a different price. The costs of

 petroleum component vary with type. However, with the passage of the Biofuel Act of 2006, Petron and otheroil firms in the country are required by law to mix bioethanol to all types of gasoline that will be distributed in

the local market. With this new development, Petron wants to determine how to blend the three components and

 bio-ethanol into the two gasoline-bioethanol fuel products in such a way as to maximize profits.

The regular gasoline-bioethanol blend fuel can be sold for PhP35.00 per liter and the premium gasoline-

 bioethanol blend can be sold for PhP38.00 per liter. For the current product introduction production-planning period, Petron can obtain four components at the costs and in the quantities shown in Table 1.

The specifications for the new fuel products restrict the amounts of each component that can be used in each

gasoline-bioethanol blend product. The product specifications are shown in Table 2. Initial introduction of the

new products to the market requires Petron to produce at least 300,000 liters of regular gasoline-bioethanol

 blend.

Table 1. Petroleum component cost and supply

Petroleum component Cost (PhP)/li Amount available (li)

Component 1 (C1-Bioethanol) 13.00 200,000

Component 2 (C2) 15.00 200,000

Component 3 (C3) 18.00 400,000

Component 4 (C4) 20.00 400,000

Table 2. Specifications for gasoline types

Product Specifications

Regular-bioethanol blend gasoline At most 10% component 1

At most 20% component 2

At least 40% component 3

At most 20% component 4Premium-bioethanol blend gasoline At most 05% component 1

At most 20% component 2

At least 40% component 3

At most 30% component 4

Solution:

Let xij = liters of component i used in gasoline j

wherei = 1, 2, 3, 4 and

 j = r  if regular-bioethanol blend gasoline

= p if premium-bioethanol blend gasoline

Decision variables (8):

 x1r  = liters of C1 in regular-bio-ethanol blend gasoline

 x2r  = liters of C2 in regular-bio-ethanol blend gasoline

 x3r  = liters of C3 in regular-bio-ethanol blend gasoline

 x4r  = liters of C4 in regular-bio-ethanol blend gasoline

 x1p = liters of C1 in premium-bio-ethanol blend gasoline

 x2p = liters of C2 in premium-bio-ethanol blend gasoline

 x3p = liters of C1 in premium-bio-ethanol blend gasoline

 x4p = liters of C1 in premium-bio-ethanol blend gasoline

Total of each type of gasoline-bioethanol blend produced:

Regular gasoline-bioethanol blend = x1r  + x2r  + x3r  + x4r  

Premium gasoline-bioethanol blend = x1p + x2p + x3p + x4p 

8/17/2019 Blending Problem

2/3

IE 141: Operations Research 1

Sample Problem

Total of each component used:

C1 = x1r  + x1p  C2 = x2r  + x2p  C3 = x3r  + x3p  C4 = x4r  + x4p 

Constraints:

Available components:

 x1r  + x1p ≤ 200,000 

 x2r  + x2p≤ 200,000 

 x3r  + x3p  ≤ 400,000

 x4r  + x4p ≤ 400,000 

Product specifications:

Regular gasoline-bioethanol blend:

 x1r  / ( x1r  + x2r  + x3r  + x4r ) ≤ 0.10 → + 0.90 x1r   –  0.10 x2r   –  0.10 x3r   –  0.10 x4r  ≤ 0 

 x2r  / ( x1r  + x2r  + x3r  + x4r ) ≤ 0.20  → –  0.20 x1r  + 0.80 x2r   –  0.20 x3r   –  0.20 x4r  ≤ 0 

 x3r  / ( x1r  + x2r  + x3r  + x4r ) ≥ 0.40 → –  0.40 x1r   –  0.40 x2r  + 0.60 x3r   –  0.40 x4r  ≥ 0 

 x4r  / ( x1r  + x2r  + x3r  + x4r ) ≤ 0.20 → –  0.20 x1r   –  0.20 x2r   –  0.20 x3r  + 0.80 x4r  ≤ 0 

Premium gasoline-bioethanol blend:

 X 1p / ( x1p + x2p + x3p + x4p) ≤ 0.05 → +0.95 x1r   –  0.05 x2r   –  0.05 x3r   –  0.05 x4r  ≤ 0 

 x2p / ( x1p + x2p + x3p + x4p) ≤ 0.20  → –  0.20 x1r  + 0.80 x2r   –  0.20 x3r   –  0.20 x4r  ≤ 0 

 x3p / ( x1p + x2p + x3p + x4p) ≥ 0.40 → –  0.40 x1r   –  0.40 x2r  + 0.60 x3r   –  0.40 x4r  ≥ 0 

 x4p / ( x1p + x2p + x3p + x4p) ≤ 0.30 → –  0.30 x1r   –  0.30 x2r   –  0.320 x3r + 0.70 x4r ≤ 0 

Regular-bioethanol production:

 x1r  + x2r  + x3r  + x4r  ≥ 300,000 

Objective function:

Maximize profit.

Profit = Total Revenue –  Total Cost

= (price of regular-bioethanol blend x quantity of regular-bioethanol blend)

+ (price of premium-bioethanol blend x quantity of premium-bioethanol blend)

- {[( cost of C1) x (qty of C1)] + [( cost of C2) x (qty of C2)]

+ {[( cost of C3) x (qty of C3)] + [( cost of C4) x (qty of C4)]}

Maximize Z = [35( x1r  + x2r  + x3r  + x4r ) + 38( x1p + x2p + x3p + x4p)]

-  [13( x1r  + x1p) + 15( x2r  + x2p) + 18( x3r  + x3p) + 20( x4r  + x4p)]

Simplifying

Z = 22 x1r  + 20 x2r  + 17 x3r  + 15 x4r  + 25 x1p + 23 x2p + 20 x3p + 18 x4p 

The linear programming model with 8 decision variables and 14 constraints is:

8/17/2019 Blending Problem

3/3

IE 141: Operations Research 1

Sample Problem

Maximize Z = 22 x1r  + 20 x2r  + 17 x3r  + 15 x4r  + 25 x1p + 23 x2p + 20 x3p + 18 x4p 

s.t.

 x1r  + x1p ≤ 200,000 

 x2r  + x2p≤ 200,000 

 x3r  + x3p  ≤ 400,000

 x4r  + x4p ≤ 400,000 

+ 0.90 x1r   –  0.10 x2r   –  0.10 x3r   –  0.10 x4r  ≤ 0 

 –  0.20 x1r  + 0.80 x2r   –  0.20 x3r   –  0.20 x4r  ≤ 0 

 –  0.40 x1r   –  0.40 x2r  + 0.60 x3r   –  0.40 x4r  ≥ 0 

 –  0.20 x1r   –  0.20 x2r   –  0.20 x3r  + 0.80 x4r  ≤ 0 

+0.95 x1r   –  0.05 x2r   –  0.05 x3r   –  0.05 x4r  ≤ 0 

 –  0.20 x1r  + 0.80 x2r   –  0.20 x3r   –  0.20 x4r  ≤ 0 

 –  0.40 x1r   –  0.40 x2r  + 0.60 x3r   –  0.40 x4r  ≥ 0 

 –  0.30 x1r   –  0.30 x2r   –  0.320 x3r + 0.70 x4r ≤ 0 

 x1r  + x2r  + x3r  + x4r  ≥ 300,000 

 x1r , x2r , x3r , x4r , x1p, x2p, x3p, x4p ≥ 0