Question

A petroleum refinery can handle the processing of two grades of
crude oils, #1 and #2, to produce four kinds of products gasoline, kerosene, fuel oil and
residual.
The total costs include raw material costs and processing costs. The raw material costs for the
crude oils are $24/bbl for #1, and $15/bbl for #2. The processing costs are $0.5/bbl for #1, and
$1.0/bbl for #2.
The sales prices of the products are $36/bbl for gasoline, $24/bbl for kerosene, $21/bbl for fuel
oil and $10/bbl for residual.
The following is the information for the product yield (volume percent of raw material) and
maximum allowable production of each product (bbl/day):
One bbl crude oil #1 can produce 0.80 bbl gasoline, 0.05 bbl kerosene, 0.10 bbl fuel oil, and
0.05 bbl residual. One bbl crude oil #2 can produce 0.44 bbl gasoline, 0.10 bbl kerosene, 0.36
bbl fuel oil, and 0.10 bbl residual. The maximum allowable production for gasoline is 24,000
bbl/day, for kerosene is 2,000 bbl/day, and for fuel oil is 6,000 bbl/day, and no maximum
allowable production limitation for residual.
(1) Based on the data above, derive a mathematical formulation for this linear programming
problem and determine: The optimum feed schedule (bbl/day) of the two crude oils for the
maximum profit, using method of simultaneous equations. How much is the maximum profit
($/day)?
(2) Comment on how variation of raw material costs and sales prices will affect the feed
schedule and profitability on this kind of project in industry.

Answer 1

(1).

Cost (per bbl)	Crude oil #1	Crude oil #2
Raw material	24	15
Processing	0.5	1

(per bbl)	Sale price	Crude oil #1	Crude oil #2	Max. production
Gasoline	36	0.8	0.44	24000
Kerosene	24	0.05	0.1	2000
Fuel oil	21	0.1	0.36	6000
Residual	10	0.05	0.1	-

Let the amounts of crude oil #1 and crude oil #2 be x and y bbl/day respectively.

Cost	Crude oil #1 (x bbl/day)	Crude oil #2 (y bbl/day)
Raw material	24x	15y
Processing	0.5x	1y
Total Cost	24.5x	16y

	Crude oil #1 (x bbl/day)	Crude oil #2 (y bbl/day)	Total produce (bbl /day)	Sale price
Gasoline	0.8x	0.44y	0.8x + 0.44y	36(0.8x + 0.44y)
Kerosene	0.05x	0.1y	0.05x + 0.1y	24(0.05x + 0.1y)
Fuel oil	0.1x	0.36y	0.1x + 0.36y	21(0.1x + 0.36y)
Residual	0.05x	0.1y	0.05x + 0.1y	10(0.05x + 0.1y)
			Total	32.6x + 26.8 y

Profit = Sale price - cost price

z = (32.6x + 26.8y) - (24.5x + 16y) = 8.1x - 10.8y

Maximize z = 8.1x - 10.8y  

Constraints:

Gasoline:    --------------- (1)

Kerosene:   --------------- (2)

Fuel oil:    --------------- (3)

Raw material:

Solving (1) and (2),

Substitute x from (1) in (2),

Substitute y in (1)

Similarly solving (2) and (3)

Solving (1) and (3)

The optimum values lie at one of there corner points P1, P2, or P3.

z (P1) = 8.1*26206.89 - 10.8*6896.55 = 137793.06

z (P2) = 8.1*150000 - 10.8*12500 = -13500

z (P3) = 8.1*24580.16 - 10.8*9836.06 = 92869.848

Thus, the maximum profit lies at P1

Feed Schedule: Crude oil #1 = 26206 bbl/day Crude oil #2 = 6896 bbl/day

{# Barrels can't be in fraction}

Maximum Profit = $ 137791.8

  

(2). Variation in raw material cost ans sales prices will change the profit (i.e, the objective function) and thus the feed schedule will also change acordingly.