Question

An oil company has to transport oil products from three sites (S1, S2, S3), through two refineries (R1, R2) and finally to three markets (C1, C2, C3). The supply(in 1,000 barrels per day) are given in Table 1.

Table 1. (Site supply)

Site Supply

S1 1,500

S2 750

S3 1,200

Table 2 below shows the costs in $1,000 of transporting 1,000 barrels of crude oil between sites and refineries. It also has the costs of refining 1,000 barrels of crude oil at each refinery, and maximum capacities (in 1,000 barrels per day) of each refinery. Table 3 below shows the costs in $1,000 of transporting 1,000 barrels of refined fuel between refineries and customers. It also gives customer demands (in 1,000 barrels per day).

a) Given that all customer demands must be fulfilled, while trying to maintain costs at the lowest possible level, formulate the adequate model to solve the problem.

Table 2. Site to refinery costs, refinery costs, and refinery capacities

From/to R1 R2

S1 20 7

S2 12 10

S3 15 8

Refining Costs 7 6

Capacity 1000 1200

Table 3. Refinery to customer costs and customer demands

From/to C1 C2 C3

R1 6 8 5

R2 4 10 7

Demand 500 800 900

Note: Assume there is no waste in the refining process, so that the same quantity that enters goes out.

Answer 1

Let the variables be as defined below:

From/To	R1	R2
S1	S11	S12
S2	S21	S22
S3	S31	S32
From/To	C1	C2	C3
R1	R11	R12	R13
R2	R21	R22	R23

Thus: cost of transporting crude oil between oil sites and refineries = (20*S11)+(7*S12)+(12*S21)+(10*S22)+(15*S31)+(8*S32)

Cost of refining at R1 and R2 = 7*(S11+S21+S31) + 6*(S12+S22+S32)

Cost of transporting crude oil between refineries and customers = (6*R11)+(8*R12)+(5*R13)+(4*R21)+(10*R22)+(7*R23)

Objective function = cost of transporting crude oil between oil sites and refineries+Cost of refining at R1 and R2+Cost of transporting crude oil between refineries and customers

This has to be minimized.

Constraints:

First lets look at the supply side constraints:

1. S11+S12<=1500

2. S21+S22<=750

3. S31+S32<=1200

Next we look at capacity constraints of the refiners.

4. S11+S21+S31<=1000

5. S12+S22+S23<=1200

Next we look at the fact that quantity received by a refinery should be = quantity supplied by a refinery.

6. S11+S21+S31 = R11+R12+R13

7. S12+S22+S32 = R21+R22+R23

Lastly we look at the demand side constraints.

8. R11+R21 = 500

9. R12+R22 = 800

10. R13+R23 = 900

Lastly all the variables >=0 i.e. non-negativity constraint.

Solving in excel, using the solver function, the following solution is obtained:

From/To	R1	R2
S1	0.00	1,200
S2	750.00	0.00
S3	250.00	0.00
From/To	C1	C2	C3
R1	0.00	512.50	487.50
R2	500.00	287.50	412.50
				Formula
Cost of transporting crude oil between sites and refineries	21,150.00			(20*S11)+(7*S12)+(12*S21)+(10*S22)+(15*S31)+(8*S32)
Cost of refining at R1 and R2	14,200.00			7*(S11+S21+S31) + 6*(S12+S22+S32)
Cost of transporting crude oil between refineries and customers	14,300.00			(6*R11)+(8*R12)+(5*R13)+(4*R21)+(10*R22)+(7*R23)
Total costs	49,650.00
Constraints
1,200.00	<=	1,500.00		S11+S12<=1500
750.00	<=	750.00		S21+S22<=750
250.00	<=	1,200.00		S31+S32<=1200
1,000.00	<=	1,000.00		S11+S21+S31<=1000
1,200.00	<=	1,200.00		S12+S22+S23<=1200
1,000.00	=	1,000.00		S11+S21+S31 = R11+R12+R13
1,200.00	=	1,200.00		S12+S22+S32 = R21+R22+R23
500.00	=	500.00		R11+R21 = 500
800.00	=	800.00		R12+R22 = 800
900.00	=	900.00		R13+R23 = 900

Thus minimized total cost is $49,650 (in thousands $).