An oil company wants to create lube oil, gasoline and diesel fuel at two refineries. There are two sources of crude oil. Consider arc 2-4. The per unit shipping cost of crude B from source 2 (node 2) to refinery 2 (node 4) is $11 and the yield is 85 percent. The following network representation depicts this problem. What is the balance of flow constraint for node 3 (Refinery 1)? A) X₁₃ + X₂₃ − .95 X₃₅ − .90 X₃₆ − .90 X₃₇ = 0 B) .80 X₁₃ + .95 X₂₃ − X₃₅ − X₃₆ − X₃₇ = 0 C) .80 X₁₃ + .95 X₂₃ − .90 X₃₆ − .90 X₃₇ ≥ 0 D) X₁₃ + X₂₃ − X₃₅ − X₃₆ − X₃₇ ≥ 0

Question

Quizplus · Accepted Answer

The balance of flow constraint for node 3 (Refinery 1) can be written as: Input flow from source 1 + Input flow from source 2 - Output flow to lube oil production - Output flow to gasoline production - Output flow to diesel fuel production = 0 Using the variable naming convention in the network representation, this can be expressed as: X₁₃ + XS1U2B213S1U2B0 - X₃₅ - X₃₆ - X₃₇ = 0 (We ignore the fact that refinery 2 is also producing gasoline and diesel fuel for this equation because it is not relevant to the balance of flow constraint for node 3.) Now we substitute the value of XS1U2B213S1U2B0 from arc 2-4, which is: XS1U2B213S1U2B0 = 0.85 * XS1U2B214S1U2B0 (Note that we use the yield of 85% for crude B.) Substituting this, we get: X₁₃ + 0.85 * XS1U2B214S1U2B0 - X₃₅ - X₃₆ - X₃₇ = 0 Multiplying both sides by 0.80 because only 80% of crude A goes to Refinery 1, we get: 0.80 * X₁₃ + 0.68 * XS1U2B214S1U2B0 - 0.80 * X₃₅ - 0.80 * X₃₆ - 0.80 * X₃₇ = 0 (Note that we use 80% instead of 0.8 to make the equation more readable.) Therefore, the balance of flow constraint for node 3 is: 0.80 X₁₃ + 0.95 X₂₃ - X₃₅ - X₃₆ - X₃₇ = 0 So, the best choice is B.

An Oil Company Wants to Create Lube Oil, Gasoline and Diesel