Transformation and Transportation OPRE 6389 Quiz Practice Questions by Metin C¸Akanyıldırım Compiled at 10:56 on Thursday 31St March, 2016
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Transformation and Transportation OPRE 6389 Quiz Practice Questions by Metin C¸akanyıldırım Compiled at 10:56 on Thursday 31st March, 2016 A. Put T before a statement if you think that statement is true. Otherwise put X. When there are compar- isons, assume that everything is constant or same, except for the compared attributes/quantities. 1. ( ) Chinamax, Seawaymax, Panamax, Aframax and Suzemax tanker sizes are all based on different canal, harbor, river and channel limitations. ANSWER X. 2. ( ) Aquifers and salt domes are both impermeable and geologically sealed gas storage options. ANSWER X. 3. ( ) Aquifer storage is the most expensive and least desirable natural gas storage option. ANSWER T. 4. ( ) Japan is the largest importer of liquefied natural gas in the world. ANSWER T. 5. ( ) Long distance shipments of natural gas is more economical with pipeline than with LNG tankers. ANSWER X. 1. A Wasington DC company buys pipeline capacity to ship gas from Henry Hub to DC at $0.6 per MMBtu. The price spread between these locations will remain positive but the magnitude of the spread can be 0.55, 0.65, 0.75, 0.85, 0.95 dollars per MMBtu, each magnitude is equally likely. What is the probability of profiting from this venture? ANSWER It profits when the spread is more than 0.6, which happens with 80%=0.2+0.2+0.2+0.2 chance when the spread is 0.65, 0.75, 0.85 or 0.95 dollars per MMBtu. A f g f g 2. A pipeline network has arcs = (5, 3), (4, 5), (8, 4), (3, 5), (8, 3), (4, 8), (8, 5), (4, 3) and supplies b3, b4, b5, b8 , where some b’s can be negative while others are nonnegative. Draw this network. Can you exactly identify sources and markets by looking at the direction of the pipelines without knowing b’s? Write a constraint in terms of b’s and flow variables xij from i to j to ensure that flows in and out of node 5 are feasible: If node 5 is a source, it should not supply more than it can; If node 5 is a market, it should receive at least its demand. ANSWER Connect f8, 4g with a bidirectional arc and connect f3, 5g with a bidirectional arc. Draw all possible directional arcs from f8, 4g to f3, 5g. When finished, your network should look like a “closed envelope”. Since directional arcs are from f8, 4g to f3, 5g, at least one of f8, 4g is a source and at least one of f3, 5g is a market. But we cannot exactly say that f8, 4g are sources and f3, 5g are markets, as for example, 4 can be a market with b4 < 0. ≤ − − − ≤ ≤ We use out f low - in f low supply: x53 x35 x45 x85 b5. When node 5 is a market, b5 0, − ≥ − ≥ alternatively you can write x35 + x45 + x85 x53 b5, that is in f low - out f low magnitude of demand. Consider the following refinery for the rest of questions: Cracking and coking facilities of a refinery respectively receive heavy oil (LiO) and residuum (R) as inputs. Each barrel of LiO, when cracked, yields 0.5 barrel of cracked oil (CO) and 0.4 barrel of cracked gasoline (CG). Each barrel of R, after coking, yields 0.6 barrel of lube oil (LO). Note that both cracking and coking create some waste so sum of output amounts is less than the input amount. Blending has no waste and it blends CO, HO and R to obtain fuel oil (FO). Output of the cracking, cooking and blending are CG, FO and LO. 3. Draw a flow diagram with 3 processes (cracking, coking and blending as blocks), show 2 inputs (LiO, R), an intermediate (CO) and 3 outputs (CG, FO, LO). ANSWER Three processes of the diagram have the following inputs/outputs: Cracking: Input LiO and Outputs CG, CO. Coking: Input R and Output LO. Blending: Inputs CO, HO, R and Output FO. 4. If inputs are LiO=100, R=80 and the two outputs are CG=12, LO=12, what is the output FO amount? ANSWER CG=12 implies input LiO=30=12/0.4 for cracking and output CO=15=30*0.5. LiO reserved for blending is 70=100-30. LO=12 implies input R=20=12/0.6 for coking. R reserved for blending is 60=80-20. Inputs into blending are LiO=70, CO=15 and R=60, their sum is FO=145. 5. If outputs are CG=20, FO=100, LO=18, what are the input LiO and R amounts, be as specific as possible? 1 ANSWER CG=20 implies input LiO=50=20/0.4 for cracking and output CO=25=50*0.5. LO=18 implies input R=30=18/0.6 for coking. The amount of CO that goes into blending is 25, the rest of output FO must come from LiO and R, so for the amounts that go into blending LiO+R=100-25=75. Adding to these amounts of LiO and R used in cracking and coking, we have LiO+R=75+50+30=155 for the inputs. Note that we cannot uniquely determine LiO and R, all we can say is LiO+R = 155, LiO ≥ 50 and R ≥ 30. 6. If outputs are CG=20, FO=100, LO=18 and CO, LiO, R blends in the ratios of 1:2:1, what are the input LiO and R amounts, be as specific as possible? ANSWER Repeating the steps above LiO+R = 155, LiO ≥ 50 and R ≥ 30. Since CO=25, the blending ratios require blended amount of LiO to be 50 and the blended amount of R to be 25. We have LiO+R = 155, LiO ≥ 50+50 and R ≥ 30+25, whose solution is LiO=100 and R=55. Blending constraint on the ratios helps us to obatin a unique solution for input amounts. 2.