# hw2solutions - ≤ 5 (Amount shipped to Hous. ≤ Hous....

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IEOR 162 HW#2 Chapter 3 – Section Review 4. OIL = thousands of barrels of purchased oil HOS= thousands of barrels of non-cracked heating oil sold HOP= thousands of barrels of heating oil processed further AFS = thousands of barrels of non-cracked aviation fuel sold AFP = thousands of barrels of aviation fuel processed further Then we should solve: max z = 40HOS + 90HOP - 40(OIL) + 130AFP + 60AFS st OIL 20 .5(OIL) = AFS + AFP .5(OIL) = HOS + HOP AFP + .75HOP 8 All variables 0 11. All variables are in millions of barrels of oil. x 11 = Oil sent each year from LA to Houston. x 12 = Oil sent each year from LA to NY. x 21 = Oil sent each year from Chicago to Houston. x 22 = Oil sent each year from Chicago to NY. y 1 = Capacity added to LA. y 2 = Capacity added to Chicago. Then a correct formulation (objective function is in thousands of dollars) is max z = 10{20x 11 +15x 12 +18x 21 +17x 22 }-120y 1 -150y 2 s.t. x 11 + x 21

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Unformatted text preview: ≤ 5 (Amount shipped to Hous. ≤ Hous. Demand) x 12 + x 22 ≤ 5 (Amount shipped to NY ≤ NY Demand) x 11 + x 12 ≤ 2+y 1 (Don't ship more from LA than is available) x 21 + x 22 ≤ 3+y 2 (Don't ship more from Chicago than is available) All variables ≥ 24. Let Xi = Number of type i DRG cases handled weekly. Then the correct formulation follows. Max z= 2000*X1 + 1500*X2 + 500*X3 + 300*X4 st 2) 7*X1 + 4*X2 + 2*X3 + X4 <= 570 3) 5*X1 + 2*X2 + X3 <= 1000 4) 30*X1 + 10*X2 + 5*X3 + X4 <= 50000 5) 800*X1 + 500*X2 + 150*X3 + 50*X4 <= 50000 6) X1 >= 10 7) X2 >= 15 8) X3 >= 40 9) X4 >= 160 All variables ≥ Chapter 4-Section1 1. max z = 3x 1 + 2x 2 s.t. 2x 1 + x 2 + s 1 = 100 x 1 + x 2 + s 2 = 80 x 1 + s 3 = 40 All variables ≥ 2. min z = 50x 1 + 100x 2 s.t. 7x 1 + 2x 2- e 1 = 28 2x 1 + 12x 2- e 2 = 24 All variables ≥...
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## This note was uploaded on 03/03/2009 for the course IEOR 161 taught by Professor Lim during the Fall '08 term at University of California, Berkeley.

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hw2solutions - ≤ 5 (Amount shipped to Hous. ≤ Hous....

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