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# HW 1 solution - Math 482 HWl(Answers and comments ’1 If...

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Unformatted text preview: Math 482 HWl (Answers and comments) ’1'. If the reﬁnery uses 51:1 barrels of light crude and 232 barrels of heavy crude, then it needs 0.31m + 0.31:2 2 900000 to produce sufﬁcient amount of gasoline, 0.2331 + 0.41132 2 800000 to produce sufﬁcient amount of heating oil, and 0.3331 + 0.2352 2 500000 to produce sufficient amount of jet fuel. In order to get the problem in standard form, we introduce surplus variables phi/2, and y;;, and get 2 : 35x1 + 309:2 —> min with respect to 031‘} “0.3% — y} I 900000, 0.2::l “0.41s — yg : 800000, 0.3351 +0.21%; — 1/3 : 500000, Int/173:2, 92,93 2 0- ._,. 9 There are several ways to get a canonical form equivalent to our problem. One of them gets 2 = 271 + 31"? — — 43:3 —> min with respect to —61r1 fan/2 +12’2’ +413 2 —.’L"1 —3:L'.’Z +3142l +4.13 2 —9, :01, (15/2, .r’z’, (£3 2 3, Note that the matrix of our ‘system’ is A z 8]. The ‘cornersl of the set of feasible solutions are .i" : (J, 0) and 51:” = (0, 3/2). The corresponding objective values are 8 and 9/2. Since 9/2 < 8. I: re minimum value ol’ the objective function is 9/2, attained at 1"” = (0. 3/2). l. The vertices of the polytope defined by the constraints are (0, 0), (0, 3/2), (4/5. 6/5), (12/7. 2/7). and (7/4. 0). The maximum value of 26/5 is obtained at the vertex (4/5, 6/5). 5. Since .r'l is unconstrained, we can replace it with If — 1'1". and solve the new system. But a hotter solution is as follows. We eliminate 1'1 from the constraints: If we substract '2 times the second constraint from the ﬁrst, we obtain —'2:rg + 31:3 : 2 \$1 — 1'15 : However since .z'l is uncOiistrained, the last equation is always satisfied; in particular, we may set .l'l : 1'3 + l tl’u‘ougliout the LP. Hence we obtain the following: 2 L: 41.2 .i- 21-3 + 1 ——>, max suliicct to A | [\J [Q (\L R + (N W c~ IV H C l (we can replace the max 4x2 + 25133 + 1 with min — 4172 + 2333 if we want to be in strict standard form). Now note that the constraint equationt 1:2 2 gm — 1 together with the restrietion \$2 3 O is equivalent to the inequality 1.53:3 — 1 2 0. Hence) we obtain 2:83:3—3 a max subject to However, we can always ﬁnd a larger value of the objective function by Choosing a larger value of \$3. Thus, the objective function is unbounded, and the maximum is not attained at any ﬁnite point. ...
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