This preview shows pages 1–6. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: WM 4 50Wb4’15 _ _ 01835321 (@0491? $z‘5WM MXa/Jv x24 '5 am/ Agra/:9 papacy x3 ,5 5‘
a 6wa ' a; '_ 5 4224/45.. g g G??C3?s€i «(keg Tag ¢?;\Qwﬁ\ 3C? \e ’u‘m the
° Nth tidal mgx _
L9) 3h. T: R EC? ﬁt} 35h Aggie (:9)
:3 2x ’3 
M , ‘ W i , Slab)‘
{fimg a maﬁa {$33 ‘ﬁo :~> Q gym , ‘
v r c l »¥\r\aJt’
m ages is?) . 3a a as exists X i=ch ‘r’
RX :2 ‘0 x5 x) the, E33¥ece Mm c: aggxiSWQW Problem #3(m)
a) Suppose the feasible region is bounded. Then maxing]: cm :2 2:121 c, * at, zil=1* 5 z 6 * 2'22} Hence, the value of the objective function at any point in F is bounded above by 6 * ELI so
the objective value at the optimal point is bounded as well and, hence, ﬁnite. 5 22; Ml * 332' S q. b)(§) We know by Farkas’ Theorem that for any value of 5:, either the given system holds or there
exists [31,2] such that yM + z * e _>_ O,y:Z + z < 0 where e = (1,1,1, . . .,1). We will Show that no
such [y, 2] can exist for any :7: 6 P, so the ﬁrst system must be feasible. to the second system of equations. Since the columns
> ——z for every Suppose we have a feasible solution, [31, z}
of M correspond to extreme points of F, the ﬁrst set of equations tell us that yp, extreme point, p, of F. So, consider the following LP: min{y:r : x E F We know that an optimal solution exists and there is one that is some extreme point of F, since F is bounded.
Let 15, be such an optimal solution. The optimal objective value ypi satisﬁes yM + z 2 0. Further, since 1? E F, yfz Z ypi 2
value.) Hence, ya? + z 2 3,117, + 2 Z vz + z z
This contradicts the fact that [y,z] was a feasible solution,
Therefore, by Farkas’ Lemma, the original system must have a solution 2 —2 because we know that y
~—z (Since pi is the minimizer, 51': cannot have a better objective 0. 80, y does not satisfy yf: + z < 0.
so no such feasible solution exists. for any i: e F. Note: You CANNOT use any information from the original system (i.e. the requirement that A Z 0) to prove the new system (2)
systems is feasible (
feasible. You need to sh Problem #4
(a)(12 pts) Deﬁne:
= light crude oil purchased in Dryden mLD
mHD
93L?“ 93HT
and 9 ft
kft
jft
9f
kf
if f €{D(ryden), T(rumansburg)} ll H H ll ll gasoline at f
kerosene at f
jet fuel at f H H is infeasible. Farkas’ Theorem states that exactly one of the and it was proven in last week’s homework), so clearly they both can’t be
ow directly that the second system is infeasible, like above. heavy crude oil purchased in Dryden
light crude oil purchased in Trumansburg
heavy crude oil purchased in Trumansburg gasoline shipped from f to t
kerosene shipped from f to t
jet fuel shipped from f to t t E {B(inghamton), R(ochester), S(yracuse)} Let these be our decision variables, all nonnegative.
Then let’s add the following constraints:
Purchase Constraints 1'LD + $11 S 2, 000,000
xHD + :L'HT S 1,000,000 Conversion Constraints 90 = .40:L‘LD + .32xHD
kD : 20:ch + .4093HD
jD 235181,]: + 20:81—10
91‘ :2 .4013LT + 32ng
[CT 2: .ZOSCLT + .40.??qu
jT = .3532LT.+ .QOLBHT Demand Constraints 903 + 9BR + 9135 S gr)
9TB + 9TB + 9T8 S 9T
9133 + 9TB Z 200, 000
91312 + QTR Z 300, 000
905 + 9T8 2 500, 000
[£193 + 19133 + km S kD
kTB + kTR + krs _<. kT
kDB + kTB Z 400, 000 16133 + kTR 2 200,000 sz + [0119 2 100,000 3173 + Jim + J'Ds S jD jTB + jTR'+ jars S jT jDR + java 2 600,000 3133 + Jim 2 300, 000
The objective function Cost = Purchasing cost + Shipping cost :
minimize 25(17LD + CELT) + 20(IL‘HD
jDR) + 4001135 + 16133 +jDs) + ~80(9TB + kTB + adv; + IIIHT) + ~50(9DB + 16173 + jDB) + .75(gDR + 16012 +
l+ 45(9TR + kTR + jTR) + 50(9Ts + kTs +st) s) involve only gasoline, kerosene or b) (8 pts) Most of the constraints (the demand constraint
each involving only one end product, jet fuel. Hence, a decomposition scheme into sub—problems,
may solve the system faster. Problem #5
a)(2) Note the columns corresponding to [51,32,>\1,u1] form an identity matrix in the initial tableau,
1 1/3 40/23 —1 0 0 0 1 '1  . . . “1 ”
so B Wlll be then correspondmg columns 111 the ﬁnal tableau. Hence, B —. 0 1/6 1/3 _1/2
0 ~1 / 6 2/3 1/2 nd by taking the negative reduced costs for the appropriate variables; in this b)(2)[7r,a] can be fou
= [0,1/3,68/3,24/3l Case) l7rv 0] :: lcsla 6827 CA1 7 CAM] c)(3) A2, A3, and M are all basic, and we know 7r 2 cBB‘l, so CB 2 7rB = [0, 9,24,22]. Therefore, lumen, cm] = [9, 24, 22}. d) (2) New objective function is: (c2 — WA2)£L‘2
c2 = [3,0], 7r = [0,1/3], 0 1
2 
A * l 1 4 l
So, the LP to be solved is given by:
min 8/3X4 + 1/3X5
s.t. X4 + X5 3 3 X4 +2X5 S4 e)(3) Let p be the extreme point given. The coefﬁcients for the next extreme point in the original master program can be calculated by
O 1 2 1
2 A ._ ..
A ~l1—1ll1l—l1l
Hence, the entire column is [1, 1,0, 1]T. f) (2) To correspond to this tableau, the column would be 1 1/3 40/3 ~1 1 1/3 ,_1 T_ 0 0 0 1 1 _ 1
B [1’1’0’1] "‘ 0 1/6 1/3 —1/2 0 " 1/3
0 —1/6 2/3 1/2 1 1/3 g)(3) 0/33 = Gus '“ “Aus
cm. = c225 = 13,0112,11T = .
cps =6 [0,§,§§,2§] [1,1,0,1}T =6—(0+§—+0+8) = ~§ so.
Therefore, this proposal will not improve the solution to the restricted master problem. h)(3)
max —a1
3.1:. 0A1 + 0M1 + a1 = 3
0A1 + 0m + $2 = 5
A1 = 1
#1 = 1
A1, #1, 31, S1 2 0 ...
View
Full
Document
 Summer '07
 SHMOYS/LEWIS

Click to edit the document details