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¥L= y.\:)(3 :M5XL:X;=‘ Kg? :0 tl called The computational procedure of Example 11.2 is
g in 19 used in this form for solving LP problems by G. Dantzi the method as follows: “adjacent extreme point” the simplex glgorithm; it was ﬁrst
47. We can display the progress of 7‘4 Note that the simplex algorithm is an method. 35 Geometric observations: Recall from the previous examples with ﬁnite optimum solutions,
that some “corner point” is always optimal. We now formalize this notion and establish it rigorously.
We say that a set? S; R11 is (329113;, provided X1,X2ES&DdOSA_<_1=>[1\X1+(1A)X2]ES. . Geometrically, the stipulation is that the entire line segment joining any two points of S must also lie in S. E.g., convex sets nonconvex sets 11.9 Theorem: The feasible region S = {x: Ax : b, x 2 0} {or an LP problem is a convex set.
Proof: Let x1, x2 e S, 0 5 A 5 1 and deﬁne x = Ax1 + (l  A) x2. We must show 3: E S. We have XIES=Ax1=b=$A(/\x1)=,\b _
}=> Ax: A(Ax1 + (1  A)x2) = Ab + (1 ~A)b = b.
x2 e 3 => Ax2 = b => A((lA)x2) = (14):; ~ Also, x1es=>xlzo=>ix120
2 2 }=>x=ixl+(1A)x220.
x 68:”: 20=>(1—,\)x220 Thus we have shown that Ax = b, x 2 0. Hence, it E S. B When x = Ax1 + (1  /\)x2 and 0 S A S 1, we say that x is a convex combination of x1 2 and x . An extreme point or vertex, what we’ve been calling a “corner point”, is a point of S which cannot be represented as a convex combination of two other points of S, i.e., a point that does not lie
on the interior of any line segment within S. Algebraically, this means that when x is an extreme point of S, then 36
x1, x2 E S
0<A<1 }=>xl=x2=x.
x = Ax1 + (l  A)x2
Why are extreme points important for LP? The intuitive reason is that for x = Axl + (1  a\)::2
with x1, in2 extreme, then cx = Mcxl) + (1  A) (cx2). : 2 I.e., ex is between ex1 & cx2. Thus either can1 or ex is better than cx. Hence look only at extreme points. To show this precisely we study the relationship between basic feasible solutions and
extreme points (algebraic insight H geometric insight). Suppose i = (EB, EN) = (B°1b, 0) is basic
feasible for Ax = b, x _>_ 0. Now if i“: = «\x1 + (1 — A)x2, where x1 & x2 are feasible solutions & 0 < A < 1, then EN = 0 = Axllq + (1  A)x12q. Since x119, Jig; 2 0, we must have xllq = x2N = 0.‘ But every solution for Ax = b must satisfy xB = B'lb  B'leN. Substituting xllq = 0, x12q = 0 resp. for xN shows x113 = x% = B’lb. So i = x1 = x2 and we have shown that:
(1) Every basic feasible solution is an extreme point of the polyhedron of feasible solutions. Now for the converse  Let T: be an extreme point and let A + be the submatrix of A consisting of columns Aj
for which 5% > 0. If these columns are linearly independent, then either they form a basis of A or
else we can add additional independent columns of A, say A0 (corresponding variablesare at level 0
in i), to get a basis B. I.e., B = [A+§ A0] and the equations require BXB + NiN =A+XA+ +A0§A +NXN =b.
>9 0 =0 0 20
Thus i corresponds (not necessarily uniquely) to a basis B « i.e, i is a basic feasible solution.
Alternatively, if A + has dependent columns, we can find a vector x ¢ 0 so that Ax = 0 and xjgé0=>Aj isacolumnof A+=>ij>0. Hence for small a > 0 we have that i + e x and )‘r  e x are & feasible. Check this:
A(i :tcx)=AJ'c ich=Ai=b and i ingO since xjylzo =>ij>0 and eissmall.
But then i 2 £6? + e x) +  e x) so i could not be extreme. (See Example 11.11 below.)
This contradiction says the case in which A + has dependent columns cannot arise. Thus we have shown:  37 (2) To any extreme point i there corresponds a basis B so that i is the basic feasible solution relative to basis B. Combining (1) 8L (2) we have proved: 11.10 Theorem: Basic feasible solutions for {Ax = b, x Z 0} (algebra) H extreme points of the polyhedron of feasible solutions. (geometry) , D 11.11 Example: ; 1‘; 2¥3=3 max x1 + x2 s.t. 2x +x +x :3
1 2 3 ax “1'3
x1+2x2 +x4=3 ‘5 1 z
xj20,Vj. Here the point (x1, x2) = (1, 1) is extreme & corresponds to the basis B = I: f g :l’ 1!:13 = (X1, X2) With DOHbBSiS N = 1, KN = (x3, x4). But consider the solution 1’: = (g, %, 0, which is feasible, but not basic (too many positive ij’s). Here the corresponding columns of A for positive ij’s are dependent: E275]w~[f]2[é]+i?]=[3} I.e., in above argument, take it = (x1, x2, x3, x4) = (1, —2, 0, 3). Now let 6 = i, say. 3 6
+ (30109“)
Theniicx=(,12,0,%)iﬁ(1v2,0,3)  i1040) ‘ D “MOI We note that the correspondence of Theorem II.10 is not oneto—one; in fact, many bases can
correspond to the same extreme point  see Example 11.14 below. The same reasoning with the 6idea above can be used to show that: 38 11.12 Theorem: If an LP has a ﬁnite optimal solution, it has an optimal basic feasible solution. :1: Proof: Suppose x is optimal, but not basic. Thenasaboveﬁnd x:;’:0 sothat Ax=0 Q4 xj¢0=>xf>0.
Now cx>0 => x*+cx better than x*
}=>cx=0. & cx<0=>x*cx betterthanx* So now choose 6 small, but just large enough to allow some to drop to zero when we replace
in“ <— x*  c x. (Note that if x S 0, we replace x* «— x*+ e x, and the same argument applies.)
Still have x* Z 0, but at least one less positive component now in x*, and x* is still optimal. Thus we iterate the argument until the columns Aj corresponding to x; > 0 are linearly independent. El Comments: (1) This shows that, except when optimum is unbounded or problem is infeasible, our intuition is correct  only need consider the extreme points in order to optimize. (2) Same idea can be used to improve in)! nonbasic solution. In Example 11.11, take i = (l 1 3 g) 2’ 2’ 2’ 2
& let x = (1, —2, 0, 3), as before. Here cx = —1, so we want to look at 5':  c x to improve the objective. Now i  c x Z 0 so long as c s % (see lst component). So let 6: l to get 1':  6 x 2
better & basic feasible. ll
A
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I 49
Duality Theory. We motivate this topic by considering once again the activity analysis model of Example 1.4. Given production activities j = 1,..., n 8; resources i = 1, ..., m. bi 2 units of resource i available; cj = unit proﬁt for activity j; aij 2 consumption rates.
Denote xj z: level at which jth activity is to operate.
To maximize proﬁt, solve max cx '
s.t. Ax 5 b PRIMAL
x Z 0. Now suppose 1 unit of resource i brings yi dollars on the open market. At which market prices will I sell (rather than use) my resources?
First, yi Z 0 for 1 5 i g m is required. Second, for each unit of activity j we now have two choices for the resources it consumes — sell these at prices yi per unit, or put them into production at proﬁt cj. m .
So market prices will be attractive when 2 yiaij 2 cj, for each j. Thus we are interested in
. i= . .
prices y so that y 3 0 and yA _>_ c. ‘Note that here the components of y are assocrated With the rows, not columns, of A. So the market value of our existing resources is determined by solving min yb
s.t. yA _>_ c DUAL
y 2 0» Note that the yi’s here play the same role as shadow prices:
yi = per unit price for selling resource i at proﬁt
shadow price for resource i = per unit cost we should be willing to pay for purchasing additional units of resource i. 50 Thus we are led to consider the dual pair of LPs given by: PRIMAL (P) DUAL (D) max ex min yb s.t. Ax 5 b s.t. yA 2 c r
x Z 0 y 2 0 Note that we now treat these LPs in “symmetric form” with S primal constraints, but later show ﬂ LP has a natural dual and the theory we develop here will apply generally. III. 2 Theorem (Weak Duality): x (P)  feasible and y (D)—feasible => ex 3 yb. .
Proof: ex 5 (yA)x = y(A.x) g yb. El
(xgo,yAzc Cy20,Axgb III. 3 Corollary (Unboundedness): (P) unbounded => (D) infeasible
and (D) unbounded => (P) infeasible.
Proof: If (P) is unbounded, then any feasible solution to the dual would bound (P)’s value from above  hence none can exist. Similarly for (D) unbounded r,» (P) infeasible. El 111.4 Corollary (Optimality): x* (P)  feasible, y* (D)  feasible & ex* = y*b => x*, y* are optimal
in (P) 86 (D), respectively. M: For any (P)  feasible x, Weak Duality implies ex 5 y*b. But, by assumption, we now have
ex* = y*b. Thus ex 5 ex* for any (P) — feasible x — hence x* is (P)optimal. We establish (D)
optimality for y* similarly. U The simplex algorithm shows that all possibilities for (P) and (D) are summarized by the table: 51 We’ve shown XXX cannot happen. Further, to see that (1), (2), (3) are possible, consider examples: (1) (2) (3)
max x1 min yl max x1 min  yl max x1 + x2 min y1 y2
s.t. 0x1 5 1 3.1;. y10 2 1 s.t. x1 5 1 s.t. yl 2 1 s.t. x1 — x2 _<_ 1 s.t. y1  y2 2 1
3120 3’120 x120 3’120 —X1+x251 y1+y221
' 31,3‘220 yle220 Note that in each remaining case, at least one of the two LPs l_13_s an optimal solution.
Suppose (P) has an optimal solution 3:“. Then we showed that (P) must have a basic optimal solution, say relative to basis B, with x* = (x*B, x113) = (B‘lb, 0), and: Starting tableau Basis B tableau
1....xn ‘ slacks I x slacks .
1 ‘ 1 i 1
Ami"0J9 1:133. éJEB.B 4'58133
A j I i b ‘ B'lA ' 3'1 l B'lb
l I : l
where B‘lb Z 0, c—cBB'lA _<_ 0 and 413131 _<_ 0. We also know that ex* = optimal objective function value = cBB'lb. Consider the solution y* = cBB'1 for First note y* is {Lasiblg for (D) because : y* = cBB’1 2 0 and c  cBB'lA S 0 => y*A 2 c.
In (D) the gig; of y* is y*b = cBB‘lb. Since cx* = cBB'lb = y*b, the Optimality Corollary 111.4 shows that y* must be an optimal
solution for Thus the simplex algorithm shows that when (P) has an optimal solution x*,», (D) cannot be infeasible.
By symmetry (write (D) like (P) & apply above argument) it follows that when (D) has an optimal
solution y*, (P) cannot be infeasible. All this is summarized by: 111.5 Theorem (Strong Duality): For dual LPs (P) & (D), exactly one of the following four possibilities occurs:
(1) (P) is unbounded and (D) is infeasible;
(2) (D) is unbounded and (P) is infeasible;
(3) both (P) and (D) are infeasible; (4) both (P) and (D) have optimal solutions with the same objective value. Cl \ am I 31 was; K‘ l
c s \ v 2c» * \ 1‘\ 76 A. GAL L wkw ‘ wL \\
\ ( =3 ix
\i C 75 ‘ *f l: \\
l c =0 E i
E ‘ ' \
l i 4:) l
,1; 1' K}
. . I ﬁ 0 7, D ’1 0 7/ 0
An immediate consequence of thrs result is: / 4:) x
. ' a
‘2} : 0 4M». ‘ WILth V ‘\»> ,mmmwlm. III.6 Theorem (Complementary Slackness): Suppose x is (P)—feasible & y is (D)feasible. Then x 85
yareMéy(h—Ax)=0=(yAC)X. _ frog: If x, y are optimal in (P) & (D) then ex = yb, by Strong Duality. But Weak Duality
(III2) => at S (yA)x = y (Ax) 5 Yb So cx = yb implies all g in this last relation are actually 2. Thus cx = yAx = yb => y(b  Ax) = 0 =(yA c)x. (<=) Y(b  Ax) = 0 => yb = y(Ax) = (yA)x
=> yb = ex.
(yA  c)x = 0 => (yA)x = cx
The Optimality Corollary 111.4 now implies x,y are optimal in (P), D We interpret complementary slackness as follows. The condition y(b  Ax) = 0 says that m n
‘2 " E z 0 J WM ith dual variable 2 0 by feas. dith primal slack Z 0 by feas.
Since pairwise all terms are 2 0, We must have: For each i, either the ith dual variable or the ith primal slack is equal to zero. This is called complementarity. Alternatively, for each i,
a I I I n
positive slack 1n constraint 1 of (P) => bi — 2 a..x. > 0 => yi= 0. 5:1 11 J Similarly for the dual: For each j, m
positive surplus in constraint j of (D) :2 yiau  c. > 0 z) x. = 0. i=1 1J J J
111.? Example: (P) max x1 + 7x2 — 5x3 + 14x4 (D) min 24y1 + 4y2
s.t. 3x1 + 4x2 + 5x3 + 6x4 5 24 s.t. 3y1 — y2 2 1
‘X1+X2‘2X3 3‘1: X2: x3: X4 2 0 5y1 " 2Y2 2 '5 3’1, 3'2 2 0' 53 Consider the dual feasible solution y = (y1, y2) = (1, 4). We can use complementary slackness to prove y is (D)—optimal as follows:
{ yl > 0 => in every optimal (P)solution, ﬁrst constraint holds at =. Y2 > 0 =# similarly for (P)— constraint #2.
Also, { 4y1 + y2 = 8 > 7 => x2 = 0 in every optimal (P)solution.
5y1  2y2 =  3 > 5 => x3 = 0 in every optimal (P)—solution.
Thus we can consider only x1 & x4 to get: 3x1 + 6x4 = 24
}<¢x1=2,x4=3. This primal solution is feasible and has value 40; since y = (1, 4) and x = (2, 0, 0, 3) both have value
40, both are optimal. El Duality is symmetric: “The dual of the dual is the primal”. Why? Careful now, using transposes: (P) max ch (D) min be
s.t. AXT _<_ bT s.t. yA Z c
x 2 0 y 2 0 (All vectors are row vectors — transpose to get col. vectors.) Rewrite (D) as max (b)yT take min z(cT) <¢ max czT
s.t. (~AT)yT S cT (IEval s.t. z(—AT) 2 b > s.t. A2T S b7 yZO z_>_0 z>0. 57 III.9 Algorithm (Dual Simplex):
Step 0: Find an initial basic solution to the primal which is dual feasible  say tableau is E K .— 3.0 .— ..
0"!
0 Step 1: Select the exit variable in row r by the rule: If b N 0, STOP, the current basic solution is optimal. Otherwise, pick 1:3! to exit, Where b, = min bi < 0. 1 , ’ .
Step 2: Select the entry variable in column k by the rule: If id _>_ 0 for all 5, STOP, the primal . , E.
problem is infeasible. Otherwise, pick xk to enter, where n—l‘l; = . _in 0{ a < at J: rJ rj Step 3: Pivot on ark & return to Step 1.  U 111.10 Example: Infeasibility is demonstrated here in Step 2 of the dual simplex procedure. 9 In Example 111.1 use r.h.s. ( 8 ) to get new ﬁnal tableau: .
1 Pivot on (:r) Says that x1 + x2 + x3 + x6 = 1
but for x1, x2, x3, x6 2 0 this is clearly impossible. D MM LHC‘ 4* 3KZ+ 313 at 3x,+ 1L+¢>c3sci
3%: "' ZXz‘i' liygglo
1: i" 752.4” ;53 5‘4 XIJXL "L"; 710. 58
To see that the dual simplex procedure is ﬁnite, we can argue as before. Again we have only a ﬁnite number of bases, so barring dual degeneracy, objective increases strictly at each step and the procedure must terminate. We could also argue ﬁniteness using the least index rule, interpreting the procedure here as if we were executing the usual simplex algorithm on the dual LP.
Some applications of the dual simplex method: (1) Sensitivity analysis when r.h.s. change 2 loss of (P)—feasibility. (2) Raoptimization after entering a new constraint into an optimal tableau  useful in
“integer programming”, studied in OMIE 321. (3) Solving problems for which there is a natural starting dual feasible solution, but no natural starting primal feasible solution  can avoid use of artiﬁcials.
We’ve seen examples of (1) and we discuss (2) in detail later. 111.11 Example: An example of (3) above. min 9y1 + 10y2 + 4y3
s.t. 3y1 + 3y2 + y3 _>_ 4 Y1+2Y2+ 3’3 23 Y1, Y2: 3’3 2 0
1
max 9y1  10y2  4y3
s.t. ~3y1  3y2  y3 + y4 = —4
'Y1'2Y2‘ y3 +Y5=’3 y1,y2,y3,y4.y5 _>_ 0 Basis y1 y2 y3 y 4 y5 rhs y4 out, yl in y5 out, y2 in yl out, y3 in optimal 59 Note that there is an important drawback in Example III.11 —— we don’t discover a feasible solution for the primal until optimality; the procedure is a socalled dual method. More on Sensitivity: Addition/deletion of variables and constraints °
Variable addition: Given an optimal tableau, we want to consider the effect of adding a new column
to the original problem. Don’t start over, just add new column as nonbasic and continue via primal simplex if cj  cBB'lAj > 0, else old solution is still optimal. 1
J B Aj Example: (Initial and ﬁnal tableaus only) New. ‘
actrvrt updated new activity now let x7 enter basis
Sc continue via primal 4 simplex algorithm. Variable deletion: If xj is to be deleted from the ﬁnal tableau and it is nonbasic, simply drop Xi and tableau remains optimal.
Now suppose x is basic, say basic in row i. We want to get 1: out of the basis. We consider two J J
cases: 1. Row i has a negative entry, say in < 0. Then do min ratio test to determine '6 E
ii]: = aging; 0 :5. Pivot on Eik retains dual feasiblity. This is clear for a“ 2 0, and when a” < 0 it follows because of the min ratio selection of index k. 60 2. Row i has no negative entry.
3.. an > 0 for some E sé j. '6 E
Then do diﬁerent min ratio test to get :7;— = min 33? .
1k 5_ la
If ,
#3
Pivot on Eik retains dual feasibility  check this as in case 1. ' b. Eijzl is the only nonzero row i entry. Then xj enters into no other equation  just drop it. E.g., to delete x2 in previous example, pivot on 525 = 1 and x5 enters basis for x2  drop 'X2  but primal feasibility is lost — continue via dual simplex algorithm. Constraint addition: We Wish to add a single constraint to the optimal tableau of a problem. Don’t
start over from scratch — simply write the constraint in updated form & let its slack become basic. If the new solution is no longer feasible, reoptimize using the dual simplex method. Example: Add restriction x1 + x2 5 3 to previous problem. Note that the old optimal basis had x1 = x2 = 2, so the constraint gill be violated. Thus in updated form, the tableau will be infeasible => use dual simplex method. 2 updated form of constraint is
[*X3 + 0x5 " 1X6 + X7 = ‘1]
Enter into tableau. & in ﬁnal tableau: x1 = 2  2x3  x5 + 2x6 X1+x2$3¢>x1+x2+x7=38zx720 }
x2=2+x3+x5~3x6 Dual Simplex
pivot x4 out, x3 in 61
Remarks: (1) Row addition can make the original primal problem infeasible. If so, this will be indicated by an unbounded dual solution (since we know dual is feasible). (2) Row addition of equality constraint may require introduction of an artiﬁcial variable and a return to Phase I  treat this in the obvious way. Constraint deletion: For (g) — type constraint. _
Assume slack for this constraint is basic in ﬁnal tableau  if not, we pivot to force the slack basic as ,
described in the following three cases, then simply delete this row and the column corresponding to its
slack. (1) If the slack column has some positive entry, do a min ratio primal test and thus pivot to preserve primal feasibility  drop row and basic slack and reoptimize using prim...
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