THEOREM: For fesible liner progrm in its stndrd form, the optimum vlue of the objective over its nonempty fesible region is () either unbounded or (b) is chievble t lest t one extreme point of the fesible region. We will see wht is ment by stndrd form very shortly More generlly, the bove theorem nd the grphs we sw tell us tht every LP is in exctly one of the following sttes: 1. Fesible with unique optimum solution (cluse b of theorem) 2. Fesible with infinitely mny optim (cluse b of theorem) 3. Fesible, with no optimum solution becuse the objective is unbounded (cluse of theorem) 4. Infesible, nd hence with no optimum solution 2018, Jynt Rjgopl
Assume for now tht we hve fesible LP nd tht the objective is bounded (i.e., we re in Sttes 1 or 2). Then the following re true: A. The LP hs t lest one optiml corner point (or extreme point) 1. If in Stte 1, exctly one extreme point is optiml 2. If in Stte 2, t lest two djcent (neighboring) extreme points re optiml B. The number of extreme points is finite C. If the objective function t some extreme point is s good s or better thn t ll of its djcent extreme point, then this extreme point is optiml for the LP This suggest simple lgorithmic pproch to finding the optimum of fesible nd bounded LP 2018, Jynt Rjgopl
STEP 0 (Initiliztion): Find n initil extreme point nd mke it the current cndidte (if one cnnot be found the LP is in stte 4, i.e. it is infesible so STOP). STEP 1(Stopping Criterion Check): Is the objective t the current extreme point t lest s good or better thn it is t ll of its djcent (neighboring) extreme points? If so this must be the optiml extreme point (vi C) so STOP. If not, go to Step 2. STEP 2(Itertive Step): One (or more ccurtely, t lest one ) of the djcent extreme points is better so mke it the current cndidte. Then return to Step 1. QUESTIONS: 1) How to find n initil extreme point? 2) Wht is the lgebric chrcteriztion of n extreme point? 3) Wht is the lgebric chrcteriztion of djcent extreme points, i.e., how to move from n extreme point to one of its neighbors (in Step 2)? 2018, Jynt Rjgopl
The SIMPLEX Method - Development Every LP is in exctly one of the following sttes: 1. Fesible with unique optimum solution. 2. Fesible with infinitely mny optim. 3. Fesible, with no optimum solution (becuse the objective is unbounded). 4. Infesible, nd hence with no optimum solution. Assume we re in Sttes 1 or 2. Then the following re true: A. The LP hs t lest one optiml corner point (or extreme point). 1) If in Stte 1, exctly one extreme point is optiml. 2) If in Stte 2, t lest two djcent (neighboring) extreme points re optiml. B. The number of extreme points is finite. C. If the objective function t some extreme point is s good s or better thn t ll of its djcent extreme point, then this extreme point is optiml for the LP.
It follows then tht we cn use the following itertive lgorithmic procedure: STEP 0 (Initiliztion): Find n initil extreme point nd mke it the current cndidte (if one cnnot be found the LP is in stte 4, i.e. it is infesible so STOP). STEP 1 (Stopping Criterion Check): Is the objective t the current extreme point t lest s good or better thn it is t ll of its djcent (neighboring) extreme points? If so this must be the optiml extreme point (vi C) so STOP. If not, go to Step 2. STEP 2 (Itertive Step): At lest one of the djcent extreme points is better so mke it the current cndidte nd go to Step 1. QUESTIONS: 1) How to find n initil extreme point? 2) Wht is the lgebric chrcteriztion of n extreme point? 3) Wht is the lgebric chrcteriztion of djcent extreme points, i.e., how to move from n extreme point to one of its neighbors (in Step 2)?
Stndrd Form of Liner Progrm Three conditions must be met: 1) All constrints must be stted s equlities of the form g k (x)=b k, where g k (x) is liner function of x. 2) The RHS for ech constrint must be nonnegtive, i.e., ll b k 0 3) All vribles must be nonnegtive, i.e., x i 0 for ll i. Thus the LP (with m constrints nd n vribles) looks like Min (or Mx) c 1 x 1 + c 2 x 2 + + c n x n st 11 x 1 + 12 x 2 + + 1n x n = b 1 21 x 1 + 22 x 2 + + 2n x n = b 2 : m1 x 1 + m2 x 2 + + mn x n = b m x 1, x 2,, x n 0. where b 1, b 2,, b n 0. More compctly, let c=[c 1 c 2... c n ] nd A= : 11 21 m1 12 22 : m2......... 1n 2n : mn x1 x 2 x= : x n b1 b 2 nd x= : b m
Then the LP my be restted s Min (or Mx) cx, st Ax=b, x 0, where b 0. This finl system of constrint equtions of the LP in stndrd form is clled the AUGMENTED SYSTEM. Consider such n ugmented system of n vribles in m equtions, where m n. Note tht the ugmented system does not include the nonnegtivity conditions! Also note tht this system hs three possible sttes: (1) no solution, (2) exctly one solution, or (3) infinitely mny solutions. Fesible solution: Any solution to the ugmented system tht lso stisfies nonnegtivity is clled fesible solution E.g. Consider the following constrints nd the corresponding ugmented system: x 1 + 2x 2 120 x 1 + 2x 2 + S 1 = 120 x 1 + x 2 90 x 1 + x 2 + S 2 = 90 x 1 70 x 1 + S 3 = 70 x 2 50 x 2 + S 4 = 50
Note tht n=6 nd m=4. 1) An (infesible) solution is [x 1 =70, x 2 =50, S 1 =-50, S 2 =-30, S 3 =0, S 4 =0] 2) A fesible solution is [x 1 =20, x 2 =20, S 1 =60, S 2 =50, S 3 =50, S 4 =30] Bsic Solution: Suppose we fix (n-m) out of the n vribles t zero, nd try to solve the system of m equtions in the remining m vribles. If solution to this exists, then it is clled bsic solution. Solution (1) on the previous pge is n exmple of bsic solution. Bsic Fesible Solution: A bsic solution tht lso stisfies nonnegtivity is clled bsic fesible solution (BFS). An exmple of BFS is [x 1 =20, x 2 =50, S 1 =0, S 2 =20, S 3 =50, S 4 =0]. Note tht (1) is not BFS even though it is bsic solution! Augmented System One Solution No Solution Infinitely mny solutions If it is lso fesible ( 0) Originl LP is infesible then the originl LP hs Fesible Infesible exctly one fesible solution - it is lso optiml! Bsic Nonbsic Bsic Nonbsic
FACT: Ech BFS corresponds to n extreme point of the fesible region. In bsic solution, the (n-m) vribles tht re chosen to be fixed t zero re clled the nonbsic vribles nd the remining m vribles re clled bsic vribles. Note tht we cn choose (n-m) out of the n vribles in n n n! = = different wys. Thus we cn get n m m ( n m)! m! mximum of this mny different bsic solutions (some of which will lso be bsic fesible solutions) in prctice ll of these my not exist since in some cses, the resulting m x m system my not hve solution! n 6! In our exmple, n=6 nd m=4 so tht = = 15 n m (6 4)!4! The 15 combintions of nonbsic vribles re: 1) x 1, x 2 2) x 1, S 1 6) x 2, S 1 3) x 1, S 2 7) x 2, S 2 10) S 1, S 2 4) x 1, S 3 8) x 2, S 3 11) S 1, S 3 13) S 2, S 2 5) x 1, S 4 9) x 2, S 4 12) S 1, S 4 14) S 2, S 4 15) S 3, S 4 Try to locte ech of these on the grphicl representtion!
X 2 Mx 20X 1 +10X 2 S 3 =0, X 1 =70 st X 1 +2X 2 120 (Constr. 1, S 1 ) X 1 +X 2 90 (Constr. 2, S 2 ) 90 X 1 70 (Constr. 3, S 3 ) S 2 =0, X 1 +X 2 =90 X 2 50 (Constr. 4, S 4 ) X 1, X 2 0 50 II S 1 =S 4 =0 S 2 =S 4 =0 S 3 =S 4 =0 S 4 =0, X 2 =50 III 10 I S 1 =S 2 =0 X * 1 =70 IV S 1 =S 3 =0 X * 2 =20 Fesible Region Opt. Vl. =1600 V S 2 =S 3 =0 VI S 1 =0, X 1 +2X 2 =120 10 20 30 40 50 60 70 80 90 100 110 120 X 1 1., : BASIC SOLUTIONS (13 of them). Note tht there should be ( 4 6 )=15 of these, but only 13 exist becuse X 1 =70 is prllel to the X 2 -xis nd X 2 =50 is prllel to the X 1 -xis. 2. : BASIC FEASIBLE SOLUTION: (6 of the bove 13, numbered I, II, III, IV, V, VI) 3. : Contour of the objective function corresponding to vlue of 1600 4. : Set of ll Fesible solutions to the LP