Solutions to exam in SF1811 Optimization, Jan 14, 2015

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1 Solutons to exam n SF8 Optmzaton, Jan 4, O------O -4 \ / \ / The network: \/ where all lnks go from left to rght. /\ / \ / \ 6 O------O (a) Let x = ( x 3, x 4, x 23, x 24 ) T, where the varable x j stands for the flow n the lnk from node to node j, and let c = ( c 3, c 4, c 23, c 24 ) T. Then the total cost for the flow s gven by c T x. The flow balance condtons n the nodes can be wrtten Ax = b, where 3 6 A = and b = 4 5 Fnally, the gven drectons of the lnks mply the constrants x. (It s recommended to remove the last row n A and the correspondng last component n b to get a system wthout any redundant equaton.).(b) The four dfferent spannng trees are shown n the followng fgure, together wh the unque lnk flows whch satsfy Ax = b O------O O------O O------O O O \ / \ / \ / 5\ / \ / 3\ / \/ -\ / \/ /\ \ / /\ 6/ \ \ / 4/ \ / \ \ / / \ O O O------O O------O O------O Spannng Spannng Spannng Spannng tree tree 2 tree 3 tree 4 The lnk flows n the frst spannng tree are calculated as follows: The only way to satsfy the supply constrant n node 2 s to let x 23 = 6. Then the only way to satsfy the demand constrant n node 3 s to let x 3 = 2. Then the only way to satsfy the supply constrant n node s to let x 4 = 5. Then the demand constrant n node 4 s also satsfed.

2 The lnk flows for the other spannng trees are calculated n a smlar way. We thus have the followng four basc solutons: x = ( 2, 5, 6, ) T, correspondng to spannng tree number, x = ( 4,,, 6) T, correspondng to spannng tree number 2, x = ( 3,,, 5) T, correspondng to spannng tree number 3, and x = (, 3, 4, 2) T, correspondng to spannng tree number 4. All these four solutons satsfy Ax = b, but only the last two satsfy x. Thus, the basc solutons correspondng to spannng trees 3 and 4 are feasble basc solutons, whle the basc solutons correspondng to spannng trees and 2 are nfeasble basc solutons..(c) For a gven feasble basc soluton, the smplex multplers y for the dfferent nodes are calculated from y 4 = and y y j = c j for all lnks (, j) n the correspondng spannng tree. For the feasble basc soluton correspondng to spannng tree 3, we get y 4 =, y 2 = y 4 + c 24 = c 24, y 3 = y 2 c 23 = c 24 c 23, y = y 3 + c 3 = c 24 c 23 + c 3. The reduced cost for the only non-basc varable s then gven by r 4 = c 4 y + y 4 = c 4 c 24 + c 23 c 3. Thus, f c 3 c 4 c 23 + c 24 < then r 4 >, and then x = (3,,, 5) T s the unque optmal soluton to the consdered problem. For the feasble basc soluton correspondng to spannng tree 4, we get y 4 =, y 2 = y 4 + c 24 = c 24, y = y 4 + c 4 = c 4, y 3 = y 2 c 23 = c 24 c 23, The reduced cost for the only non-basc varable s then gven by r 3 = c 3 y + y 3 = c 3 c 4 + c 24 c 23. Thus, f c 3 c 4 c 23 + c 24 > then r 3 >, and then x = (, 3, 4, 2) T s the unque optmal soluton to the consdered problem. If c 3 c 4 c 23 + c 24 = then both the above feasble basc solutons are optmal solutons, and then every convex combnaton of these two solutons,.e. x = t (3,,, 5) T + ( t) (, 3, 4, 2) T, where t [, ], s also an optmal soluton snce the constrants and the objectve functon are lnear. As an example, the followng three solutons (correspondng to t = /3, /2 and 2/3) are optmal solutons for the case that c 3 c 4 c 23 + c 24 = : x = (, 2, 3, 3) T, x = (5, 5, 25, 35) T and x = (2,, 2, 4) T. 2

3 2.(a) We have an LP problem on the standard form where A = 2 3 4, b = mnmera c T x då Ax = b, x, ( ) 8 and c 4 T = (4, 4, 2, 4, 4). The startng soluton should have the basc varables x and x 5, whch means that β = (, 5) and ν = (2, 3, 4) The correspondng basc matrx s A β =, whle A 4 ν =. 3 2 The values of the current basc varables are gven by x β = b, where the vector b s calculated from the system A β b = b,.e. ) ( ) ) ( ) 4 ( b 8 ( b =, wth the soluton b = =. 4 b2 4 b2 2 The vector y wth smplex multplers s obtaned by the system A T β y = c β,.e. ( ) ( ) ( ) ( ) 4 y 4 y =, wth the soluton y = =. 4 4 y 2 Then the reduced costs for the non-basc varables are obtaned from 2 3 r T ν = c T ν y T A ν = (4, 2, 4) (, ) = (, 2, ). 3 2 Snce r ν2 = r 3 = 2 s smallest, and <, we let x 3 become the new basc varable. Then we should calculate the vector ā 3 from the system A β ā 3 = a 3,.e. ) ( ) ) ( ) 4 (ā3 2 (ā3.5 =, wth the soluton ā = =..5 ā 23 The largest permtted value of the new basc varable x 3 s then gven by { } { } b t max = mn ā 3 > = mn ā 3.5, 2 =.5.5 = b. ā 3 Mnmzng ndex s =, whch mples that x β = x should no longer be a basc varable. Its place as basc varable s taken by x 3, so that β = (3, 5) and ν = (2,, 4) The correspondng basc matrx s A β =, whle A 2 ν =. 3 4 The values of the current basc varables are x β = b, where the vector b s calculated from the system A β b = b,.e. ) ( ) ) ( ) 2 4 ( b 8 ( b 2 =, wth the soluton b = =. 2 b2 4 b2 y 2 ā 23 3

4 The vector y wth smplex multplers s obtaned from the system A T β y = c β,.e. ( ) ( ) ( ) ( ) 2 2 y 2 y =, wth the soluton y = =. 4 4 y 2 Then the reduced costs for the non-basc varables are obtaned from 3 r T ν = c T ν y T A ν = (4, 4, 4) (, ) = (3, 4, ). 3 4 Snce r ν the current feasble basc soluton s optmal. Thus, x = (,, 2,, ) T s an optmal soluton, wth optmal value c T x = 8. y 2 2.(b) If the prmal problem s on the standard form mnmze c T x subject to Ax = b, x, the correspondng dual problem s: maxmze b T y subject to A T y c, whch becomes maxmze 8y + 4y 2 subject to 4y 2 4, y + 3y 2 4, 2y + 2y 2 2, 3y + y 2 4, 4y 4. Ths dual problem can be llustrated by drawng the constrants and some level lnes for the objectve functon n a coordnate system wth y and y 2 on the axes. (The fgure s omtted here.) It s well known that an optmal soluton to ths problem s gven by the vector y of smplex multplers for the optmal basc soluton n (a) above,.e. y = (, ) T. Alternatvely, ths can be seen from the fgure (whch s omtted here). Check: It s easy to verfy that y = (, ) T satsfes the dual constrants, wth dual objectve value 8y + 4y 2 = 8 = the optmal value of the prmal problem. Thus, y = (, ) T s an optmal soluton to the dual problem. 4

5 2.(c) If the second constrant n the prmal problem s removed, the correspondng dual problem becomes maxmze 8y subject to y 4, y 4, 2y 2, 3y 4, 4y 4. The optmal soluton to ths problem s clearly y =, wth the optmal value 8y = 8. But then the optmal value of the reduced prmal problem must also be = 8. Snce the optmal soluton x = (,, 2,, ) T from (a) above s feasble also to the reduced prmal problem, and stll has the objectve value c T x = 8, t follows that x = (,, 2,, ) T s an optmal soluton also to the reduced prmal problem! (But not a basc soluton. Two optmal basc solutons are now x = (,, 4,, ) T and x = (,,,, 2) T, wth objectve values = 8.) 2.(d) If the frst constrant n the prmal problem s removed, the correspondng dual problem becomes maxmze 4y subject to 4y 4, 3y 4, 2y 2, y 4, y 4. The optmal soluton to ths problem s clearly y =, wth the optmal value 4y = 4, and then the optmal value of the reduced prmal problem must also be = 4. But the optmal soluton x = (,, 2,, ) T from (a) above stll has the objectve value c T x = 8 > 4, so t can not be an optmal soluton to the reduced prmal problem! (Two optmal basc solutons are now x = (,, 2,, ) T and x = (,,,, ) T, wth objectve values = 4.) 5

6 3.(a) The objectve functon s f(x) = 2 xt Hx + c T x, wth H = 2, c =. LDL T -factorzaton of H gves H = LDL T = Snce there s a negatve dagonal element n D, the matrx H s not postve semdefnte, whch n turn mples that there s no optmal soluton to the problemen of mnmzng f(x) wthout constrants. (Wth e.g. d = (,, ) T, f(t d) = t 2 when t.) 3.(b) We now have a QP problem wth equalty constrants,.e. a problem of the form mnmze 2 xt Hx + c T x subject to Ax = b, wth H and c as above, A = [ ] and b =. The general soluton to Ax = b,.e. to x x 2 + x 3 =, s gven by x x 2 = + v + v 2, for arbtrary values on v and v 2, x 3 whch means that x = s a feasble soluton, and Z = s a matrx whos columns form a bass for the null space of A. After the varable change x = x+zv we should solve the system (Z T HZ)v = Z T (H x+c), provded that Z T HZ s at least postve semdefnte. [ We have that Z T 2 HZ = 2 4 The system (Z T HZ)v = Z T (H x + c) becomes ], whch s postve semdefnte (but not postve defnte). [ ]( ) ( ) 2 v =, 2 4 v 2 ( ) 2t wth the solutons ˆv(t) =, for arbtrary values on the real number t, whch mples t t that ˆx(t) = x + Zˆv(t) = 2t, for t IR, are the (nfnte number of) optmal solutons. t Note that f(ˆx(t)) = for all t IR. 6

7 3.(c) Agan, we have a problem on the form: mnmze 2 xt Hx + c T x subject to Ax = b, ( ) wth H and c as above, A =, and b =. x The general soluton to Ax = b s now x 2 = v, for v IR, whch mples that x 3 x = s a feasble soluton, and z = form a bass for the null space of A. After the varable change x = x + zv, we should solve the system (z T Hz)v = z T (H x + c), provded that z T Hz s at least. We have that z T Hz = 6 >, so the system (z T Hz)v = z T (H x + c) becomes 6v =, wth the unque soluton ˆv =, so that ˆx = x + z ˆv = s the unque optmal soluton. 7

8 4.(a) The Lagrange functon for the consdered problem s gven by L(x, y) = 2 (x q)t (x q) + y T (b Ax), wth x IR n and y IR m. The Lagrange relaxed problem PR y s defned, for a gven y, as the problem of mnmzng L(x, y) wth respect to x IR n. Snce L(x, y) = 2 xt I x (A T y + q) T x + b T y + 2 qt q, and the unt matrx I s postve defnte, the unque optmal soluton to PR y s gven by x(y) = A T y + q. Then the dual objectve functon becomes ϕ(y) = L( x(y), y) = 2 (AT y + q) T (A T y + q) + b T y + 2 qt q = = 2 yt AA T y + (b Aq) T y. [ 4.(b) From now on, A = ( 4 6 Then AA T = and b Aq = 4 3 ] ( 6, b = 3 ) ( ) 2 = 4 ) ( 4 and q = (, 2, 2, ) T. so that the dual objectve functon becomes ϕ(y) = 2y 2 2y y y 2. Alternatve calculaton of the dual functon, wthout usng the results from (a): The consdered problem s: mnmze f(x) subject to g (x) and g 2 (x), where f(x) = 2 (x ) (x 2 2) (x 3 2) (x 4 ) 2, g (x) = 6 x x 2 + x 3 x 4 and g 2 (x) = 3 x x 2 x 3 + x 4. The Lagrange functon then becomes: L(x, y) = 2 (x ) (x 2 2) (x 3 2) (x 4 ) 2 + y (6 x x 2 + x 3 x 4 ) + y 2 (3 x x 2 x 3 + x 4 ) = = 2 (x ) 2 (y +y 2 ) x + 2 (x 2 2) 2 (y +y 2 ) x (x 3 2) 2 (y 2 y ) x (x 4 ) 2 (y y 2 ) x 4 + 6y + 3y 2. Mnmzng ths wth respect to x gves: x (y) = +y +y 2, x 2 (y) = 2+y +y 2, x 3 (y) = 2+y 2 y, x 4 (y) = +y y 2, so that x(y) = (+y +y 2, 2+y +y 2, 2+y 2 y, +y y 2 ) T, and then the dual functon becomes ϕ(y) = L( x(y), y) = 2 ( x(y) ) 2 (y +y 2 ) x(y) + 2 ( x(y) 2 2) 2 (y +y 2 ) x(y) ( x(y) 3 2) 2 (y 2 y ) x(y) ( x(y) 4 ) 2 (y y 2 ) x(y) 4 + 6y + 3y 2 = 2 (y +y 2 ) 2 (y +y 2 ) (y +y 2 ) (y +y 2 ) 2 2(y +y 2 ) (y +y 2 ) (y 2 y 2 ) 2 2(y 2 y ) (y 2 y ) (y y 2 ) 2 (y y 2 ) (y y 2 ) 2 + 6y + 3y 2 = = 2y 2 2y y y 2, as above. ), 8

9 The dual problem then becomes: D: maxmze ϕ(y) = 2y 2 2y y y 2 subject to y and y 2, whch decomposes nto the two separate problems D : maxmze 2y 2 + 4y subject to y, and D 2 : maxmze 2y 2 2 y 2 subject to y 2. Clearly, the optmal soluton to the frst problem s ŷ =, whle the optmal soluton to the second problem s ŷ 2 =. Thus, ŷ = (, ) T s the unque optmal soluton to D, wth ϕ(ŷ) = 2. 4.(c) Let ˆx = x(ŷ) = (+ŷ +ŷ 2, 2+ŷ +ŷ 2, 2+ŷ 2 ŷ, +ŷ ŷ 2 ) T = (2, 3,, 2) T. Then Aˆx b = (6, 4) T (6, 3) T, so ˆx s a feasble soluton to the prmal problem. Further, the prmal objectve value s f(ˆx) = 2 (ˆx q)t (ˆx q) = 2 (,,, )T 2 = 2. Snce ˆx s feasble to P and f(ˆx) = ϕ(ŷ), we conclude that ˆx s an optmal soluton to P. x 4.(d) Snce f(x) T = x 2 2 x 3 2, g (x) T =, g 2(x) T =, x 4 the KKT condtons become: x (KKT-): x 2 2 x y + y 2 = x 4 (KKT-2): 6 x x 2 + x 3 x 4 and 3 x x 2 x 3 + x 4. (KKT-3): y and y 2. (KKT-4): y (6 x x 2 + x 3 x 4 ) = and y 2 (3 x x 2 x 3 + x 4 ) =.. Wth x = ˆx = (2, 3,, 2) T and y = ŷ = (, ) T, we get that 2 (KKT-): =. OK! 2 (KKT-2): (KKT-3): (KKT-4): = and =. OK! and. OK! = and ( ) =. OK! Thus, ˆx and ŷ satsfy the KKT condtons, and snce the consdered problem s a convex problem, we can conclude (agan) that ˆx s a global optmal soluton. 9

10 5.(a) Change notaton and let the varable vector be called x,.e. x = (x, x 2, x 3 ) T = (x, y, r) T. m Then f(x) = 2 h (x) 2 = 2 h(x)t h(x), where = h (x) = (x a ) 2 + (x 2 b ) 2 x 3 and h(x) = (h (x),..., h m (x)) T. The gradent of h s gven by ( h (x) = x a (x a ) 2 + (x 2 b ) 2, x 2 b (x a ) 2 + (x 2 b ) 2, ), and h(x) s the m 3 matrx wth these gradents as rows. Wth the gven data, we get that f(x) = 2 (h (x) 2 + h 2 (x) 2 + h 3 (x) 2 + h 4 (x) 2 ), where h (x) = (x 5) 2 + x 2 2 x 3, h 2 (x) = x 2 + (x 2 6) 2 x 3, h 3 (x) = (x + 4) 2 + x 2 2 x 3, h 4 (x) = x 2 + (x 2 + 5) 2 x 3. The startng pont should be x () = (,, 5) T. and then h(x () ) =, f(x() ) = and h(x () ) = In Gauss-Newtons method, we should solve h(x () ) T h(x () )d = h(x () ) T h(x () ), 2 d.5 whch becomes 2 d 2 =, wth the soluton d () =.5. 4 d 3 We try t =, so that x (2) = x () + t d () = x () + d () = (.5,.5, 5) T. Then h(x (2) ) = ( , , , ) T = 2 ( 82, 22, 82, 22 ) T 2 (,,, )T, so that f(x (2) ) = 2 h(x(2) ) T h(x (2) ) 8 ( ) = 2 < = f(x() ), Thus, the choce t = s accepted, and x (2) = (.5,.5, 5) T s the next teraton pont..

11 5.(b) Let (x, x 2 ) = the locaton of the (common) center of C and C 2, z = the square of the radus of the small crcle C, and z 2 = the square of the radus of the large crcle C 2. Then the problem can be formulated as follows n the varables are x, x 2, z and z 2 : mnmze πz 2 πz subject to (x a ) 2 + (x 2 b ) 2 z, =,..., m, (x a ) 2 + (x 2 b ) 2 z 2, =,..., m. 5.(c) For each gven pont (x, x 2 ) T IR 2, the correspondng optmal values of z and z 2 n the above problem are clearly ẑ (x, x 2 ) = mn{(x a ) 2 + (x 2 b ) 2 } and ẑ (x, x 2 ) = max{(x a ) 2 + (x 2 b ) 2 }. Thus, the above problem can be wrtten mnmze π max{(x a ) 2 + (x 2 b ) 2 } π mn{(x a ) 2 + (x 2 b ) 2 }, whch s a problem n just the two varables x and x 2. (However, the objectve functon s not dfferentable, so ths s not a correct answer to 5.(b).) But mn{(x a ) 2 + (x 2 b ) 2 } = mn { 2a x + a 2 2b x 2 + b 2 }, and x 2 + x mn max {x 2 2a x + a 2 + x 2 2 2b x 2 + b 2 } = {(x a ) 2 + (x 2 b ) 2 } = max{x 2 2a x + a 2 + x 2 2 2b x 2 + b 2 } = x 2 + x max { 2a x + a 2 2b x 2 + b 2 }. Therefore, max{(x a ) 2 + (x 2 b ) 2 } mn{(x a ) 2 + (x 2 b ) 2 } = max{ 2a x + a 2 2b x 2 + b 2 } mn{ 2a x + a 2 2b x 2 + b 2 }, so the above problem can be wrtten mnmze π max{ 2a x + a 2 2b x 2 + b 2 } π mn{ 2a x + a 2 2b x 2 + b 2 }, whch may equvalently be wrtten mnmze πw 2 πw subject to w + 2a x + 2b x 2 a 2 + b 2, =,..., m, w 2 + 2a x + 2b x 2 a 2 + b 2, =,..., m, whch s an LP problem n the varables x, x 2, w and w 2.

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