HW9_Solutions

HW9_Solutions - fimek ‘1 for E4004. LPNs. 6.16 Sohm...

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Unformatted text preview: fimek ‘1 for E4004. LPNs. 6.16 Sohm (Mina dqnmm pmqmmmrnfi. J; (a; PM W sh»er paw W W afic‘rn at M002, A. Damn frcvr Length of vhz SW? Paw from A to U. using mam 7 edqu. i=l , f.(B)=2. f.co)=‘l, f1(3)=uo, f,cp;=lo. f.(c>=oo_ 1:), ](2.43): min (051 f.(C), madam), Eb-rflJE), F%+f,(p))= F] , (D ,F) We): mfn( Bedlam. Do+f.(D>, Fc+ f.(P>)= 5. (B) . fut»:— mrn( owfim, bv+f.(9>)= 1> (9) guy): mrn( 9E+ 19(3), FE+F.(F))': p (3,1?) 1MP): mrn( up {43), OH {1, (a), ewfim): ‘? (a), 1:3,, {5(8): mm ( 0% {21(0), bé+fsti>x fb+fikthE+fitP>= 8 (C) jaw): WMBc-ffum, bcfifitbx Fawn): I7 (F) fgw): mrn(w+f.cc), BD+ flue»): :8 (0) {5(5): mrn(ge+fi(la>, 3FE+ fun): H (F) fun: mm (EFJr fin», (Inna), EF+ 11(6)): I’; (0) «=4, 114(3): rm'n(3;+‘?2.(0), Io+fw>>, Mimi), 7+fim>= >0 (C) {14(6):an ( n, m. 1: )=H (a) f4u>>= mrncao, l3)$!5 (9) 344(5): min( «8, I5)=15. (F) f4(F)=mrn(ng,25, 15):» (E) f: 5, #03): mm ( c4, :8, >5,>o>=(4, (C) f;(c>= mFAQ», 4., M): J. (p) fut»: rm‘nU4, go): 24 (0) {4(5): minUo, m: :5 (1:) {’NF): MDT "1/ m: :7 (E) E; shortest LUMij shown [min- shortest pm tree: Ma): 2 A—ue . 1141(6): 5 Aagac fl(m:7 MD {HEF‘O A-vE ¥,(F)=‘i Aw»? (b) PM W sfumcut pm tree ma m‘qm an: mm C. Dm 3am): bu)qu of flu slwrtm pm from a tv u, using mam“ I adage/s. 12'. f.(A)= oo, was J; ,f.(1>>= (J) , {49:00 , fat): 8. 1:), Wu»: mfmlimfim, 5mm), em 12m, mqmm: 5 (a) 1am»): mmMB’rWM, DB—rfiw), Ewfim, F0+€um= :5 (F) f>(D)=m;‘n(1ArD+fl-(A)l lawfwen: n, (m we): mrnmafim). 33+ 10.02.). FE+fi<F>>= [0 (F) {149: mm (AFJrfMA). Wham, "EF-f‘fitEH: 10 (3,15) 1:1), qu): mrn(n7, )«o, )0, >0): :7 (g) {3(3): mrnq, >5, yo, :7 )=7 (A) ’ fi<p>=mrn( p, >5): [2 (A) {1(E)=mm(15, 25, 1;): 12, (F) , fJF): mmua )>.12)=|2 (E) [:4, fim): mrnUf, “1.2;, 22): ‘1 (B) 104(3): mm, >2, :2, 1‘1): H1 (A,F>,f4(p)= mrn(24, m: '7 (9) fans): mrnb‘], F1, «4 )= [4 (F) , {Zapammu‘h (4, l4)=l4.(B,E) . i=5, M): mm(1l, 14, >4, >47: 1!, (8) {5(9): mm(l|,l7,14, 1|): ll (A),f;(D>=mrn(Ib,a7>=rb (A) 125(13): minklfi, 1‘1, Ib)‘: 16(13), fHFF mrn(lfi,>b,1b)=rb (E). ,-—> shortut Lu)qu shortest pm: ghor‘tm W two fJA)‘ 5 CegaA ‘ f1de 1; Cat; g 10w)? :2 cabaAab 121(9): (0 cape ficF>= 8 cal: 9. Use dalmmrc pmljrAmmfnfij {79 gall/a a knapsaok Problbm in which W knapsaob Can hold up to 1); Ma and flu imms 1er Coner m cu famous: item waqmtm lgwfrt l ), [1 2 5 7 >5 1; 7 50 Answer fad) max bwefat of units (7f 00%th m cpw 071 Jam ’0 th )2 X. it of {tans u b.=1> 0;), W=1),. = ))‘ want fin)». 0 fuel): {9m an d. _ 9 - 7: @ fuoh— £333 .bsxw fwd ax». bah. 9 €40): f6“): "‘= 5(6): 0 , Xr= 0. fan): fs(8)=~-- =12wM= 50, >0.= I. @ fz(d)= szXIXT ‘Ja; (d-Cfi‘fl’. '5 ,(o)= fuo): 0 , funsfsm: o, fibwfiuwo, flapfowwo, fd4)=)0)(47=0, X50. f:(5)=ma><?fi(5),15+fi(0)7c= :5, fab): max ?f»((7), >5+ 105m} = >5, X: =l {50): mm i fam, 25+ {’50) ’,= 50, {22(8): max H?qu 25+fsm’. = 50. X210 {54): mm ? fu‘h, 15+fi£4fl= 50. X, = 0. film): ww ffwm, 25+ 105(5) , 50+ {26(0)} = 50‘ X15 0, 2 f;(u)=m”\><?]as(m. 25+ “6), 50+f5m1= 5o. x;=o, 7, f¢(«2)=mm‘(fa(lz>, 254f5I—{M 50+fi(z)’u= 75. xt=l 30mm: ynaxéfiun, 25+ {15(3), 5044145)}: 75. mm SI blyv 4' . $ 11.01;): max 9.103(9), [2+ am7), 24+ ,(7). 3b+fi<4>, 48+f>(1>,=75, XI: 0. O uni-ts of item I => Total law'sz = 75. )7“ Conswtu vhb following pram: max 5x. +4x,+ >00, Sit. 4X.+ bx” YX; 5 8 XI. Xx, )0, 7,0 integer m» Sm WS 025 0 knapsauc problem. ‘Hm, knapsack can led H) to 8010. 17mm am 170 b@ wnsrdmal. Mm 00qu < vb) benefit I 4 g; 2 ’2 4 1/ 7 > (b) Construot a MW fLo modal corrupondrnéj to «bias pmblsm. a) 7 v, Noduu pairs ( item Consrdared, wuqhx. of knapsack). Arts is); c if we can mow, fawn .' t0 bid Win51 cumin number of 17M items conSIdmo‘ in , and mm considered in — char in} = I . Baunds= All upper baunds v . Lowe; bounds 0. "ExtUna‘ FKOW= S’fart [I], “find I’d. Arc [mph = baasz added If moving from i w J‘. Ara F(ow= I: 060%}? the, amp from 1' 1:0 J‘. 0- Wwisz. MAX W pod-A» from start; to and (Lorre/spends to an Optima! knapsad: solution. A‘nodner possTbkL mode/l rs an m mm: pmjz- . Nada/st me, i =5 , units of wag/d: m (aft for J;-i+l,---’). Ami-bji'fia'm we, 1-» swap—J): jumbst 1»qu wawtfarsmjej, Wu «C (puts (7" might (in spent on ma 1. Emma's: All uppu bounds I, Mam bands 0. EXUZ/rnal Flaw 9mm D3 , and ["3. Are Length = bwef’nr added from H» 3. Amflnw= Izacwrv vha Stapfnm in‘,o:o-b6mwrs¢ Max Luau“ pawx fmm 9mm +0 and arrayed; 170 an opfimal knapsack golm‘on. (a) 80W, m problem u/sfrwj dljnamfo pmajrammrncj. ffidh my bwefu‘t if d units ma sperm 0/) items “b,'t+u~--%. Yr- n of item! , bn= 5, ct=4 110:8, T=2; Xy— 5?? of Franz , 1794, Gz=)) want, f,(8>. X»: t? a? Mm); , b5=1, OF) (Dfudwo «far all 01. max Q 115(01): :22); fbaxzfi ffld‘ C4Xa))) = )00‘0‘ b5)“ b 124(0): fun = 0, X»: 0 , fim= fa)»: 2, x5=n {15(4): {25(5): 4 . x51“: , f5<e>=fm>= 6. xa,=% , {25(8)=8, ><+=4 Q) 111 (d )= $33 ? bx. + fud'bx.) ’.. a ;(o>= fun): 0, fun: fwmo, fun: {flow} , x; gm»: max H50), 4+fato)’. = 4 , x1: ll 0 {film—~— max Ham), 4+ {huff 4 , XL: 0, I f;(5>= max Hus), 4+ {am}: b , x,= l. fab): max Had», Arm“st 8+f5w>1= 8 , X;= 2. fin): max “25m, 4+fln,(4), 8+fwfl= 8, X}: I , 2. f,(3)= WW flaw). 4+ $45), 8+f5m‘,= go, xyc >_ @ fad): 4m? ? M. + fud- com} ~$ f.(8)= max? {11(8), 5+f.(4>_ u0+f,(oz’;= I0, X),= 0, 2. 0 Watts of Item I Dr 2 units of item u 2 units of item 2 I M517 of item 3; % Tomi loanefrv :2 :0. Than is, far flu; m‘qmw LP pro/ohm, vhz apn‘maL go/ufim Xi: 2, Xt=0/ X550, 3=f0. 09 E a 3:, 3+ § N is x.=0. >642, Xa=‘ .‘Z=IO 0T ...
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This note was uploaded on 02/08/2012 for the course IEOR 4004 taught by Professor Sethuraman during the Fall '10 term at Columbia.

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HW9_Solutions - fimek ‘1 for E4004. LPNs. 6.16 Sohm...

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