2 Dynamic Programming 2.1 Basic attributes of Dynamic Programming Basic attributes of Dynamic Programming 2.2 A DP approach for the Knapsack Problem

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1 2 Dyamic Programmig I what follows, we itroduce the techique Dyamic Programmig It geeralizes the basic priciples of recursio Basic ideas of recursive algorithms The etire problem is separated ito smaller predefied subproblems These predetermied smaller subproblems are solved separately Subsequetly, the resultig partial solutios are combied to a complete solutio to the origial problem This geeratio of a complete solutio is determied by a recursive formula Typical examples of recursive algorithms Quicksort Recursive versio of biary search Recursive versio of merge sort 2.1 Basic attributes of Dyamic Programmig Dyamic Programmig geeralizes the priciples of recursive algorithms Here, the particular subproblemsthat are eeded to costruct a solutio to the larger origial problem are ukow Therefore, i order to solve the origial problem, all smaller subproblemsof a specific size have to be solved ad all possible combiatios (followig a recursive formula) have to be assessed while the best oe is implemeted However, if a geerated combiatio leads to a ifeasible solutio, this alterative is pealized by ifiite costs Therefore, this alterative is excluded from further cosideratios Wirtschaftsiformatik ud Operatios Research 72 Wirtschaftsiformatik ud Operatios Research 73 Basic attributes of Dyamic Programmig Thus, i various applicatios, the derived subproblemsmay overlap ad, therefore, parts have to be solved multiple times durig the optimizatio process This leads to wasted computig time A couterstrategy is to explicitly eumerate the distict subproblems ad solve them i the right order Moreover, the process ca be accelerated by explicitly itegratig domiatio rules ad/or bouds fathomig existig partial solutios. These extesios are frequetly deoted as bouded Dyamic Programmig approaches 2.2 A DP approach for the Kapsack Problem I what follows, we itroduce a first example of a very simple but somehow iterestig Dyamic Programmig approach It solves the well-kow Kapsack Problem (KP) to optimality However, sice this problem is prove to be (biary) NP-hard, we caot expect a strogly polyomial solutio approach The Kapsack Problem has may applicatios as a problem model or deals as a importat subproblem for other applicatios/optimizatio problems First of all, we briefly itroduce the problem KP itself Wirtschaftsiformatik ud Operatios Research 74 Wirtschaftsiformatik ud Operatios Research 75 1

2 2.2.1 The Kapsack Problem Give is a set of items ad a fixed capacity limit C Each item has a weight w ad a price p Sought: a subset of items which maximizes the total price (obective) ad whose sum of weights does ot exceed the capacity limit C Each item may oly be icluded oce Cosequetly, a biary variable x idicates whether a item belogs to the sought subset or ot The problem is ofte illustrated by a example where a bag or a kapsack with a fixed maximum weight limit has to be packed with items of differet prices ad weights The 0-1 Kapsack Problem is defied as: = 1 = 1 { } max z = p x s.t. w x C x 0,1 = 1,..., Wirtschaftsiformatik ud Operatios Research A pseudo-polyomial DP for the KP It has bee show that the Kapsack Problem is (biary) NP-hard However, its structure allows quite efficiet solutio approaches Amog them, the most efficiet algorithms for solvig the Kapsack Problem are i particular DP approaches Applied to the Kapsack Problem, items may be cosidered iteratively oe by oe i order to decide about a potetial assigmet Cosequetly, by defiig f i (C ) as the maximal price attaiable for assigig the items 1,,i, (i, C C), we calculate p 1 if 0 w1 C f ( 1 C ) = 0 if 0 C < w1 { 1 ( ) 1 ( ) } ( C ) max f + i C, fi C wi pi if C wi 0 f ( i C ) = fi 1 otherwise Wirtschaftsiformatik ud Operatios Research 77 Observatios Clearly, the algorithm iteratively aalyzes whether a assigmet of a item is favorable or ot Cosequetly, the recursive formula always compares the cases of assigig or ot assigig a item Thus, if assiged, the weight of the item accordigly reduces the capacity of the kapsack ad provides a additioal beefit Otherwise, the elemet is left outside ad the problem is reduced to the optimal assigmet of the remaiig items 1,,-1 Thus, sice we have to calculate. C fuctio values, the total complexity is restricted by O(. C) Pseudo-polyomial ruig time A total ruig time of O(. C) seems to be moderate ad somehow polyomial. Or ot? What about the factor C? It is the capacity of the kapsack ad, therefore, ot restricted at all Ufortuately, it ca grow expoetially i the umber of elemets Note that i the proof of NP-Completeess of the Kapsack Problem, we eed a expoetially growig kapsack However, if the kapsack is of moderate size (this applies to most applicatios), the algorithm is quite efficiet Wirtschaftsiformatik ud Operatios Research 78 Wirtschaftsiformatik ud Operatios Research 79 2

3 A simple example Let us cosider a small example Maximize Z = 4 x + 7 x + 5 x s.t. 4 x1 + 5 x2 + 3 x3 10 x { 0,1} p We sort the items accordig to their efficiecy e = w Thus, we obtai the followig table Number of item Price Weight Efficiecy / /3 Observatios Clearly, the parameter C i the recursive formula gives some potetial for optimizatio Sice ot all partial kapsack capacities are reasoable, i.e., it exists a smaller kapsack obtaiig a idetical total weight, we may improve the efficiecy of the defied calculatios by makig use of the followig scheme Specifically, we commece the ivestigatio with a empty kapsack ad defie each possible assigmet by a tuple (p i, w i ) of total price ad total weight 0 Thus, we obtai R = {( 0,0) } Additioally, we have the followig resultig sets t t 0: {(, ) ((, ) ( ) ) } 1 t t R p w p w R > = p p, 1 t w wt R w C Wirtschaftsiformatik ud Operatios Research 80 Wirtschaftsiformatik ud Operatios Research 81 Further observatios We may reduce the resultig sets by erasig the domiated tuples I.e., a tuple (p,w) is domiated by some tuple (p,w ) if it holds that p p ad w>w or if it holds that p<p ad w w Wirtschaftsiformatik ud Operatios Research 82 Applied to our example We commece with the iitial trivial set 0 R = The, we resume with 0 R = 1 R = 2 R = {( 0,0 )} x 1 =0 x 1 =1 {( 0,0 ),( 4,4 )} x 2 =1 x 2 =0 x 2 =1 x 2 =0 {( 0,0 )} {( 0,0 ),( 7,5 ),( 4,4 ),( 11,9 )} Wirtschaftsiformatik ud Operatios Research 83 3

4 Resume with the set 2 R = 3 R = Best path Applied to our example {( 0,0 ),( 7,5 ),( 4,4 ),( 11,9 )} x 3 =0 x 3 =1 x 3 =0 x 3 =1 x 3 =0 x 3 =0 x 3 =1 {( 0,0 ),( 5,3 ),( 7,5 ),( 12,8 ),( 4,4 ),( 9,7 ),( 11,9 )} ( 0,0) ( 0,0 ) ( 7,5) ( 12,8 ) Possible improvemets Improvemets ca be obtaied by itroducig additioal bouds that restrict the umber of cosidered states withi the calculatio process Additioally, domiace criteria may be applied Specifically, isights ito the problem structure attaied by empirical studies reveal that i most cases oly a small subset of items aroud the critical oe are ecessary to cosider Simulatios with extremely large istaces show that much smaller umbered items are always part of the optimal solutio (i these experimets) much larger items are ever part of the optimal solutio (i these experimets) Best bouded DP approaches make widely use of this sigificat pheomeo Wirtschaftsiformatik ud Operatios Research 84 Wirtschaftsiformatik ud Operatios Research A DP approach for the atsp I what follows, we itroduce a somewhat much more complex DP approach for the asymmetric Travelig Salesma Problem with time widows Therefore, we agai briefly itroduce the problem first The Travelig Salesma Problem (TSP) Most simple model for mappig pure distributio or pick up problems i local traffic Problem defiitio: Give N as the umber of customers to be visited Directed weighted graph G=(V,A,C) with V={1,,N}, A set of arcs, ad C=[c i, ] the matrix of costs Sought Hamiltoia cycle i G of miimal total costs A Hamiltoia cycle is a cycle passig through each ode of G exactly oce Wirtschaftsiformatik ud Operatios Research 86 Wirtschaftsiformatik ud Operatios Research 87 4

5 The Travelig Salesma Problem The TSP has bee show to be a NP-hard problem (Wegeer (2005) pp.53-62) If costs are symmetric, e.g., the costs for travelig betwee two locatios i the etwork do ot deped o the directio of this travel, a symmetric TSP (stsp) arises; otherwise a asymmetric TSP (atsp) is cosidered If the triagle iequality is always fulfilled or Euclidea distaces are used, the problem is termed as the Euclidea TSP or the TSP with the triagle iequality Mathematical problem defiitio I the followig, we give a exact mathematical defiitio of the TSP that correspods directly to a possible solutio mappig Subsequetly, we give the defiitio for the asymmetric case Later o i this course, we will also cosider the special case of a symmetric TSP. Due to its simplified structure, it allows for the applicatio of specific methods Wirtschaftsiformatik ud Operatios Research 88 Wirtschaftsiformatik ud Operatios Research 89 Asymmetric case: Parameters ad variables Asymmetric case: Restrictios Parameters i, = { } ( 1 ; ( )) : Costs for usig the arc ( ) N : Number of odes (customers), V 1,..., N c i N i i, Variables { } i,, i ( ) i i 1. i 1,..., N : x = 1 x = 1 Each ode has two eighbors N 2.For all proper subsets S { 1,..., N} with 2 S : xi, 1 2 i S V S ( Subcycle prevetio) { } { } 3. i, 1,..., N : 0 x 1 4. i, 1,..., N : x is a iteger i, i, i, ( ( )) x 1 i N;1 N i : Biary variable that is oe if ad oly if the Salesma Path uses the arc ( i, ) Wirtschaftsiformatik ud Operatios Research 90 Wirtschaftsiformatik ud Operatios Research 91 5

6 Asymmetric case: Obective fuctio Applicatio of the TSP plaig milk rus milk ru startig at the depot N N ( 1 i, N ) i = 1 = 1( i ) Miimize F x = c x i, i, i, log distace trasport local area Source local area Destiatio milk ru edig at the depot Use of available vehicles, carriers or exteral parters Wirtschaftsiformatik ud Operatios Research 92 Wirtschaftsiformatik ud Operatios Research 93 Excursus: Trasportatio etworks I Excursus: Trasportatio etworks II Raster structure: areas Hub&Spoke structure: areas MultipleHub Structure: areas Regioalhub Structure: areas Hub Hub Hub Hub Source or destiatio Hub Source or destiatio Source or destiatio Hub Source or destiatio Wirtschaftsiformatik ud Operatios Research 94 Wirtschaftsiformatik ud Operatios Research 95 6

7 Excursus: Trasportatio etworks III Excursus: Trasportatio etwork IV Feederhub Structure: areas Mixed Structure: areas Exteded mixed structure: areas Source Hub Hub Hub Hub Source or destiatio Hub Hub Source or destiatio Hub Destiatio Hub Source or destiatio Wirtschaftsiformatik ud Operatios Research 96 Wirtschaftsiformatik ud Operatios Research A DP approach for the TSPTW I what follows, we itroduce a sophisticated exact solutio approach to the Travelig Salesma Problem This approach was origially proposed by Dumas et al. (1995) It cosiders the TSPTW, i.e., the Travelig Salesma Problem with time widows Here, each customer comes alog with a time widow that restricts delivery time The time widows [a i,b i ] are hard, i.e., delivery time at customer i is required to be i the cotiuous iterval [a i,b i ] Additioally, there is a service time s i at each customer i The followig solutio algorithm bases o a Dyamic Programmig cocept Notatios Basic defiitios G = ( N A) { }, : Network N = 1,2,..., : Set of customers A = {( i, ) ( N N) a + s + t b } { } { 1,2,..., 2, 1} N = N = ( i i + 1) A i i i, A path i G is defied as a ode sequece i,..., i such that each arc, 1 k Wirtschaftsiformatik ud Operatios Research 98 Wirtschaftsiformatik ud Operatios Research 99 7

8 Notatios I what follows, we itroduce a somewhat modified otatio This otatio is used i order to itroduce the Dyamic Programmig approach Parameters t : Travel time from customer i to i, s : Service time at customer i i [ a b ], : Time widow at customer i i i Variables t : Poit i time whe service starts at customer i i We defie for a path goig from customer i to customer { } m ax t i + ti, + s i, a if ti + s i + t i, b t = otherwise Observatio If a customer is visited too early, waitig time is icurred at this locatio This results from the lower boud a i the calculatio of t Note that the cosidered problem is somewhat modified It seeks the fidig of a tour from 1 to, but ot a cyclical tour However, the origial problem ca be mapped by doublig ote 1 The, is set to +1 Wirtschaftsiformatik ud Operatios Research 100 Wirtschaftsiformatik ud Operatios Research Dyamic Programmig formulatio We itroduce F(S,,t) Least cost of a path startig at ode 1, passig through every ode of S exactly oce, ad edig at ode at time t or later We assume that ode belogs to set S Additioally, S is a subset of set N Servicig ode is therefore possible at time t Recursive equatio The fuctio F(S,,t) ca be computed by makig use of the followig equatio S N : S : t, with a t b : ( ) F S,, t = mi F S, i, t + c + + i, t t si ti, ai t bi ( i, ) A i S { } We iitialize this calculatio by { } 2,..., 1 : ({ 1, },, ) F t { ( { } ) } ( ) { } c1, if 1, A ad max a1 + s1 + t1,, a t b = otherwise Wirtschaftsiformatik ud Operatios Research 102 Wirtschaftsiformatik ud Operatios Research 103 8

9 Determiatio of optimal solutio The optimal solutio to a istace of the TSPTW ca be idetified by mi mi,, (, ) i A ai t bi { F ( N i t ) + ci, t b ti, si} Observatios Clearly, the Dyamic Programmig approach geerates shortest paths i a forward maer Specifically, i step s=1,,-1, a path of legth s is costructed I each step, a shortest path is sought o a state graph whose odes are the states F(S,i,t) Note that the umber of states F(S,i,t) (a i t b i ) is coutable if all travel times as well as a i adb i are itegers Amog the states, we oly coserve the Pareto optimal states Note that for two give states F(S,i,t 1 ) ad F(S,i,t 2 ), the secod state ca be elimiated if t 1 t 2 ad F(S,i,t 1 ) F(S,i,t 2 ) Wirtschaftsiformatik ud Operatios Research 104 Wirtschaftsiformatik ud Operatios Research 105 Cogitio The fuctio F(S,i,t) is stepwise decreasig as a fuctio of t over the iterval [a i,b i ] Clearly, after elimiatig the domiated states, the remaiig states ca be ordered by icreasig time ad decreasig cost value Thus, we obtai the two-dimesioal labels (t,f(s,i,t)) of (S,i) Therefore, we itroduce FIRST(S,i) as the first, i.e., lowest, time value of that ordered list of o domiated states Thus, FIRST(S,i) represets the fastest feasible arrival at ode i by visitig all other odes of set S beforehad Post feasibility tests The complexity of the described approach is maily drive by the determiatio of suitable sets S Thus, the umber of states to be explored grows expoetially Cosequetly, it is very promisig to itroduce rules ad tests i order to elimiate ifeasible states, i.e., states that caot be exteded to a complete feasible state Hece, these rules are applied i order to elimiate partial paths that do ot comply with ecessary orderigs resultig from existig time widow restrictios I order to measure earliest arrival times ad latest departure times, we itroduce the abbreviatios EAT ad LDT Wirtschaftsiformatik ud Operatios Research 106 Wirtschaftsiformatik ud Operatios Research 107 9

10 EAT(i,) EAT(i,), LDT(i,) Determies the earliest arrival time at ode if ode i is visited before Clearly, this value ca be computed by solvig a Shortest Path Problem from ode i Specifically, the path starts at time a i ad satisfies the time widow restrictios at all odes from ode i to ode LDT(i,) Additioal remarks o EAT(i,), LDT(i,) Note that the values EAT(i,) ad LDT(i,) ca be overestimated or uderestimated, respectively, by usig a ucostraied time path. Sice this is much faster to compute i practice, it is frequetly applied Additioally, if travel times satisfy the triagle iequality, o shortest paths eed to be computed. Hece, i this case, EAT(i,)ca be replaced by a i + s i + t i, Determies the latest departure time from ode i such that t is feasible This ca be computed by solvig a costraied Shortest Path Problem, where the path starts at ode at time b ad uses the reverse arc directio Wirtschaftsiformatik ud Operatios Research 108 Wirtschaftsiformatik ud Operatios Research 109 BEFORE() I what follows, we compute for a ode the set of odes BEFORE() Withi this set, there are all odes that have to be executed before ode Specifically, it is checked whether the precedig visit of ode prohibits a timely service of ode k Cosequetly, we ca compute BEFORE() as follows {, k} ( ) ( ) BEFORE = k N EAT k > b Agai, if the triagle iequality holds for travel times, EAT(,k) > b k ca be replaced by a + s + t,k > b k Feasible extesios I what follows, we say that a state (S,i,t) ca be feasibly exteded towards if it holds: I.e., it holds that: t + s + t b i i, ( S { } { a t + si + ti, } ) State,,max, ca be created if it holds: t + s + t b i i, Wirtschaftsiformatik ud Operatios Research 110 Wirtschaftsiformatik ud Operatios Research

11 Post feasibility test 1 Give the states (S,i,t) for all a i t b i If the smallest time value to begi service at ode i is greater tha the latest feasible departure time toward, i.e., for allit holds that: (, ) + > mi (, ) FIRST S i s LDT i i S The the states (S,i,t) for all a i t b i ca be fathomed (deleted) sice they do ot admit feasible extesios towards ay ode Test 1 This global test elimiates the states (S,i,t) for all t by oly examiig the earliest time to begi service at ode i If, for this time, a feasible arrival at ay other ode is impossible, the all the states (S,i,t) are elimiated I this case, the states to be treated ext are those with a ew edig ode i S\{i} but a idetical set S If, fially, all edig odes i S are explored, a ew set S is examied Wirtschaftsiformatik ud Operatios Research 112 Wirtschaftsiformatik ud Operatios Research 113 Post feasibility test 2 Give the states (S,i,t) for all a i t b i ad give ode ( does ot belog to S) Additioally, we have (i,) out of A If BEFORE ( ) S the o feasible extesio exists towards Cosequetly, ode caot be a successor of i Thus, differet successors have to be tested istead If the ew label Post feasibility test 3 ( S i t ) t ai t bi ( ) ( S i t ) t + si LDT ( i ) k k S ( i k) A i i ( ) ( ) ( S i t ) t t Give the state,, for a fixed, ad give ode, S. Additioally, i, A. If,, ca be exteded toward, i.e.,,, but caot be exteded further to some,,,, i.e., t + s + t, + s > LDT, k, the caot succeed i for S, i, t ad the states,,, are ot exteded towards. ( S { },,max { a, t + s i + t i, }) is ot created, a ew ode l is cosidered because o other labels ca be feasible for time values greater tha t Wirtschaftsiformatik ud Operatios Research 114 Wirtschaftsiformatik ud Operatios Research

12 Computatioal experimets The algorithm was coded i C Experimets were coducted o a Hewlett-Packard workstatio HP9000/730 By makig use of specific preprocessig rules, the arc set A was reduced cosiderably (Lagevi et al. (1990)) Time widows were also reduced by applyig the rules proposed by Desrochers et al. (1992) Specific set of problems was tested It cosists of symmetric Euclidea problems (cf. Lagevi et al. (1990)) Wirtschaftsiformatik ud Operatios Research 116 Set of problems Customer coordiates are uiformly distributed betwee 0 ad 50 ad travel times equal distaces Time widows are geerated aroud the times to begi service at each customer of a secod earest eighbor TSP tour Each side of a time widow is geerated as a uiform radom variable i the iterval [0,w/2], where w=20, 40, 60, 80, ad 100 Clearly, for a give problem size, problem difficulty icreases with the time widows width Cosequetly, the umber of overlappig time widows icreases Hece, the umber of predetermied ode sequeces decreases sigificatly Wirtschaftsiformatik ud Operatios Research 117 Complexity of problem istaces Esurig tight time widows is a hard ob. Therefore, tight time widows make the problem hard to solve to optimality?! I believe it is ust the opposite. Tight time widows reduce the feasible solutio space Complexity of problem istaces Complexity icreases sigificatly with problem size because the geographical area remais costat Hece, as icreases, the desity of the poits i this area icreases I tur, this decreases the ability of the time widows to reduce the umber of possible tours Wheever time widows start overlappig, expoetial complexity prevails ad icreases the umber of tours to be explored sigificatly I what follows, we provide the geerated results accordig to problem sizes (umber of customers) ad time widow widths Wirtschaftsiformatik ud Operatios Research 118 Wirtschaftsiformatik ud Operatios Research

13 Direct cogitios The preseted algorithm was able to solve problems with up to 200 odes ad fairly wide time widows As assumed, the CPU time icreases with time widow width ad with problem size For arrow widow widths, the complexity does ot icrease expoetially. This allows for solvig larger sized problems i reasoable time For istace, a 250 ode problem with w=20 was solved i less tha 10 secods Usig the logarithm i base 10 of the CPU time (see Table I), it becomes obvious for a give problem size that the behavior of the algorithm is expoetial Wirtschaftsiformatik ud Operatios Research 120 Wirtschaftsiformatik ud Operatios Research 121 Impact of the post feasibility tests Coclusios As illustrated by table II, the impact of the post feasibility tests was examied additioally Clearly, test 2 is the most beeficial oe Its absece icreases the umber of examied states sigificatly Thus, for problem size =60, istaces with widow width w=60 caot be solved i reasoable time Test 1 is the least powerful because much of its work was already performed by reducig the width of the time widows i the preprocessig phase For the geeratio of further problem istaces, we fixed w at 60 (fairly wide time widows are geerated) The umber of commo arcs betwee the secod earest eighbor TSP tour ad the optimal tour stayed at approximately 20% eve for 800 odes problems Wirtschaftsiformatik ud Operatios Research 122 Wirtschaftsiformatik ud Operatios Research

14 Costat poit desity problems Results Owig to the post feasibility rules, problems of size 800 could be solved i approximately 650 secods However, larger size problems face memory limitatios Wirtschaftsiformatik ud Operatios Research 124 Wirtschaftsiformatik ud Operatios Research A small example We cosider the followig simple example: 4 7 C = i s i a i b i T = Matrix C: c i, := Cost for travelig from ode i to ode Matrix T: t i, := Travel time for travelig from ode i to ode Nodes start at idex 1 Wirtschaftsiformatik ud Operatios Research 126 Task ad start of algorithm The TSPTW problem starts at customer/ode 1 Calculate the optimal tour ad give the costs as well as the duratio of the tour Problem starts ad eds at ode 1 Customer 1 is doubled to customer 4 N = {1,2,3,4}, N ={1,2,3}, i {1,...,3} : c = c, t = t, i,4 i,1 i,4 i,1 EAT( i,4) = EAT( i,1), LDT( i,4) = LDT( i,1) Wirtschaftsiformatik ud Operatios Research

15 Calculatio of EAT(i,), LDT(i,), BEFORE() Iteratio 1: s=1 i EAT(i,) LDT(i,) Note: The LDT(i,1)-values are ot required ad, therefore, ot computed. Geerate the states (S,i) of all sets S with S = 2 S (S,i) 2 3 {1,2} (3,4) {1,3} (10,7) (t,f(s,i,t)), t=3=max{a 2,a 1 +s 1 +t 1,2 }=max{3,1} Keep oly pareto-optimal states: i BEFORE() {} {} {2} (S,i) 2 3 {1,2} (3,4) {1,3} (10,7) Wirtschaftsiformatik ud Operatios Research 128 Wirtschaftsiformatik ud Operatios Research 129 Post feasibility tests Test 1: Partial solutio ({1,3}, 3, 10) is ot extedable towards =2 remove this state from the available states Test 2 ad Test 3 do ot elimiate ay further states Resultig available partial solutio: (S,i) 2 3 {1,2} (3,4) Wirtschaftsiformatik ud Operatios Research 130 Iteratio 2: s=2 Exted every available partial solutio to states with S =3 Available set of states oly cosists of the state ({1,2}, (3,4)). Exted this state by addig ode 3: (S,i) 3 {1,2,3} (13,10), (14,10), (15,10) (t,f(s,i,t)), with t=13=max{a 3,3+s 2 +t 2,3 }=max{10,3+2+8} Keep oly pareto-optimal states: (S,i) 3 {1,2,3} (13,10) Wirtschaftsiformatik ud Operatios Research

16 Example Calculatio of optimal solutio All odes of N have bee added to the available states Calculatig the optimal solutio by the formula mi mi,, (, ) i A ai t bi { F ( N i t ) + ci, t b ti, si} Oly oe state available Calculate costs of optimal solutio by F({1,2,3},3,13) + c 3,4 = = 12 Duratio: 13 + s 3 + t 3,4 = 13 + s 3 + t 3,1 = = 23 Optimal solutio: , Costs: 12, Duratio: Exactly solvig the Lie TSP by DP A specific case of the Travelig Salesma Problem occurs wheever all customers are located alog a straight lie ad the startig positio x* of the salesma is also o this lie E.g., there is a sigle road that coects all these customers ad alog which supply is executed Therefore, each customer i is located at positio x i ad it holds: c i = x -x i We additioally itroduce time widows [r i,d i ] ito the problem defiitio. Specifically, r i determies the release time ad d i a due date at the locatio of customer i Moreover, at each locatio i, a hadlig or service time h i is give Obective fuctio is defied as the miimizatio of the maximum completio time (i.e., the makespa) Wirtschaftsiformatik ud Operatios Research 132 Wirtschaftsiformatik ud Operatios Research 133 Lie TSP TSP o a lie? Who eeds this??? Hey! There are various importat applicatios Applicatios: Ilad ship alog the Rhie Cargo ships have to supply ilad ports alog the Rhie Idividual time restrictios have to be obeyed at each port There may be ports that permit a delivery before the release date Latest date of supply (hard time widow) Differet directio-depedet travel speeds (i flow/agaist flow directio) Miimizatio of the tour legth fulfillig all due dates Duisburg Düsseldorf Leverkuse Cologe Maiz I ll show you three of them! I flow directio: faster trasportatio possible Agaist flow directio: slower trasportatio Wirtschaftsiformatik ud Operatios Research 134 Wirtschaftsiformatik ud Operatios Research

17 Applicatios: Delivery alog a coast lie Customer locatio X with three delivery requests, 1. request 1 (r 1 =9:00am, d 1 =11:00am) 2. request 2 (r 2 =2:00pm, d 2 =5:00pm) 3. request 3 (r 3 =3:00pm, d 3 =5:30pm) JiT supply of mixed-model assembly lies Statios are supplied by specific tow trais Tight time widow costraits to be obeyed at the statios arise from the lauched variat sequece Tow trais Statio 1 Statio 2 Statio 3 Statio 4 Floater Overlappig areas Worker Mixed-model assembly lie Wirtschaftsiformatik ud Operatios Research 136 Wirtschaftsiformatik ud Operatios Research

18 Proof of Theorem Clearly, sice processig times ad release dates are zero, it costs othig to serve a customer whe travelig by Therefore, it is assumed that each customer is immediately served the first time its locatio is visited by the vehicle Specifically, if the vehicle has visited locatios a ad b with a b, the all obs whose locatios belog to the iterval [a,b] have bee serviced I what follows, we assume that ob umbers are sorted accordig to their locatios, i.e., it holds x 1 x 2 x 3 x Moreover, we assume that the startig positio x*=x i* coicides with the locatio of some customer Note that such a artificial customer ca be geerated without causig additioal costs Proof of Theorem Moreover, i order to esure feasibility, we assume that x i* - x i d i for all i; otherwise the problem is ot solvable at all We fix some pair of obs (i ad ) with 1 i i* We cosider all schedules where ob i is timely visited for the first time (i.e., ot after the due date) ad, before that, all obs that are located i the iterval [x i, x ] were visited timely Based o these schedules, we itroduce two additioal abbreviatios V - (i,) Earliest poit i time where such a costellatio is possible V + (i,) Earliest poit i time i a schedule where ob is timely visited for the first time ad, before that, all obs that are located i the iterval [x i, x ] have bee visited timely as well Note that if those time poits do ot exist, these abbreviatios are set to ifiity Wirtschaftsiformatik ud Operatios Research 140 Wirtschaftsiformatik ud Operatios Research 141 Proof of Theorem Proof of Theorem Observatios Clearly, we kow ( ) ( ) * + * * i < : V i, = x x 1 Aalogously, we ca coclude that It holds: ( ) ( ) < :, = 3 * * * i i V i i x x i Moreover, it holds: ( ) = ( ) * + * i < i : V i, i 2 Moreover, it holds: > * * i V i ( ) = ( ) :, 4 (1) results from the fact that we ca directly use the direct path to x (2) directly follows sice we start at positio i* ad caot serve customer i before Furthermore, we coclude V + (i*,i*)=0 (3) agai results from the fact that we ca directly use the direct path to x i (4) directly follows sice we start at positio i* ad caot serve customer before Furthermore, we obtai V - (i*,i*)=0 Wirtschaftsiformatik ud Operatios Research 142 Wirtschaftsiformatik ud Operatios Research

19 Additioal abbreviatios We itroduce U + (i,) as the miimum poit i time where we ca reach customer after servig all customers i, i+1,, -1 Thus, we apply the followig recursive formula * + : (, ) + { V ( i ) + x x 1 V ( i ) + x xi}( ) i < i < U i = mi, 1,, 1 5 Clearly, (5) results from the followig cogitios If we wat to visit customer for the first time, we either come from -1 or we come from customer i Otherwise, we would ot have visited all customers i, i+1,, -1 before Wirtschaftsiformatik ud Operatios Research 144 Additioal abbreviatios Aalogously, we itroduce U - (i,) as the miimum poit i time whe customer i is reached after we have served all customers i+1, i+2,, Thus, we apply the followig recursive formula * : (, ) + { V ( i + ) + xi + 1 xi V ( i + ) + x xi}( ) i < i < U i = mi 1,, 1, 6 Clearly, (6) agai results from the followig cogitios If we wat to visit customer i for the first time, we either come from customer or we come from customer i+1 Otherwise, we would ot have visited all customers i+1, i+2,, - 1, before Wirtschaftsiformatik ud Operatios Research 145 Cosequece Based o these abbreviatios, we ca itroduce the followig useful recursios Specifically, it holds: ad additioally: ( ) ( ) + + * + U i, if U i, d i < i < : V ( i, ) = 7 otherwise ( ) ( ) * U i, if U i, d i < i < : V ( i, ) i = 8 otherwise ( ) ( ) Proof of Theorem Cosequetly, by applyig these formulas, we ca geerate V + (1,) ad V - (1,) Clearly, the miimum of both is the completio time of the optimal tour table I.e., mi{v + (1,), V - (1,)} provides the miimal completio time of the vehicle tour This provides us with a Dyamic Programmig procedure Iitially, we calculate the values V + (i*, ) for >i*, ad V - (i, i*) for i*>i Subsequetly, we ca apply the formulas (7) ad (8) i order to get the followig values Fially, V + (1,) ad V - (1,) are computed Wirtschaftsiformatik ud Operatios Research 146 Wirtschaftsiformatik ud Operatios Research

20 Computatioal effort We have to calculate two arrays (V+ ad V-) of customer pairs i,, with i< Thus, we have to geerate iteratively O(²) values The geeratio of each value requires costat time Specifically, we ust have to evaluate two differet values whose geeratio agai takes costat time All i all, we obtai a total ruig time of O(²) The optimal tour ca be obtaied by storig a flag for each pair (i,) that idicates whether the plus or the mius case was favorable No release times or deadlies Further results Zero processig or hadlig (service) times Trivial case Geeral processig or hadlig (service) times Trivial case Release times oly O(²) NP-complete Pseudo-polyomial ukow Deadlies oly O(²) NP-complete Pseudo-polyomial ukow Geeral time widows Strogly NPcomplete Strogly NP-complete Wirtschaftsiformatik ud Operatios Research 148 Wirtschaftsiformatik ud Operatios Research Dyamic Programmig i Schedulig I what follows, we cosider a very sophisticated example of a Dyamic Programmig approach that oce solved a ope research questio Is it possible to provide a pseudo-polyomial solutio approach for fidig a optimal schedule for oe stage scearios while miimizig the total sum of tardiess? Fortuately, the aswer was yes, but required the uderstadig of a specifically desiged solutio approach It is based o Dyamic Programmig Job Shop Schedulig Basics I what follows, we cosider schedulig problems I.e., we state the followig problem Give M machies or resources, N obs to be produced Each ob comprises a predetermied set of operatios to be executed o the resources Sought A productio sequece of all N obs for each machie Determiatio of the timetables Cosequetly, we have to decide about The sequece of the competig obs o the machies ad the resultig timetable Wirtschaftsiformatik ud Operatios Research 150 Wirtschaftsiformatik ud Operatios Research

21 Assumptios Productio program is give Lot sizes are give Process sequece of each ob is give Operatig times are give No operatio of the obs ca be processed simultaeously o more tha oe machie Every machie ca process at most oe ob at each poit of time All N obs ad its data are available (static problem) at the begiig of the plaig horizo There are ever bottleecks cocerig trasports ad storage No maiteace ad reparatory activities O each machie, setup times are idepedet of the realized operatio sequece Give ad sought Give MS: Machie sequece matrix PT: Matrix of the processig times Sought JS: Job sequece matrix TT: Timetable plaig matrix with ( ) a tm, 1 m M;1 N Poit of time i which the processig of ob starts at machie m [ TU] Wirtschaftsiformatik ud Operatios Research 152 Wirtschaftsiformatik ud Operatios Research Mathematical model Mathematical model Restrictios Variables a m, ( ) ( ) t 1 N;1 m M : see above y 1 m M;1 N;1 k N : Biary variable defiig the m,, k sequece of obs, i.e., Machie sequece restrictios ( ) derived from the matrix MS : a a { 1,..., 1 } : { 1,..., } : [ ] [ ] [ + ] m M N t + p t m, m, m 1, y m,, k 1 if ob is processed o machie m before ob k = 0 otherwise [ m] defies here the idex of the machie that executes the mth operatio of ob Wirtschaftsiformatik ud Operatios Research 154 Wirtschaftsiformatik ud Operatios Research

22 Mathematical model Restrictios I case of the ob sequece restrictios, the formulatio depeds o the structure of the foud solutio But we have to esure that two obs are ever processed simultaeously o oe machie ad, therefore, a arbitrary sequece of those obs has to be realized Therefore, there are the followig two possible cases: ( 1 ) ( 2 ) ( k ) First case before : t + p t a a m, m, m, k a a m, k m, k m, ( k ) Secod case before : t + p t Both possibilities have to be cosidered i the model! Mathematical model Restrictios ( ) Job sequece restrictios depeds o the chose solutio : a a { 1,..., 1 } :, { 1,..., } : m, m, m,k ( 1 m,, k ) a a { 1,..., 1 } :, { 1,..., } : m M k N t + p t + y C m M k N t + p t + y C m,k m,k m, m,, k C defies a umber that is larger tha each defiitio of M N = m, m= 1 = 1 a the timetable variables t, e.g., C p m, Wirtschaftsiformatik ud Operatios Research 156 Wirtschaftsiformatik ud Operatios Research 157 Mathematical model Domais { 1,..., } :, { 1,..., }( ) : { 0,1} m M k N k y { } { } m 1,..., M : 1,..., N : t 0 a m, m,, k Obective fuctios The model defied above ca be see as a geeral startig poit for so-called Job Shop Schedulig problems It abstais from the defiitio of a particular obective fuctio but ca be exteded by a specific applicatio-depedet defiitio I literature, a huge set of differet obective fuctios is proposed. These fuctios maily ifluece the efficiecy of applied solutio procedures I what follows, we will give some examples of wellkow obectives Wirtschaftsiformatik ud Operatios Research 158 Wirtschaftsiformatik ud Operatios Research

23 Miimizatio of total makespa This obective fuctio miimizes the duratio for producig the total productio quatities, i.e., it pursues the miimizatio of the maximum completio time over all processed obs { } [ M ] { [ M] { }} Miimize Z = t = max t 1,..., N with: 1 max, 1,..., N : t : Poit of time i which the last, processig of ob is completed Miimizatio of machie waitig times Sum of all machie waitig times for all used resources M N Miimize Z2 = tmax pm, m= 1 = 1 Uused capacity of machie m M N = M tmax pm, m= 1 = 1 Note: M N Sice pm, ad M are costats, Z1 ad Z2 are equivalet m= 1 = 1 Wirtschaftsiformatik ud Operatios Research 160 Wirtschaftsiformatik ud Operatios Research 161 Miimizatio of total completio time This obective tries to miimize the total sum of all idividual completio times Therefore, we compute the sum of completio times over all processed obs N Miimize Z3 = t[ M], = 1 This obective is equivalet to the miimizatio of the sum of waitig times of all obs Miimize Z = N M w 4 m, = 1 m= 1 Miimizatio of total lateess Here, we wat to miimize the total lateess over all N obs to be produced i the cosidered productio system Cosequetly, a compesatio betwee early ad late deliveries is o loger possible N Miimize Z6 = max { t[ ] d,0 M, } = 1 with: d : Due date of ob Wirtschaftsiformatik ud Operatios Research 162 Wirtschaftsiformatik ud Operatios Research

24 Miimizatio of maximum lateess By usig this obective, we somehow try to balace the lateess i the foud solutio amog the differet obs Thus, we try to miimize the maximum lateess of a ob i the foud solutio Miimize Z = { { [ ] } 7 max max t d, 0 { 1,..., N M, }} with: d : Due date of ob Mi. of sum of weighted completio times Here, each ob obtais a idividual weight ratig its completio time i the productio system Thus, we receive a combied weighted sum of ob completio times Miimize Z = w t[ ] with: N 8 M, = 1 w : Weight of product Wirtschaftsiformatik ud Operatios Research 164 Wirtschaftsiformatik ud Operatios Research Schedule classes I the followig, we itroduce some basic terms accordig to specific types of schedules I the schedulig theory, a distictio is frequetly made betwee Sequece, Schedule, ad fially Schedulig policy Wirtschaftsiformatik ud Operatios Research 166 Basic terms Sequece Correspods to a specific permutatio of obs to be processed o a give machie Schedule Usually correspods to a allocatio of obs withi a more complicated settig of machies that could allow for preemptio of obs by other obs that are released at later poits i time. Comprises timetables Schedulig policy Ofte used i stochastic settigs; a policy prescribes a appropriate actio for ay of the states the system may be i. I determiistic cases, usually oly sequeces or schedules are of importace but ca be exteded by rule defiitios Wirtschaftsiformatik ud Operatios Research

25 Defiitio: No-delay schedules A feasible schedule is called o-delay if o machie is kept idle whe there is a operatio available for processig No-delay schedules These schedules are ot allowed to comprise uforced idleess of machies i the productio process Here, i most cases, we cosider o-delay schedules sice otherwise a improvemet possibility seems to be straightforward HOWEVER: There may be some special costellatios for o-preemptive models where it pays to have periods of uforced idleess. This results from some specific effects of o-delay schedules (c.f. Piedo, M. (2012) pp.22) Wirtschaftsiformatik ud Operatios Research 168 Wirtschaftsiformatik ud Operatios Research Defiitio: Active schedules A feasible schedule is called active if o operatio ca be completed earlier by startig earlier or by chagig the process sequece o machies without delayig ay other operatio Lemma: Attributes of active schedules A o-delay schedule is always active Wirtschaftsiformatik ud Operatios Research 170 Wirtschaftsiformatik ud Operatios Research

26 Proof of the Lemma Let us assume there is a o-delay schedule that is ot active The, we kow there is a machie m where shiftig a operatio of ob i to a earlier positio at poit of time t results i a earlier completio without delayig the other operatios But, if this is true, we kow that durig the processig of the schedule o machie m, there is a costellatio at poit of time t where the cosidered machie m is idle but ca process ob i istead This cotradicts the assumptio that the schedule is o-delay Attributes of active schedules Note that the reverse is ot ecessarily true I.e., there are some active schedules that are ot odelay Example: Schedule is active but ot o-delay Machie 1 Machie 2 Machie Wirtschaftsiformatik ud Operatios Research 172 Wirtschaftsiformatik ud Operatios Research Defiitio: Semi-active schedules Lemma: Cosequeces A feasible schedule is called semi-active if o operatio ca be completed earlier without alterig the processig sequece o ay of the machies A active schedule is always semi-active The proof is trivial ad follows immediately from the defiitio Wirtschaftsiformatik ud Operatios Research 174 Wirtschaftsiformatik ud Operatios Research

27 Attributes of semi-active schedules Note that the reverse is ot ecessarily true I.e., there are some semi-active schedules that are ot active Example: Schedule is semi-active but ot active Machie 1 Machie 2 Machie What is the best schedule class? I hate waitig. Therefore, I always would go for o-delay! Be careful! Eve for mootoous obective fuctios, waitig ca make sese.if it is for istace - a ob with high weight that would be delayed otherwise Wirtschaftsiformatik ud Operatios Research 176 Wirtschaftsiformatik ud Operatios Research 177 Schedule class hierarchy The cosidered sigle-stage problem Class of o-delay schedules Optimal schedules Class of active schedules Class of semi-active schedules I what follows, we cosider a problem that was very challegig ad received a eormous amout of attetio util its status has bee completely clarified i 1990 It is the sigle-stage schedulig problem pursuig the miimizatio of total tardiess N obs have to be scheduled at a sigle stage Each ob possesses a idividual soft due date Obective is the miimizatio of total tardiess (without weights) It has bee prove i 1990 by Du ad Leug that this problem is already biary NP-hard Hece, due to the theory of NP-Completeess, it is coectured that this rules out the existece of strogly polyomial algorithms However, there is a very efficiet DP algorithm that works i pseudo-polyomial time (itroduced by Lawler (1977)) Wirtschaftsiformatik ud Operatios Research 178 Wirtschaftsiformatik ud Operatios Research

28 2.5.3 The DP for the cosidered problem Clearly, by aalyzig the problem i detail, it becomes obvious that there is the followig geeral simple domiace criterio applicable (Emmos (1969)) Lemma: If p i p ad d i d, the there exists a optimal schedule that sequeces ob i before ob Proof: The proof is trivial sice it follows directly from the cogitio that preferrig i agaist does ot have egative side effects Wirtschaftsiformatik ud Operatios Research 180 Proof of domiace criterio Specifically, if we have a schedule with the two obs i opposite sequece, we exchage them Cosequetly, the itermediate completio times of obs scheduled betwee the two obs are ot icreased The completio times of obs scheduled before behid both obs are ot affected at all Due to the fact that ob i is more urget tha ob, we coclude that the exchage caot icrease the total obective value This completes the proof Wirtschaftsiformatik ud Operatios Research 181 Due date sesitivity of the optimal schedule We cosider optimal schedules ad ask how far we ca postpoe the due date of a ob without affectig the optimality of a solutio to the origial problem For this purpose, we defie First istace P with obs ad processig times p 1,,p as well as due dates d 1,,d, Optimal solutio S with maximum completio time C (k) of ob k Secod istace P with obs ad processig times p 1,,p as well as due dates d 1,,max{d k,c (k)},,d, Optimal solutio S with completio time C (k) of ob k Due date sesitivity of the optimal schedule Lemma: Every sequece that is optimal for the secod istace is also optimal for the first oe Proof: We defie V (S) ad V (S) as the total tardiess of a schedule S for the due date structure of the istaces P ad P. Let V (S )=V (S )+A k ad V (S )=V (S )+B k be the total tardiess uder the respective istaces. Clearly, if it holds that C (k) d k, both sets of due dates are the same ad we have idetical optimal solutios. Therefore, the propositio follows. Wirtschaftsiformatik ud Operatios Research 182 Wirtschaftsiformatik ud Operatios Research

29 Proof of Lemma Thus, we assume that it holds C (k)>d k V (S )=V (S )+A k ad V (S )=V (S )+B k I this case, we have: A k =C (k)-d k Moreover, it holds that B k =max{0, mi{c (k),c (k)}-d k } Clearly, we coclude the followig: mi{c (k),c (k)} C (k) B k A k Sice S is optimal for P, we obtai V (S ) V (S ) Cosequetly, we obtai V (S ) V (S ) Therefore, sice S is optimal for P, we obtai V (S )=V (S ) ads is optimal for P This completes the proof Assumptios I what follows, we assume (without loss of geerality) All ob processig times are differet from each other Jobs are reumbered i sequece of o-decreasig due dates, i.e., it holds that d 1 d 2 d Moreover, we defie p k =max{p 1,p 2,,p } Cosequetly, the kth urget ob has the largest processig time Clearly, due to Lemma , we kow that there is a optimal schedule where all obs of set {1,,k-1} will be scheduled before ob k However, what about the remaiig obs? Wirtschaftsiformatik ud Operatios Research 184 Wirtschaftsiformatik ud Operatios Research 185 Cosequece Lemma: There exists a iteger δwith0 δ -k such that there exists a optimal schedule S i which ob k is preceded by all obs with k+δad followed by all obs with>k+δ Proof: Let C (k) deote the latest completio time of ob k i a optimal schedule of the origial problem (due dates d 1,,d ) S is optimal for the modified problem (with due dates d 1,,max{d k,c (k)},,d ) ad satisfies the restrictio of Lemma Proof of Lemma Let C (k) be the completio time of ob k uder schedule S Clearly, by applyig Lemma , schedule S is also optimal for the origial problem istace P We therefore coclude that C (k) max{c (k),d k } This is true sice C (k) is assumed to be the latest completio time of ob k i all optimal schedules for P Moreover, we kow that all obs with a due date later tha max{c (k),d k } are processed after ob k (if this would ot be the case, we ca reassig this ob after ob k without deterioratig the obective fuctio value) Wirtschaftsiformatik ud Operatios Research 186 Wirtschaftsiformatik ud Operatios Research

30 Proof of Lemma Clearly, additioally, ob k is preceded by all obs with a due date earlier tha max{c (k),d k } (this results from a direct applicatio of Lemma ) Cosequetly, we set δ to the maximum iteger such that d k+δ max{c (k),d k } This completes the proof Cosequeces of fidigs Ad what is wo? No oe kows the optimal solutios beforehad. Therefore, the δ value is ukow This is ust the reaso why we apply Dyamic Programmig. We eumerate all possible values Wirtschaftsiformatik ud Operatios Research 188 Wirtschaftsiformatik ud Operatios Research 189 Cosequeces of fidigs The Dyamic Programmig approach utilizes a procedure that geerates a optimal schedule for the set of obs 1,,l with a ob k that has the largest processig time Due to the results derived above, we kow that we have to eumerate possible positios of ob k while this ob, fortuately, splits the other obs ito two idepedet sets, amely the followig ob sets with their positios i the correspodig schedule obs 1,,k-1,k+1,,k+δ are processed (i some order) firstly ob k processed after them ad, fially, the obs k+δ+1,,l that are processed i some order lastly This leads to the followig recursive computatio Wirtschaftsiformatik ud Operatios Research 190 The recursive formula of the DP (,, ) = { i < k} ( ( ) ) ( ) J l k i i l p p V J, l, k, t : Miimal total tardiess for processig the obs of set J, l, k i It holds that: a optimal sequece startig at time t (, ) = 0 ad ({ }, ) = max{ 0, + } V t V t t p d V ( J (, k + δ, k ), t ) + max 0, t + p k + p dk J(, k + δ, k ) Completio time of ob k,,, = mi, δ + V J ( k + δ + 1, l, k ), t + p J(, k + δ, k ) Completio time of ob k V ( J ( l k ) t ) { } ( ) k i ( ) V ({ } ) with k J, l, k such that p = max p i J, l, k Optimal solutio is obtaied by solvig: 1,...,,0 Wirtschaftsiformatik ud Operatios Research

31 2.5.4 A simple example Jobs p d We cosider a example with 5 obs that are already sorted accordig to urgecy Clearly, we have to solve V({1,2,3,4,5},0) For this purpose, we ask for the ob with largest processig time It is ob 3. Hece, obs 1 ad 2 are scheduled before ob 3 We test the positios 3,4,5 for ob 3 (i.e., we set δ to 0,1, ad 2) This leads to the followig result Wirtschaftsiformatik ud Operatios Research 192 V Solvig the example ({ 1,2,3,4,5 },0) V ( J ( 1,3,3 ),0) + max{ ,0} + V ( J ( 4,5,3 ),347) mi V ( J ( 1,4,3 ),0) max{ ,0} V ( J ( 5,5,3 ),430) V ( J ( 1,5,3 ),0) + max{ ,0 } + V (,560) V ({ 1,2 },0) V ({ 4,5 },347) mi V ({ 1,2,4 },0) 164 V ({ 5 },430) V ({ 1,2,4,5 },0) V (,560) = = + + The remaiig smaller problems ca be solved directly Clearly, V({1,2},0)=0 with both sequeces (1,2) ad (2,1) Moreover, V({1,2,4},0)=0 with (1,2,4) or with (2,1,4) V({4,5},347)= =317 with the optimal sequece (4,5) Wirtschaftsiformatik ud Operatios Research 193 Solvig the example Moreover, we obtai V({5},430)= =223 Fially, we have V({1,2,4,5},0)= = =76 with the optimal schedules (1,2,4,5) or (2,1,4,5) By isertig these values, we obtai V ({ 1,2,3,4,5 },0) V ({ 1,2 },0) V ({ 4,5 },347) = mi V ({ 1,2,4 },0) V ({ 5 },430) = mi = mi 387 V ({ 1,2,4,5 },0) 294 V (,560) Therefore, we obtai the optimal schedules (1,2,4,5,3) ad (2,1,4,5,3) with total tardiess 370 Complexity of the DP The worst case complexity of the approach ca be derived directly by estimatig the umber of occurrig problems to be solved Each problem is solved i time O() (Testig all possible δ- values) Moreover, we have O(³) sets V(,l,k) ad Σp time assigmets i the problems to be cosidered Cosequetly, the umber of problems is upper bouded by O(³. Σp ) The overall ruig time is therefore bouded by O( 4. Σp ) Hece, we have a algorithm with a pseudo-polyomial ruig time Wirtschaftsiformatik ud Operatios Research 194 Wirtschaftsiformatik ud Operatios Research