Chapter 15 MSM2M2 Introduction To Management Mathematics

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1 Chapter 15 MSM2M2 Itroductio To Maagemet Mathematics (15.1) Combiatorial Methods (15.1.1) Algorithms & Complexity At the heart of ay combiatorial method is a algorithm; a process which is (supposed) to yield the most efficiet or profitable solutio so some situatio. A simple ad obvious way to do this is as follows. I a attempt to solve this, the followig could be used. Algorithm 1 (The Greedy Algorithm) Choose the most attractive feasible optio available. However, it is easy to costruct situatios where this does ot work a umber of coditios about the problem must be satisfied i order the the Greedy Algorithm to work. I search of a better solutio to the oe supplied usig the Greedy Algorithm, it may be possible to fid all permutatios the choose the best. However, eve powerful computers may ot be able to solve eve simple problems give may years. Two areas of iterest are apparet, does a give algorithm fid a correct solutio. how efficiet is the algorithm. I a mathematical sese, the efficiecy ca be expressed i terms of how may elemetary operatios additio, multiplicatio, etc. must be performed i order to achieve the result. However, the iput must also be take ito accout as clearly a algorithm that takes 10 operatios o 3 iputs is less efficiet tha oe that takes 10 operatios o 30 iputs. Example 2 Compute the umber of elemetary operatios eeded to evaluate the polyomial fuctio p(x) = a x + a 1 x a 1 x + a 0 usig the method 1. by calculatig the powers of x, multiplyig by the coefficiet a i, the addig the terms. 2. by startig with the lowest power of x, ad savig the powers of x for use i calculatig the ext term, the addig the terms. 3. usig Horer s Method, p(x) = (... ((a x + a 1 ) x + a 2 ) x + + a 1 ) x + a 0 Proof. Solutio 1. Evaluatio usig this method ivolves 1

2 2 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS i 1 multiplicatios for the power of x i the ith term ad aother for the coefficiet, givig a total of i = 1 2 ( + 1). i=0 additios i order to combie the resultig + 1 terms. This gives a total of 1 2 ( + 1) + = Evaluatio usig this method ivolves The term with x 0 requires o multiplicatio. The term with x 1 requires oe multiplicatio, the coefficiet. Subsequet terms require a multiplicatio by x ad by the coefficiet i.e. two each. There are therefore ( 2) = 2 1 multiplicatios. additios i order to combie the + 1 terms. This gives a total of Evaluatio usig Horer s method ivolves oe multiplicatio for each power of x these are the xs outside the brackets of which there are 1 ad the first x which is multiplied by its coefficiet, givig multiplicatios. wheever a x appears it is multiplied by somethig the added to the ext coefficiet. There are therefore additios. This gives a total of 2 operatios performed. Fidig the umber of operatios is easy eough i this case, ad the umber of iputs + 2 is obvious. What is ot obvious, though, is how to combie the two. Defiitio 3 The computatioal complexity of a algorithm A, cc(a), is the umber of operatios performed o a iput of legth L. This is expressed as a fuctio of L so that cc(a) = f (L). I Example ( 2 ) the computatioal complexities ca ow be calculated ad are 1. (L 2) (L 2) L L 4. Clearly Horer s method is the most efficiet for a iput of ay give legth. Fuctios for the computatioal complexity may vary eormously for the same result but usig differet processes. It is of course desirable to achieve the computatioal complexity of lowest value. This is typically doe by reducig the power of the expressio, or i the liear case by reducig the values of the coefficiets. (15.1.2) O Notatio It is coveiet at this poit to itroduce the apparetly bizarre O otatio. Cosider the set of real fuctios defied o R + which are evetually positive, i.e. F + = { f f : R + R with f (x) > 0 x x 0, x 0 R +} For g(x) F + the followig defiitio is ow made. Defiitio 4 The fuctio O(g) for a fuctio g which is evetually positive i.e. g F + is give by the set O(g) = { f (x) F + c > 0 x 0 > 0 x x 0 f (x) c g(x) }

3 15.1. COMBINATORIAL METHODS 3 Note that whe f O(g) it is commo to write f = O(g) or eve f (x) = (g(x)). O(g) is the evetually positive fuctios whose value at x does ot exceed c times g(x), evetually. It is readily deduced that.if f (x) g(x) the O( f ) O(g), i.e. f = O(g). As fuctios ca be composed, it follows that some similar process ca be performed o the sets O. Defiitio 5 { } 1. O( f ) + O(g) = f (x) + g(x) f O( f ) ad g O(g) { } 2. O( f ) O(g) = f (x) g(x) f O( f ) ad g O(g) 3. f + O(g) = { f (x) + g(x) g O(g)} 4. f O(g) = { f (x) g(x) g O(g)} The abuse of otatio f (x) = (g(x)) is exteded so that becomes the same as. It is readily show from the defiitio that if f O(g) the O( f ) O(g), ad ideed if f g the O( f ) O(g). Alog the same lie of thought, the followig ca be show from the defiitios above. Theorem 6 For f, g F + ad k R +, 1. O( f + g) = O( f ) + O(g) 2. O( f g) = O( f ) O(g) 3. k O( f ) = O( f ) 4. g f O(g + f ) = O( f ) 5. g f g + O( f ) = O( f ) 6. f (x) ε > 0 x x 0 f + a O( f ) (a > 0) The virtue of O otatio ca be see whe cosiderig computatioal complexity fuctios. Rather tha the actual fuctio, a simplified form of it its O set is usually of iterest as it gives a idicatio of the level of complexity. For Example 2, ( ) O (L 2) (L 2) 2 = L 2. O(3L) = L. O(2L) = L. So while Horer s method is the simplest, its computatioal complexity is comparable to that of the secod method, while the first method is particularly more complex. (15.1.3) Sortig A particular class of algorithm relate to sortig, perhaps most commoly puttig a sequece of umbers ito ascedig order. First of all the Merge operatio is defied. Defiitio 7 Suppose that ( ) ( ) x 1, x 2,..., x p ad y1, y 2,..., y q are sequeces which are already sorted. The Merge operatio defies the sequece ( ) z 1, z 2,..., z p+q as follows. 1. If x 1 y 1 the set z 1 := x 1. Repeat comparig the first elemets of ( ) ( ) x 2, x 3,..., x p ad y1, y 2,..., y q. 2. If x 1 > x 1 the set z 1 := y 1. Repeat comparig the first elemets of ( ) ( ) x 1, x 2,..., x q ad y2, y 3,..., y p.

4 4 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS 3. The process termiates whe oe of the sequeces becomes empty. The remaider of the other sequece is the copied oto the ed of the ew sequece to give ( z 1, z 2,..., z p+q ) I mathematical formulae, merge will be deoted by the biary operatio symbol. The Merge operatio takes two elemetary operatios for each iteratio, of which there is oe for every elemet of both of the sequeces beig merged. I the worst case p = q, so sice the fial elemet is copied, the computatioal complexity of Merge must be 2(p + q 1). Mergig ca oly be doe o sorted sequeces, which begs the questio as to how to sort. Algorithm 8 (MergeSort) Take a usorted set of elemets, treat each elemet as a sequece ad apply Merge o pairs of the sigle elemet sequeces. Repeat applyig Merge util a sigle sorted sequece is obtaied. The MergeSort algorithm ca be expressed diagrammatically. ( 1, 2,..., k ) ( 1, 2,..., k ) }{{}... ( 1, 2,..., k ) ( 1, 2,..., k ) }{{} ( 1, 2,..., 2k ) ( 1, 2,..., 2k )... }{{} ( 1, 2,..., k ) for some N Suppose that iitially there are = 2 r elemets to sort, for some r N. Iitially there are 2 sorted sequeces the sigle elemet sequeces. After the first iteratio there are 2 sorted sequeces, ad there were applicatios of Merge. 2 After k iteratios there are 2 k sorted sequeces, ad there were applicatios of Merge. 2k Whe there is oe sorted sequece the process termiates, at which time it must be the case that = 1. This 2k gives the umber of iteratios to be k = log 2. Now, the computatioal complexity of Merge is 2(p + q 1), ad at each stage there are elemets beig merged. Therefore cc ( all merges ) < 2 ad so cc (Mergesort) < 2 log 2 = O( log 2 ) (15.1.4) Graph Theory Most of the results covered here are also covered i (??). Ideed, it is ecessary to kow the same results. Defiitio 9 A graph is a ordered pair, Γ = (V, E) where V is a o-empty fiite set, ad E is a set of two-subsets of V. The elemets of V are called vertices or odes, ad the elemets of E are called edges or arcs. Graphs ca be readily draw, as is show i Fig.??. This graph has V = {1, 2, 3, 4} ad E = {{1, 2}, {3, 4}, {1, 4}}. It does ot matter how the graph is draw i particular the edges do ot have to be straight ad two graphs that look differet but are i fact the same graph are called isomorphic. There are, however, the limitatios that there caot be ay loops, ad edges ca oly meet at a vertex. Also, may of the graph theory results require that the graph is fiite, i.e. V N. Defiitio 10 For a graph with vertices u, v ad edges e, f,

5 15.1. COMBINATORIAL METHODS 5 if u is oe of the vertices at the ed of e, the e is icidet with u. u ad v are adjacet if the edge {u, v} is i Γ. e ad f are adjacet if they are icidet with the same vertex. Defiitio 11 For a graph Γ = (V, E) with V = ad E = m the, Γ is complete if ay vertex is adjacet with ay other vertex. Such a graph is writte K, ad clearly m = ( 1) i =. i=1 2 Γ is trivial if m = 0 i.e. E =. Γ is sparse if m = O(). If Γ is ot sparse the it is dese. Some such graphs are show i Fig.2 K 4 A sparse graph A dese graph Figure 2: Differet kids of graph Clearly complete graphs are dese. Defiitio 12 A graph Γ = (V, E) is bipartite if there exists V 1 V ad V 2 V, both of which are o-empty, such that V 1 V 2 =, V 1 V 2 = V ad o two vertices of V 1 are adjacet ad o two vertices of V 2 are adjacet. A complete bipartite graph where all vertices of V 1 are coected to all vertices of V 2 but o vertices i V 1 is deoted as K pq where V 1 = p ad V 2 = q. Defiitio 13 A graph Γ = (V, E ) is a subgraph of the graph Γ = (V, E) if V V ad E E. It is commo to write Γ Γ. Defiitio 14 The degree of a vertex is the umber of edges icidet with that vertex. So if Γ = (V, E) ad v V the deg Γ v = {u {u, v} E}. Sice each edge is icidet with exactly two vertices, it is obvious that deg v = 2 E. A formal proof ca be give by iductio, or by cosiderig v V the icidece matrix of a graph. Defiitio 15 For a graph Γ = (V, E), a path i Γ is a sequece of vertices (v 1, v 2,..., v k ) such that v i is adjacet to v i+1 for 1 i k 1, ad v i = v j i = j. Iformally put, a path is a sequece of distict o-adjacet vertices. A u v path is a path which starts at u ad eds at v. A graph is called coected if there is a u v path for all u, v V. Defiitio 16 A cycle is a path (v 1, v 2,..., v k ) for which k 3 ad v 1 = v k. A graph with o cycles is called acyclic, ad a coected acyclic graph is a tree. Theorem 17 If Γ is a tree, the there is at least oe vertex which has degree 1.

6 6 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS A equivalet to this theorem is to say that if all vertex degrees are at least 2, the Γ cotais a cycle, ad ideed the proofs are to all itets ad purposes, the same. Proof. Suppose that the result is false, so that Γ is a tree ad all vertex degrees are at least 2. Cosider the path of logest legth, say P = (v 1, v 2,..., v k ) which must exist sice Γ is a fiite graph. Sice deg v k 2 there must exist some vertex v i = v k 1 which is adjacet to v k. But sice P was the path of logest legth, it must be the case that j < k 1 such that v i = v j ad hece (v i, v i+1,..., v k, v i is a cycle i Γ. Cotradictio sice Γ is a graph. Theorem 18 If a graph Γ = (V, E) is a tree, the E = V 1. Proof. The result clearly holds whe V = 1. Suppose that V 1 ad that the result holds. Cosider addig a vertex v to Γ to give the graph Γ. I order ot to create a cycle ad hece maitai the status of the graph as a tree, v must be coected to Γ by exactly oe edge. Observe that E = V 1 E 1 = ( V 1) 1 E = V 1 So Γ is a tree ad the property holds. The followig extesio to Theorem 17 ca ow be made. Theorem 19 If Γ = (V, E) is a tree, the at least two vertices have degree 1. Proof. Suppose the result is false. If o vertices have degree 1, the sice Γ is a tree it must be coected so all vertices have degree at least 2. This cotradicts Theorem 17. The remaiig case is that oe vertex, say u, has degree 1, ad all others have degree at least 2. For the vertices of degree at least 2, deg v 2 V \ {u} v V v = u 2 E 2 V \ {u} 2 ( V 1) E V 1 However, this excluded the vertex of degree 1. Addig this i, E + 1 V 1 E V But this is a cotradictio sice by Theorem 18 E = V 1. Hece Γ is ot a tree, ad hece by cotradictio the theorem must hold.

7 15.1. COMBINATORIAL METHODS 7 So far it has bee implicit that all the vertices of the graph are coected together by edges i some way. However, this eed ot be the case, as a graph may have may differet coected compoet. Defiitio 20 A graph Γ 1 is a coected compoet of the graph Γ if Γ 1 is a subgraph of Γ. Γ 1 is coected. Γ 1 is a subgraph of Γ 2 is a subgraph of Γ, the Γ 1 = Γ 2. So a coected compoet is part of a graph which is a graph i its ow right. A example is give i Figure 3. ad are the coected compoets of Figure 3: Coected compoets of a graph Theorem 21 Suppose that Γ is a coected graph with a cycle. If a edge of the cycle is removed, the the resultig graph is still coected. Proof. Let e = {v 1, v 2 } be a edge of the cycle i Γ, ad let Γ \ {e} = Γ. Sice Γ is coected, for ay two vertices u 1 ad u k there exists a path P betwee them. If P does ot traverses e, the P is a path i Γ. Suppose that the cycle i Γ be (v 1, v 2, v 3,..., v, v 1 ). Sice a cycle ca be started at ay poit, this ca be doe. If P traverses e, say P = (u 1, u 2,..., v 1, v 2,..., u k ). The (u 1, u 2,..., v 1, v, v 1,..., v 3, v 2,..., u k ) is a path coectig u 1 ad u k i Γ. Either way, ay two vertices of Γ are coected, hece the result. Theorem 22 If Γ is a acyclic graph, the E = V p(γ) where p(γ) is the umber of coected compoets of Γ. Proof. Sice Γ is acyclic, the for each coected compoet, E 1 = V i 1. Summig these up produces the required result. Theorem 23 For a graph Γ the followig are equivalet. 1. Γ is a tree. 2. Γ is acyclic ad E = V Γ is coected ad E = V For ay two vertices, there exists a uique path betwee them. The proof of this theorem is quite straightforward, ad it follows directly from (4) that if a ew edge is added to Γ the a uique cycle is created. If ay edge of this cycle is ow removed, the the graph is oce agai a tree. Defiitio 24 If Γ is a coected graph, the a subgraph Γ is a spaig tree for Γ if it is a tree ad has the same vertex set as Γ. It is clear to see that the spaig tree of a graph with vertices has 1 edges.

8 8 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS Theorem 25 A graph Γ is a spaig tree for a graph Γ if ad oly if Γ is a miimal coected spaig subgraph of Γ. A miimal coected spaig subgraph is the spaig subgraph with the least edges. Proof. Let V = ad let Γ be a arbitrary coected spaig subgraph of Γ. If Γ is acyclic, the it is a spaig tree so E(Γ ) = 1. If Γ is ot acyclic, the remove a edge from oe of the cycles. The graph is the still coected by Theorem 21. This process ca cotiue util there are o cycles i the resultig graph, Γ say. So E(Γ ) = 1. Theorem 26 A graph Γ is a spaig tree for a graph Γ if ad oly if Γ is a maximal acyclic subgraph of Γ. Proof. Let Γ be a acyclic subgraph of Γ. By Theorem 22, E(Γ ) = V(Γ ) p(γ) V(Γ) 1 So whe Γ is coected ad so spas Γ it must have E(Γ ) = V(Γ) 1 i.e. be a tree. Theorem 27 (Cayley s Theorem) The graph K has m m 2 spaig trees, where m 2. This is readily verified by cosiderig a few examples, but the proof is ot give. Defiitio 28 Comp x is the set of all vertices of a graph which are i the same coected compoet as the vertex x. Some Basic Algorithms Eough theory has bee covered at this poit to itroduce aother algorithm. Algorithm 29 (Search) For a graph Γ = (V, E) with a vertex u V, the algorithm Search idetifies all vertices of the coected compoet cotaiig u i.e. fids Comp u. The algorithm is 1. Defie the set P such that whe ever a vertex x is marked, the operatio P := P {x}. Iitially P =. 2. Defie the set Q := {u} ad mark u. 3. Cosider ay x Q ad set Q := Q \ {x}. 4. For all y / P such that {x, y} E, mark y, ad set Q := Q {y}. Uless y is already marked. 5. If Q = the the algorithm has fiished. Otherwise repeat from (3). For the proof of a algorithm, it is required to show its correctess ad its computatioal complexity. Proof. The correctess of Search is obvious. For the computatioal complexity each step is cosidered i tur. 1. No operatio is performed here. 2. Oe operatio i assigig Q, ad aother i markig u. Therefore Every vertex will be processed exactly oce, therefore these two operatios will be performed Comp u times, cotributig 2 Comp u operatios. 4. There are three operatios here, ad this step will be performed oce for each edge {x, y} E u, givig a total of 3 E u.

9 15.1. COMBINATORIAL METHODS 9 5. This compariso is performed oce for each vertex, sice oly whe all vertices have bee processed will Q =. Therefore Comp u. Summig all these up ad sice E u Comp u 1, cc( Search ) = Comp u + 3 E u E u sice Comp u E u + 1 = O ( E u ) It ca be said, therefore, that all the coected compoets of a graph ca be idetified i O( E ) time. A obvious applicatio of Search is to the questio of coectedess. This may be formulated as Give ay two vertices of a graph, are they coected i.i. does there exist a path betwee them. Sice Search fids coected compoets, it ca provide a solutio to this problem. The Miimal Spaig Tree Problem Each edge of a graph may be give a weight by some fuctio w : E R. It is the of iterest to fid the spaig tree of miimal or maximal weight. The greedy algorithm, Algorithm 1 has already bee stated, ad i this cotext is iterpreted as attractive meas the edge of least weight. feasible meas a edge that will ot create a cycle. Theorem 30 (The Greedy Algorithm For The Miimal Spaig Tree Problem) Let Γ = (V, E) be a coected graph, ad w : E R be a weight fuctio. Let (V, F) be a subgraph of Γ. The Greedy Algorithm works for the miimal spaig tree problem i that if (V, F) is a subgraph of a miimal spaig tree, the so is (V, F {e}) where e E \ F is chose by the Greedy algorithm. Proof. Let e = {u, v} be a edge of miimal weight such that (V, F {e}) is acyclic, i.e. a edge chose by the Greedy Algorithm. Observe that u ad v must be i differet coected compoets of (V, F), sice otherwise e would create a cycle. Suppose that T = (V, E ) is a miimal spaig tree. i. If e E the (V, F {e}) T ad the statemet holds. ii. If e / E the (V, E {e}) cotais a uique cycle, σ say. Now, e jois coected compoets, so sice it also creates a cycle the must exist a edge f which also jois the same coected compoets. f caot be i (V, F) sice if it were, these coected compoets would already be coected together. Hece take (V, F { f }) which must be acyclic, so at this stage the Greedy algorithm could have chose e or f. Sice e was chose, w(e) w( f ). Let T = (V, (E {e}) \ { f }) which must also be a spaig tree for Γ because a edge i a cycle has bee added ad aother removed. Now, w(t ) = w(t) + w(e) w( f ) w(t). But T was a miimal spaig tree, hece w(t ) w(t). The oly possibility is that w(t) = w(t ), ad hece e is a edge of a miimal spaig tree. Hece this secod case holds.

10 10 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS Both cases hold, so the theorem is prove. This has show that the Greedy Algorithm is correct, but what remais to be show is its computatioal complexity. Theorem 31 The computatioal complexity of the greedy algorithm for the miimal spaig tree problem is O( E 2 ). For a graph Γ = (V, E) with weight fuctio w : E R the greedy algorithm is formulated as. 1. Set F := ad p := Use the greedy algorithm to fid e E \ F such that w(e) = mi (w( f ) f E \ F). 3. If (V, F {e}) is acyclic, the set F := F {e} ad p := p + 1 ad cotiue. Otherwise, just cotiue. 4. If p = V 1 the the stop. Otherwise repeat from (2). Proof. The umber of operatios at each state is cosidered i tur. 1. I the iitialisatio there are oly 2 operatios. 2. From all the edges available i E \ F it is ecessary to fid the oe with the least weight. Each of these edges, say e 1, e 2,..., e k is assessed i tur. a 1 r := 1 1 operatio if a 2 < a r r := 2 at most 2 operatios if a 3 < a r r := 3 at most 2 operatios.. if a k < a r r := k at most 2 operatios The required edge will ow have bee foud. The umber of operatios used is ow assessed. 2(k 1) + 1 = 2k 1 = O(k) = O (E \ F) = O( E ) sice E \ F E There are two other operatios i (2) where the set E is adjusted. This gives a total of 2 + O( E ). 3. It is ow ecessary to check if (V, F {e}) is acyclic ad for this it is first ecessary to show that (N, F {e}) is acyclic where e = {u, v}, u is ot coected to v i (N, F). It is easier to prove the egatio. Suppose that (N, F {e}) has a cycle σ, the e is a edge i σ. It is clear therefore that u is coected to v i (N, F). If u is coected to v i (N, F) the a path betwee them exists i (N, F). Hece (N, F {e}) has a cycle. The Search algorithm provides a way to check for coectedess betwee two vertices, so this coditio ca be verified with O( F ) operatios. But F E 1 sice otherwise the tree (V, F) would be the origial graph Γ = (V, E) ad the algorithm would have fiished. Sice there are 4 other operatios i this step, the computatioal complexity provided here is 4 + O( E 1 ).

11 15.1. COMBINATORIAL METHODS The fial step (ad hece the loop) ca be repeated at most E times. Addig up the computatioal complexities from each stage, computatioal complexity 2 + E (O E + O( E 1) + 6) E O( E ) ( O E 2) (15.1.5) Assigmet Problems Assigmet problems ivolve creatig a oe to oe mappig, say of workers to jobs, that maximises efficiecy, say, or miimises some quatity. The total umber of optios ca be represeted by a complete bipartite graph, ad the weight of the edges represets the efficiecy or whatever of the assigmet it represets. The problem ca also be represeted by a matrix, where the ij elemet is the weight of the edge {i, j}. Example 32 Three taxis are located at distaces from three customers as show i table 32. c 1 c 2 c 3 t t t Table 1: Distaces from taxis to customers Which taxi should go to which customer i order to miimise the total distace traveled by the taxis? Proof. Solutio A exhaustive method is adopted for illustrative purposes. The permutatio taxi i goes to customer j is represeted by π(i) = j, ad this is expressed usig disjoit cycles. π w(a, π) π w(a, π) (1)(2)(3) 1+4+4=8 (3)(12) 5+4+4=13 (1)(23) 1+3+2=6 (123) 5+3+2=10 (2)(13) 2+4+2=8 (132) 2+4+4=10 Table 2: Possible solutios to the taxi problem The table shows all possible permutatios, ad it is clear that (1)(23) is the best solutio to this problem. Classical Assigmet Example 32 is a istace of the classical assigmet problem, which may be formulated as follows. Defiitio 33 Give a square ( ) matrix, fid the elemets such that 1. o two of the chose elemets are i the same row or colum. 2. the sum of the elemets is miimal (maximal).

12 12 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS Where P deotes the set of all permutatios of items ad π P, It is the ecessary to fid w(a, π) = a i, π(i) i=1 mi π P w(a, π) = mi A The set ap(a) is the set of optimal permutatios, ad AP mi or AP max represet the chose solutio, depedig o whether it is of the maximal or miimal variety. Without loss of geerality, from this poit o oly the miimisatio problem will be cosidered. The maximisatio problem ca be solved by cosiderig the matrix A. The solutio to classical assigmet utilises the followig property. If a costat α is added to every elemet of a sigle row or a sigle colum of a matrix A to give A, the w(a, π) = w(a, π) + α ad so ap(a ) = ap(a). Defiitio 34 A matrix is i ormal form if a ij 0 a ij, ad A cotais idepedet zeros i.e. it is possible to fid zeros i A which do ot lie i the same row or colum. Notice that whe a matrix is i ormal form, mi A = 0 ad sice there are o egative elemets, it is simple to read off the optimal permutatio. Theorem 35 (Köig-Egervary) Where a lie is either a row or colum, the maximal umber of idepedet zeros i a matrix is equal to the miimal umber of lies required to cover all the zeros of the matrix. Algorithm 36 (The Hugaria Method ) The Hugaria method works by trasformig the matrix, say A, ito ormal form as follows. 1. (a) Subtract from every colum of A the costat that is the least elemet of that colum, givig matrix A 1 (b) Subtract from every row of A 1 the costat that is the lease elemet of that row, givig matrix A Say there are at most k idepedet zeros i A 2 this eeds to show usig Theorem 35. (a) if k = the stop. (b) if k < proceed as follows. i. Cover all zeros i the matrix by the k lies foud. Suppose that there are k r rows ad k c colums, so that k = k r + k c. ii. Of all ucovered elemets, fid the least, t. iii. Subtract t from each ucovered row. iv. Add t to each covered colum v. cosider the matrix created, A 3, ad repeat the process from 2. There is o guaratee that the umber of idepedet zeros will icrease at every iteratio, so it is importat to show that the Hugaria method is fiite. Theorem 37 Whe the matrix i questio is iteger, the Hugaria method for solvig the classical assigmet problem is fiite i that it termiates after a fiite umber of iteratios.

13 15.1. COMBINATORIAL METHODS 13 v 2 v 1 v 3 v 4 Figure 4: A digraph Proof. From the geeral step, where the ew ( ) matrix created is A, w(a, π) = w(a, π) + tk c t( k r ) = w(a, π) + t(k c + k r ) but k c + k r = k <, so < w(a, π) if A is iteger, i.e. all its etries are itegers, the sice w(a, π) < w(a, π), it must be the case that w(a, π) w(a, π) 1 But this relatioship must hold for each iteratio. Hece Iitially w(a, π) = mi A After the first iteratio w(a, π) mi A 1 After the secod iteratio w(a, π) mi A 2 After the th iteratio.. w(a, π) mi A The algorithm stops whe mi A = 0, which ca take at most mi A iteratios. Hece the algorithm is fiite ad the theorem is proved. It ca be show that the computatioal complexity of the Hugaria method is O( 2 ). (15.1.6) Digraphs Directed Graphs A directed graph is idetical to a ormal graph with the exceptio that the edges are ordered pairs of vertices, say (e, f ) istead of {e, f }. This meas that the edges have directio. The directed edges makes the possibility of loops meaigful. Some of the terms from graph theory are modified for use i a digraph. A path may ow cotai repeated vertices, although a ormal path may exist ad is ow called a elemetary path. Similarly, cycles may cotai repeated vertices, ad a ormal

14 14 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS cycle is called a elemetary cycle. For example, i the digraph i Figure 4, (v 1, v 2, v 2, v 3, v 4 ) is a path, ad (v 4, v 3, v 1, v 2, v 3, v 1, v 4 ) is a cycle. For a sizable digraph with a ot icosiderable umber of vertices it is clear that the shortest path from oe vertex to aother is ot a trivial matter. Add to this the complexities imposed by a weightig fuctio, ad the problem soo becomes rather uwieldy. For a digraph Ϝ = (V, E) with weightig fuctio d : E R, k the weight of a path P = (v 0, v 1,..., v k ) is d(p) = d(v i 1, v i ). The shortest path problem is to fid the i=1 path of miimal weight (or legth) betwee two give vertices. A solutio is foud by fidig the shortest path from the start vertex s to all other vertices. For a set of vertices W V, defie d W (v) to be legth of the shortest sv path with all itermediate vertices i W. This is the temporary shortest distace. By extedig W from {s} to V all the shortest paths ca be foud ad hece solve the problem. Theorem 38 Choose x V \ W such that d W (x) = mi {d W (v) v V \ W} i.e. the shortest sv path. Let W = W {x} the for y V \ W, d W (y) = mi {d W (y), d W (x) + d(x, y)} I order to prove this, the followig lemma is required Lemma 39 If x is a vertex selected by the process i Theorem 38, the d W (x) = d V (x). Proof. Proof Of Lemma 39 A temporary shortest distace must be greater or equal to the actual shortest distace, hece d W (x) d V (x). Suppose that d W (x) > d V (x). The there exists a path P with d(p) = d V (x) that has at least oe itermediate vertex ot i W. Let z be the first vertex of P that is ot i W, the P ca be writte as P + P where P is as sz path i W ad P is the rest of the path. Hece d(p ) + d(p ) = d(p) < d W (x). But d(p ) > 0 so d(p ) < d W (P) Now, x was the vertex (ot i W) that had the temporary shortest distace. But z / W ad d(p ) < d W (P) where by costructio P is a sz path with itermediate vertices i W which is shorter. Hece by cotradictio the lemma holds. Proof. Proof Of Theorem 38 Cosider y V \ W. d W (y) ad d W (x) + d(x, y) are the legths of two sy paths with itermediate vertices i W, ad so d W (y) mi {d W (y), d W (x) + d(x, y)} Now, let P be the shortest sy path with itermediate vertices i W, so d(p) = d W (y). The followig three cases correspod to those show i Figure 5. case 1 If x / P the P has all itermediate vertices o W ad hece d(p) = d W (y). case 2 (a) If x is the peultimate vertex of P, the let P = (s,..., x) which is the shortest sx path with itermediate vertices i W. Hece d(p ) = d W (x) ad so d(p) = d W (x) + d(x, y). (b) Let P be the shortest sy path with all itermediate vertices i W. Sice x is ot peultimate, P may be expressed as P = P + P where P is a sx path with all itermediate vertices i W. P is a sy path with all itermediate vertices i W. Say the peultimate vertex o P is z, the z W. This meas that z must have bee chose earlier, so there exists a shorter sz path tha usig P. This cotradicts that P is the shortest sy path, so i this case d W (y) = d W (y).

15 15.1. COMBINATORIAL METHODS 15 x x x W W W y y y s s s z z Case 1 Case 2(a) Case 2(b) Figure 5: Pictorial represetatio of Theorem 38. I all cases, the shortest sy path is oe of d W (y) ad d W (x) + d(x, y). Hece d W (y) = mi {d W (y), d W (x) + d(x, y)} ad so the theorem is proved. Havig show that this process of fidig shortest paths works, it is ow possible to formulate the algorithm for puttig this result ito practice. Algorithm 40 Dijkstra For a digraph Ϝ = (V, E) ad weight fuctio d : E R the shortest paths from the vertex s are foud as follows. 1. (a) Set W := {s}. (b) Set d(x) := d(s, x) if (s, x) E else d(x) =. 2. x V \ W fid x such that d W (x ) = mi{d W (x)}. 3. Set W := W {x }. 4. For all y V \ W set d W (y) := mi{d W (y), d W (x ) + d(x, y)}. 5. If W = V the stop, else repeat from 2. Proof. The correctess of the algorithm is show by Theorem 38. The computatioal complexity is calculated for each step i tur for the assigmet, ad V 1 = 1 for the checkig of vertices. 2. Fidig x ca be doe i at most V \ W steps, which is O(). 3. As assigmet ad a uio, makig A additio, a miimum, ad a assigmet. Each of these must be doe for V \ W vertices, givig 3O() = O(). 5. Oe operatio. Note that the loop will be made 1 times. Hece the total computatioal complexity is computatioal complexity = 1 + ( 1) + ( 1) (O() O()) O() + (O()) 2 = O() + O( 2 ) = O( 2 ) Hece the computatioal complexity is O( 2 ).

16 16 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS a b 9 d f c 7 e Figure 6: Fid the shortest path from a to f. See Example 41. Iteratio Temporary Shortest Distaces Vertex Chose 1 a 2 b 5 c 3 d 8 e 10 f 20 c 3 3 d 8 e 10 f 13 b 5 4 d 14 e 11 f 15 d 8 5 e 12 f 10 f 10 6 e 10 Table 3: A ru of Dijkstra s Algorithm for Example 41. Dijkstra s algorithm is relatively complicated, ad whe it has fiished the task remais to actually fid the optimal solutio. First of all a example is give Example 41 Fid the shortest path from a to f i the weighted digraph show i Figure 6. Proof. Solutio The ru of Dijkstra s algorithm is show i Table 3. The otatio a b is used to show that the temporary shortest distace to vertex a is b. Notice that the algorithm oly fiishes whe all the vertices have bee chose. The remaiig task is to actually fid the shortest path, though clearly it has legth 10. This is doe as follows. (i) The vertices goig ito f have weights 2, 10, 20, 10, ad = 8 ad oe of the temporary shortest distaces to a chose vertex, amely d is 8. Therefore d comes before f. (ii) The vertices goig ito d have weights 9, 8, ad 5. But the shortest distace to d is 8, so the precedig vertex must be a/ Oe shortest path is therefore a, d, f, though there may be more. (15.1.7) The Trasportatio Problem As the ame suggests, the trasportatio problem cocers how to efficietly move goods from producers to cosumers. However, there are may other applicatios of the problem. Th iformatio used for the trasportatio problem may be set out i tabular form, as show i Table 4. The first colum shows the amouts produced by the producers, ad the first row shows the amouts

17 15.1. COMBINATORIAL METHODS 17 b 1 b 2 b a 1 c 11 c 12 c 1 a 2 c 21 c 22 c a m c m1 c m2 c m Table 4: Iformatio for the trasportatio table. demaded by the cosumers. The cetral values, c ij, show the cost of movig a uit from producer i to cosumer j. The problem ow is to fid umbers x ij, the amout of goods to be take from producer i to customer j. The problem ca be represeted o a bipartite graph bipartite because takig goods from producer to producer or cosumer to cosumer is clearly ot a optio. Suppose the producers are A i s with productio a i ad similarly the cosumers be B j with demad b j the clearly the trasportatio c ij is equivalet to the edge {A i, B j }. m The objective is to miimise the total cost of trasportatio, i.e. x ij c ij. There are however the obvious i=1 j=1 costraits ca a producer caot exceed capacity, ad cosumers demad as much as is stated ad this demad must be met. m Model 42 The trasportatio problem may be formalised as follows. Miimise the quatity f (X) = x ij c ij i=1 j=1 subject to the costraits x ij = a i so supplies must be used (43) j=1 m x ij = b j so demads must be met (44) i=1 x ij 0 for the sake of reality (45) It is also assumed that supplies ad demads are positive i.e. a i 0 i ad b j 0 j. f (X) is called the objective fuctio, ad X represets the set of x ij s of the solutio uder cosideratio. A solutio satisfyig equatios (43), (44), ad (45) is called feasible, ad whe f (X) attais its miimum the solutio is optimal. Note that f (X) does attai its miimum sice the set of all possible solutios is fiite. Hece if a feasible solutio exists, the a optimal solutio exists. m Theorem 46 The trasportatio has a feasible solutio if ad oly if a i = b j. i=1 j=1 Proof. The theorem has two parts, so each is take i tur,

18 18 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS If a solutio X is feasible the, m a i = i=1 Hece the oly if part. m Suppose that a i = b j = t, say. i=1 j=1 the ad = = m i=1 j=1 x ij m x ij j=1 i=1 b j j=1 by feasibility property of sums by feasibility set x ij = a ib j > 0 so (45) holds t x ij = a i t b j = a i so (44) holds j=1 j=1 m x ij = b m j t a i = b j i=1 i=1 so (43) holds Furthermore, this proof has revealed a feasible solutio to be x ij = a ib j. t m Clearly there is a problem if a i = b j. I this case a dummy producer or a dummy cosumer ca i=1 j=1 be itroduced with the required capacity or demad. The costs of trasportatio to these are zero. Hece m without loss of geerality it is assumed that a i = i=1 b j. j=1 The task of fidig a feasible solutio ad i tur a optimal solutio is ow set about. Note the followig defiitios. Defiitio 47 A solutio to the trasportatio problem is (i) basic if the associated graph is acyclic. (ii) degeerate if the associated graph is ot coected. (iii) o-degeerate if the associated graph is a spaig tree. First of all a basic feasible solutio is sought. Note that a degeerate solutio ca be exteded to a odegeerate oe by addig edges of zero weight. Defiitio 48 Let X be a basic feasible solutio. The base of X is the set B (1, 2,... m) (1, 2,... ) such that (i) x ij = 0 (i, j) / B. { } (ii) The set of vertices A i B j (i, j) B form a spaig tree for the associated graph, K m. If a basic feasible solutio X is already a spaig tree the the base is uique, but if it is degeerate the it may be possible to exted to a o-degeerate solutio i may ways. A solutio ca be checked for cycles i the trasportatio table directly without havig to draw the associated graph. Sice the graph is bipartite it is readily deduced that a cycle must joi o-zero cells i the table by alteratig vertical ad horizotal lies.

19 15.1. COMBINATORIAL METHODS 19 ) Defiitio 49 The dual solutio to the trasportatio problem is the vector (α 1 α 2... α m β 1 β 2... β where α i + β j = c ij (i, j) B. By fixig β = 0 the dual solutio ca be easily foud. ) Theorem 50 (The Optimality Criterio) Let X be a basic feasible solutio, ad (α 1 α 2... α m β 1 β 2... β be the dual solutio. If α i + β j c ij i j the X is a optimal solutio. Proof. Cosider some arbitrary feasible solutio Y. m f (Y) = c ij y ij i=1 j=1 m (α i + β j )y ij i=1 j=1 = = = m m α i y ij + β j y ij i=1 j=1 i=1 j=1 m α i a i + β j b j i=1 j=1 m i=1 α i j=1 m x ij + β j x ij j=1 i=1 m = (α i + β j )x ij i=1 j=1 sice Y is feasible sice X is feasible = (α i + β j )x ij + (α i + β j )x ij (i,j) B(X) (i,j)/ B(X) = c ij x ij + (i,j) B(X) (α i + β j )x ij (i,j)/ B(X) = c ij x ij sice (i, j) / B(X) x ij = 0 (i,j) B(X) m = c ij x ij i=1 j=1 f (Y) f (X) Hece X is a optimal solutio sice its cost is at most the cost of ay other feasible solutio. Kowig the optimality criterio allows feasible solutios to be checked for beig optimal. If they are ot the it is ecessary to fid a differet oe ad check that. Ideed, a method for fidig a sigle feasible solutio has ot jet bee exhibited. Algorithm 51 (The North-West Corer Rule) The orth-west corer rule fids a basic feasible solutio as follows 1. Defie (r, s) as the top left most available cell i the trasportatio table. 2. Assig x rs := max (a r, b s ). 3. Let a r := a r x rs ad make uavailable row r if a r = Let b s := b s x rs ad make uavailable colums s if b s x rs = If o cell is available the algorithm stops, else repeats from 1. It is claimed that this process will always fid a basic feasible solutio.

20 20 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS Proof. By costructio the solutio is feasible. What remais to be show is that the solutio is basic i.e. the associated graph is acyclic. Suppose the solutio foud has a cycle, ad let A i B j be the top left most edge of the cycle. There must be edges A r B j ad A i B s with o-zero etries i the trasportatio table i order for a cycle to exist, ote r > i ad s > j. By hypothesis A i B j is chose first, ad either row i or colum j or both are made uavailable. Hece at least oe of A r B j ad A i B s caot also be chose. Hece by cotradictio the theorem is proved. Havig foud a basic feasible solutio ad foud it ot to be optimal, a method must ow be foud to move to a optimal solutio. Sice the optimality criterio is violated, i jα i + β j > c ij. It is ow sought to iclude the offedig cell i the solutio, it will the cotribute to the αs ad βs ad so help get rid of the violatio of the optimality criterio. By aalogy with the associated graph, observe that a edge ca be added ad oe from the resultig cycle removed leavig aother spaig tree. By addig a edge A k B l say, oe of the edges A r B l ad A k B s must be removed, hece let x kl = θ be the miimum of all the x ij o the cycle ad subtract θ from each of the co-liear ad co-columar edges to A k B l. But the the solutio ologer works, so θ has to be added to the vertices co-liear or co-columar to those from which θ was subtracted... The followig algorithm therefore arises x ij θ x ij = x ij + θ x ij if A i B j is a eve arc o the cycle if A i B j is a odd arc o the cycle otherwise Fially, the trasportatio problem is illustrated by meas of a short example. Example 52 Three factories have capacities 100, 70, ad 50. Four cities demad the product i the amouts 30, 50, 50, ad 90. The costs of trasport are set out i Table Table 5: Table of costs for Example 52 Fid the optimal way to trasport the goods to the cities. Proof. Solutio Applyig the orth-west corer rule, Table 6 is produced. The small umbers i the top right of each cell are the trasport costs, ad the mai etries are the x ij s α 1 = α 2 = α 3 = 3 β 1 = 4 β 2 = 3 β 3 = 1 β 4 = 0 Table 6: First stage i the solutio of the trasportatio problem Observe that the optimality criterio fails i the cell marked with a. Cosider the (1, 4) cell, the takig θ = 20 the solutio is modified to that show i Table 7

21 15.2. LINEAR PROGRAMMING α 1 = α 2 = α 3 = 3 β 1 = 3 β 2 = 4 β 3 = 1 β 4 = 0 Table 7: First modified solutio I Table 7 the optimality criterio fails agai. Now cosider the (2, 2) cell, ad let θ = 20. The results of this modificatio are show i Table α 1 = α 2 = α 3 = 3 β 1 = 2 β 2 = 3 β 3 = 1 β 4 = 0 Table 8: Secod modified solutio I Table 8 the optimality criterio holds, so the solutio is complete. (15.2) Liear Programmig (15.2.1) Formulatio & Termiology The purpose of liear programmig is to fid the best solutio to a sigle liear equatio, subject to a umber of costraits o the variables. Whe there is oly oe equatio i two ukows the problem ca be solved graphically i the plae, but for more ukows the problem requires a theoretical solutio. While liear programmig problems may appear i may forms, it is preferable to covert them to a stadard form, which ca the be solved without loss of geerality. Model 53 (The Liear Programmig Problem) Miimise the equatio c 1 x 1 + c 2 x c x subject to the costraits a 11 x 1 + a 12 x a 1 x b 1 a 21 x 1 + a 22 x a 2 x b 2 a m1 x 1 + a m2 x a m x b m. where is oe of, =, ad. Furthermore, x i 0 i This ca be expressed i matrix form, as miimise f (x) = c T x subject to the costraits Ax = b. However, clearly equality o the latter equatio is a problem sice some costraits ivolve iequality. Also, it is ot ecessarily the case that all variable will be o-egative, as is aother requiremet.

22 22 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS To overcome these problems, slack variables are itroduced. This works as follows, g i (x) b i g i (x) + x +1 = b i for some x +1 0 g i (x) b i g i (x) x +1 = b i for some x +1 0 Similarly, the o-egativity criterio ca be solved by re-defiig x k = x +1 x +2 1 k Havig stadardised the problem i this way, a umber of ew variables will have bee itroduced. It is ofte coveiet to re-defie the subscripts so they ru from 1 to some ew value of. Defiitio 54 For a liear programmig problem i stadard form, 1. f (x) = c T x is called the objective fuctio. 2. Every x R satisfyig Ax = b ad x i 0 i is called a feasible solutio. 3. The set of feasible solutios is deoted by M, ad M opt is the set of all optimal solutios. Hece M opt = {x M f (x) f (z) z M} 4. a i is the colum vector whose etries are the ith row of the matrix A. 5. A j is the colum vector whose etries are the jth colum of the matrix A. Notice that maximisatio ca be performed by miimisig the egatio of the objective fuctio. Recall that A is a m matrix. From the above defiitios it is evidet that Ax = b x j A j = b j=1 a ij x j = b j i ai T x = b i i j=1 Fidig A Solutio There is othig to say that Ax = b must have oly oe solutio if m < the there are ecessarily may solutios. However, it may be assumed that the rak of A is m sice liearly depedet rows could be removed. From this it follows ) that there must exist a o-sigular m m submatrix of A, B say. Hece B = (A j1 A j2... A jm for some 1 j m. Defiitio 55 Where B is a o-sigular submatrix of A, B = (A j1 A j2... A jm ) the, 1. The idices j 1, j 2,..., j m are defied by the fuctio B(i) = j i for 1 i m. } 2. The set B = {A j1, A j2,..., A jm is called a basis for A, as the vectors spa the colum space of A. 3. The idices of the basis, i.e. j 1, j 2,..., j m are referred to as basic. ( 4. x B = x j1 x j2... ) T x jm is the compoet of x with basic idices. ( 5. c B = c j1 c j2... ) T c jm is the compoet of c with basic idices. Usig this otatio, it is evidet that b = Ax = A j x j = A j x j + A j x j = Bx B + A j x j j=1 A B A/ B A/ B

23 15.2. LINEAR PROGRAMMING 23 Hece a basic solutio x satisfies Bx B = b ad x j = 0 for A j / B. If also the etries of x B are o-egative the it is a basic feasible solutio, ad if x B(i) > 0 the it is o-degeerate. I summary A basic feasible solutio has zero etries bar for the m etries i the base positios. If ay of these m etries are zero, the the basic feasible solutio is degeerate. Havig foud a solutio, its optimality ow eeds to be assessed. Theorem 56 (Optimality Criterio) Let z j = c T B B 1 A j ad let z = solutio ad c j z j 0 j the x M opt. Proof. Let y be a feasible solutio, so Ay = b ad y j 0 j. By hypothesis c j z j 0 j, so c z 0. Hece (c z)y T 0 ad so cy T zy T. This gives f (y) = c T y z T y = = z j y j j=1 cb T B 1 A j y j j=1 = cb T B 1 A j y j j=1 = c T B B 1 b = c T B x B = c T x = f (x) ( z 1 z 2... z ) T. If x is a basic feasible So f (y) f (x), ad sice y was ay feasible solutio, x must be optimal. Note that this optimality criterio is a sufficiet but ot a ecessary coditio. It is therefore possible for optimal solutios to exist which do ot obey this criterio. Corollary 57 If A j B the c j z j = 0. Proof. By defiitio, B 1 A B(i) = e i, oe of the stadard ordered basis vectors. Hece z j = c T B B 1 A j = (c B(1) c B(2)... c B(m) ) e i = c B(i) = c j Hece the result. Fidig A Better Solutio Fidig a solutio ad assessig its viability is all very well, but if the solutio is ot optimal the aother solutio must be foud ad tried. Ideally a ew solutio will be better tha the old oe i that the value of the objective fuctio will be earer its miimum. From this poit it will be ecessary to distiguish betwee bases, ad so the otatio B 1, B 2,... will be used. I the case of the first basis, m Bx B1 = A B1 (i)x B1 (i) = b (58) i=1

24 24 CHAPTER 15. MSM2M2 MANAGEMENT MATHEMATICS where A B1 (i) represets the ith colum of B. The matrix B 1 A has a importat role, so for coveiece its colums will be deoted by A j = (a 1j a 2j... a mj) T. Hece A = B 1 A A = BA gives BA j = A j i=1 A B1 (i)a ij = A j where A B1 (i) represets the ith colum of B. This matrix will become A i the ext iteratio. Now, sice this is leadig to fidig aother solutio, it is assumed that the optimality criterio has bee violated. Say this is so for colum l, so c l z l < 0, 1 l. I particular the above ow gives A B1 (i)a il = A l (59) i=1 Now cosider a parameter θ, ad fid the differece betwee the solutio as it stads, ad θ times the offedig colum i.e. (58) θ(59) givig m ( θa l + A B1 (i) xb1 (i) θa ) il = b i=1 The expressio o the left had side represets a ew solutio, with the existig basic solutios beig adjusted, aother (l) beig added, ad the others remaiig zero. The ew solutio is give by x B1 (i) x B1 (i) θa il 1 i m x = x l θ x j 0 j = l A j / B However, this is t a ew solutio yet: a value for θ has to be foud. If θ = 0 the x = x, the old solutio. I order to maitai feasibility, it is clear that θ 0, so start from 0 ad icrease to make θ as big as possible. The poit whe θ will be maximal is determied by the top lie of (60) whe for some k, x B1 (k) θa kl = 0. Hece θ = x B 1 (k) a kl ( ) xb1 (i) = mi a il 1 i m a il > 0 Notice that sice x B1 (k) θa kl = 0, the vector A k drops out of the basis ad is replaced by A l. However, a problem still remais: i order for B 1 to exist the ew basis vectors must be liearly idepedet, sice it is used to defie x. Theorem 61 The ew basis, B 2 = ( B \ { A B(k) }) { AB(l) } is liearly idepedet. (60) It ca be show that c T x = c T x + θ(c l z l ) from which it is obvious that the value of the objective fuctio has bee reduced. A ew ad better basic feasible solutio has ow bee foud, but there remai two issues to be addressed. Firstly, if θ = 0 the x does ot chage. The fact that θ = 0 at all meas that x was degeerate. However, the basis is still chaged by replacig oe zero by aother. It would be more correct to say basic feasible solutio rather tha just solutio.

25 15.2. LINEAR PROGRAMMING 25 Secodly, there is a problem if a il 0 i. I this case θ ad this gives c T x = c T x + θ(c l z l ) egative because colum l violates the optimality criterio. It may be possible to choose some sufficietly large value of θ so that the objective fuctio has smaller value. O the other had the liear programmig problem may have o solutio. Solvig The Liear Programmig Problem At this poit eough theory has bee covered to actually solve a liear programmig problem. This theory ca be eatly packaged ito the Simplex Method. However, it is ecessary to make a few assumptios, which will be elimiated i the followig sectio. Give a liear programmig problem i the stadard form f (x) = c T x mi Ax = b x j 0 j assume that A cotais the m m idetity matrix as a submatrix, ad b j 0 j. These assumptios may seem trivial to achieve, but meetig both coditios at the same time is rather more difficult tha meetig just oe of them. The value of the assumptio is that B = I so that A j = B 1 A j = A j. Also, x B = b. The idetity matrix i A is there because of the slack variables. If Ax > b or Ax < b the oe slack variable must be added for each of the m costraits. This produces a costrait equatio of the form Ax + Iw = b. These slack variables make o appearace i the objective fuctio, so it follows that c B(i) = 0 i. This meas that z = 0 so for the basis subscripts, the relative costs must be 0.The assorted iformatio is usually displayed i a Simplex Table as show i Table c 1 c 2... c B(1)... c B(m)... c z 0 c 1 z 1 c 2 z c z c B(1). c B(m) x B A 1 A 2... idetity matrix... A Table 9: Geeral form of a Simplex Table. Note that it is usual to omit the first row ad first colum. Note also that the quatity z 0 is calculated as c 0 z 0 = 0 cb T x, this is the same as f (x). From the Simplex Table it is a trivial matter to check the optimality criterio, ad whether there is i fact o solutio. Assumig the optimality criterio fails but there is a solutio, a ew solutio ca be costructed usig the methods described above. But there is a better way. I the equatio BA = A, B is the matrix whose colums are the vectors of the basis. The colums of A are therefore the coefficiets which produce the correspodig colums of A from the basis vectors. I order to keep the stadard form of the Simplex table, A l, the vector replacig A k i the basis is required be the kth stadard basis vector. The basis therefore eeds to be trasformed. Recall that θ = x ( ) B(k) xi a = mi kl a il 1 i m ad the chose x kl is called the pivot. I order to trasform the basis,