New Turnpike Theorems for the Unbounded Knapsack Problem

Transcription

1 New Turnpke Theorems for the Unbounded Knapsack Problem Png Hed Huang¹* and Thomas L. Morn² ¹Krannert School of Management, ²School of Industral Engneerng Purdue Unersty, West Lafayette, Indana Abstract: We deelop sharp bounds on turnpke theorems for the unbounded knapsack problem. Turnpke theorems specfy when t s optmal to load at least one unt of the best tem (.e., the one wth the hghest bang-for-buck rato) and, thus can be used for problem preprocessng. The successe applcaton of the turnpke theorems can drastcally reduce the sze of the knapsack problems to be soled. Two of our theorems subsume known results as specal cases. The thrd one s an entrely dfferent result. We show that all three theorems specfy sharp bounds n the sense that no smaller bounds can be found under the assumed condtons. We also proe that two of the bounds can be obtaned n constant tme. Computatonal results on randomly generated problems demonstrate the effecteness of the turnpke theorems both n terms of how often they can be appled and the resultng reducton n the sze of the knapsack problems. Keywords: Integer Programmng, Knapsack Problem, Number Theory, Preprocessng, Turnpke Theorems * Png Hed Huang, huang74@mgmt.purdue.edu, Krannert School of Management, Purdue Unersty, 403 W. State St., West Lafayette, IN 47907, USA. Tel Fax

2 . Introducton The knapsack problem s one of the most celebrated problems n operatons research. The unbounded knapsack problem can be stated as follows. Gen a knapsack (or backpack) wth known weght capacty and an unlmted supply of tems, each of whch has known unt alue and unt weght, how can one pack the knapsack wth ntegral amounts of the tems so as to maxmze the alue of the load carred? It s well known that ths problem s NP-hard (Garey and Johnson, 990). There are a host of algorthms for the knapsack problem and ts arous arants (Kellerer et al., 004; Martello and Toth, 990; Morn and Marsten, 976; Zukerman et al., 00). Howeer, t s oftentmes possble to drastcally reduce the sze of the knapsack problem to be soled een before applyng one of these algorthms. Ths can be ewed as a form of problem pre-processng (Nemhauser, 007). Specfcally, one such way of cuttng down the computatonal requrements of solng knapsack problems wth large, but bounded weght capactes s to employ turnpke theorems (Garfnkel and Nemhauser, 97; Glmore and Gomory, 966; Hu, 969; Shapro, 970). If the tems are ndexed accordng to the non-ncreasng order of the bang-for-buck ratos of ther unt alues to unt weghts, then for large enough weght capactes (.e., long enough trps) t can be shown that t s optmal to load at least one unt of best tem (to take the trp on the turnpke, as t were), whch has the hghest alue-to-weght rato. Turnpke theorems specfy lower bounds on what consttutes a large enough weght capacty (rght-hand-sde) and ther successe applcaton can drastcally reduce the rght-hand-sdes and the resultng computatonal requrements. For nstance, a knapsack problem wth seeral hundred tems and a carryng capacty of 00,000 can be reduced to a problem wth a carryng capacty of only 40 (see Example 4.4). The prmary goal of the present paper s to prode sharper bounds and, thus, more effecte turnpke theorems for the unbounded knapsack problem. The plan of our paper s as follows. In secton, we ntroduce notatons and then deelop the frst turnpke theorem, whch s shown to subsume a known result attrbuted to Hu (969) by Garfnkel and Nemhauser (97). Moreoer, we proe that our bound can be obtaned n constant tme. We also proe that t s a sharp bound n the sense that n general, no smaller lower bound can be found under the assumed condtons. In secton 3, we deelop the second turnpke theorem and proe t also subsumes another known result from Garfnkel and Nemhauser (97) as a corollary. The thrd turnpke theorem, whch s an entrely dfferent

3 result, s gen n secton 4. Number theoretc results are used to proe our new turnpke theorems, and examples are gen to show that our theorems are stronger than the known results. We also show that all three of our turnpke theorems specfy sharp bounds n the sense that no smaller bounds can be found under the assumed condtons and proe that two of the bounds can be obtaned n constant tme. Computatonal experments on randomly generated problems clearly demonstrate the effecteness of the turnpke theorems not only n terms of when they can be appled but also n the resultng reducton n the sze of the knapsack problems. Indeed, the three turnpke theorems yelded aerage reductons of the rght-hand-sdes of 99.97%, 98.0% and 99.93%, respectely. The paper concludes wth a dscusson n secton 5.. Prelmnares and the Frst Turnpke Theorem The unbounded knapsack problem can be stated as follows: gen an unlmted number of n tems, each of whch has unt weght a j and unt alue c j, how can one fll a knapsack wth weght capacty b n order to maxmze the alue of the load carred? More formally, one wants to fnd non-negate ntegers { x, x,, x n } n order to maxmze n j= cx j j subject to n j= ax j j b, and x 0, nteger, j where, as usual, we assume that all data are ntegral. As noted n the ntroducton, one way of reducng the computatonal requrements of solng knapsack problems wth large, but bounded, rght-hand-sde weght capactes s to employ turnpke theorems. Other dstnctly dfferent ways nclude domnance results and perodcty propertes (Kellerer et al., 004). If we ndex the tems accordng to the non-ncreasng order of ther bang-for-buck ratos, j = c j a j, so that 3, then for large enough b we can proe that t s optmal to load at least one unt of best tem whch has the hghest alue-to-weght rato, nto the knapsack. Turnpke theorems specfy a lower bound on what consttutes a large enough b. As noted n the 3

4 Introducton, successe applcaton of such turnpke theorems can oftentmes drastcally reduce the rght-hand-sde b, and the resultng computatonal requrements. In constructng an optmal soluton of a gen knapsack problem, we (nductely) want to decde f at least one unt of tem (recall that by assumpton, tem has the hghest alue-toweght rato ) should be packed nto the knapsack. Certanly f t s optmal to pack one unt of tem, then there s a reducton of the (remanng) capacty, and thus, the problem s easer to sole. Before statng and prong our frst theorem, we wll state and proe a smple Lemma. Lemma. Let q be a poste nteger and s be a real number where s q, then we hae q s < s q +, where s = largest nteger s. Proof. Let 0 t <. Thus, we hae s = y, t follows from the assumpton s q that y q. Let s y = t wth q s q + = q y q + + q t q + = ( ) q y + + q t q + y = ( y ) + q + qt q + and snce y q, we hae q ( y ) + q + qt q + = q( t) y q + and by defnton t <, we hae < y = s, or q s < s, whch completes the proof of the Lemma.. q + Note that the nteral [0, ) can be wrtten as the dsjont unon of nterals of the form q q,,.e., q q+ [0, ) = q= q q, q q+ 4

5 and any real number r wth 0 r < falls precsely nto one of the subnterals aboe. We hae the followng theorem, c c Theorem. Assume that > (.e., > ), and let the poste nteger q be unquely a a determned by the followng relaton q q q q <. + Then there s a weght hi = qa such that f b hi, t s optmal to pack at least one unt of tem nto the knapsack. c c c3 Proof. We hae > (.e., > 3 ). Clearly, f no unts of tem are a a a 3 q loaded, then the total alue of the objecte functon s z b whch leads to z < b q +, due to the assumpton that q <. Therefore, we hae + q q z < b q +. (.) We want to show that any soluton wthout tem loaded can not be optmal. We clam that q b b < c q+ a. (.) Note that the rght-hand-sde of nequalty (.) s the alue of fllng the knapsack wth as many unts of tem as possble (whch s certanly allowed). The aboe clam wll establsh our theorem. Now let us proe the clam. By defnton, j = cj aj, thus =. c a 5

6 Therefore, nequalty (.) s equalent to the followng q b b < q+ a a, and the aboe nequalty follows at once from Lemma. by takng n the assumpton, so we hae nequalty (.) holds. Combnng () and (), we hae b s = (Note that b qa a b q a. Thus, Lemma. can be employed). Therefore, b z < c a The rght-hand-sde s equal to the objecte functon alue of a way of loadng tem. So the orgnal soluton can not be optmal. Therefore, an optmal soluton ncludes at least one unt of tem. Notce that Theorem. subsumes a known result, attrbuted to Hu (969) by Garfnkel and Nemhauser (97) as a specal case that s stated n the followng corollary. c c c Corollary.3 Let us assume that > (.e., > ), then there s a weght h I ' = a a such that f b h I ', t s optmal to pack at least one unt of tem nto the knapsack. Proof. Let us use the notaton of the precedng theorem. Note that there s a unque poste nteger q that s the maxmal possble and satsfes the followng nequalty, q q q c qc = qa q c Therefore, b mples that b qa. Thus, the corollary follows from Theorem.. Remark.4: It follows from the proof of the corollary that c h I ' = qa = h. I 6

7 Here s an nterestng ssue, as ponted out by one of the anonymous referees, of fndng the number q n the statement of the precedng theorem. If we consder the sequence 3 0 < < < < and try to ft 3 4 < nto the segment q q,,.e., we compare q q+ n c a ac a c wth one by one for n< ac (snce = = ), then the process n + c a a c a c would requre a lnear tme n the sze of ac, and we would end up n a stuaton wth a tghter bound but a longer tme (to fnd the bound). Howeer, the approprate q can be calculated n at most two steps f we use the followng lemma and, thus, our bound can be found n constant tme nstead of lnear tme. Lemma.5 Let 0< r < be any real number. There exsts a unque expresson such that r = n + r wth n beng an nteger and 0 r <. We hae poste nteger q s determned by q q r,, where the q q+ () f n = : we must hae 0< r <. Thus, there exsts a unque expresson such that r = n + r wth n beng an nteger and 0 r <, then q= n + ; () f n = and r = 0, then q = ; () all other cases, q =. Proof. Clearly, r > can be unquely expressed as n r r = + wth n (nteger) and 0 r <. Now consder case () n =. If r = 0, then assumpton. Therefore, we hae 0< r <. Agan, n r r = + wth r = n + r = contradctng to our > can be unquely expressed as r n (nteger) and 0 r <. Ths mples n < n +, so we r 7

8 hae < n r + n n. Gong back to r wth n =, we hae r = = + r + n + n n + n n + r = < =. Thus, we obtan r < + r + n + n + n +. Therefore, q= n +. n + and Consder case () n = and r = 0. We hae q q r = =, r,, thus q n r =. + q q+ Now consder case (), f n = and r > 0, then r =, thus n + r < q= ; f n >, then r =, we also obtan n + r < q=. Remark.6: In Theorem., 0< <. Thus, Lemma.5 can be appled to fnd q. Snce c a ac = =, we can smply use long dson (or contnuous fracton) to calculate n and c a a c n (f r 0 ). The exact solutons for n and n satsfy ac = n( ac ) + w= ac ( n+ r) and ac nw w wn r = + = ( + ), where 0 w < ac and 0 w < w. Example.7: Our theorem s stronger than the corollary. For consder an example wth = 00 and = 99. c 00 Hence h = I ' a a ( a ) = = 0, 99 whle 0 = <, we hae q =, thus h 00. I = qa = a 00 To achee an optmal soluton, Corollary.3 states that f b a, then we may pack 0 at least one unt of tem whereas Theorem. states that we may do so wth just b a. Therefore, for arbtrary b, we may fll the knapsack by a sutable number of unts of tem untl 8

9 00 the remanng capacty s less than a (by our theorem) or less than a (by the corollary). 0 Certanly t s more adantageous to use our theorem. In fact, hi = qa s a sharp bound n the sense that no smaller bound can be found under the condtons specfed n Theorem. as the followng proposton shows. Proposton.8 Gen the assumptons of Theorems., there are examples such that f b= h I, then t s not optmal to pack tem nto the knapsack. Proof. Consder three types of tems wth = q+, = q and 3 =, where q s any poste nteger. Ther correspondng unt weghts are a > 4q, a = qa and a 3 =. It s q easy to erfy that > and q < q q +. Hence, we hae hi = qa as shown n Theorem.. Now let b= qa. Apparently we can load one unt of tem nto the knapsack, and the objecte functon alue s z = a = ( qa )(q ) = aq aq q+. * We clam that for any other loadng, the resultng alue s strctly less than * z. Suppose we load one unt of tem, then the remanng capacty s b a = ( qa ) a < qa = a. Thus, no tem can be loaded (.e., only tem and tem 3 may be consdered). We assume that altogether there are k unts of tem loaded (t must hae k < q snce b= qa ), and the rest s flled up by a sutable number of unts of tem 3. The resultng alue s z = ka + ( b ka) 3 = ka(q + ) + ( qa ka) = kaq + ka+ qa, and t reaches the maxmum when k = q. Therefore, we hae z ( q ) aq + ( q ) a+ qa = aq aq ( a+ ) * < aq aq (4q+ ) < aq aq q+ = z Hence, t s not optmal to pack tem nto the knapsack. Thus, one can not mproe on h. I 9

10 Computatonal experments were conducted as suggested by one of the referees, n order to ascertan just how effecte our turnpke theorems are. Specfcally, how often could they be appled and what was the aerage percentage reducton n the sze of the resultng knapsack problem. Toward that end we used MATLAB 7.0 (The MathWorks Inc.) to randomly generate both c and a for =,,500 from a unform dstrbuton of ntegers from to,000. Followng Gabrel and Mnoux (00) the knapsack capacty, b, was randomly generated from a unform dstrbuton of ntegers where the range was a, a. After runnng dozens = = 0,000 randomly generated nstances, the typcal results are as follows: we ran nto 4 out of 0,000 cases where =,.e., the suffcent condtons of Theorem. were not met and among all cases, b h. Therefore, our turnpke theorem could be appled n 99.96% of the I b h cases. The aerage reducton n the rght-hand-sde, I, = 99.97%. b 3. The Second Turnpke Theorem Let S be a feasble soluton to the knapsack problem. Assume S = { j, j,..., j m } conssts of m unts of arous tems, each s denoted by j t. The alues of jt s don t hae to be unque,.e., the same type of tem may hae more than one occurrence n S. For nstance, S = {,,,,3} ndcates{tem, tem, tem, tem, tem 3}. Now let S = { j, j,..., j} conssts of frst unts of S, where m. Thus, S denotes a partal soluton wth total weght A = a. Fnally, let t= j t r be the resdue of A(mod a ), where 0 r < a and m( Hardy and Wrght, 938). Frst we hae the obous result. Proposton 3. If b 0(mod a ),.e., b s dsble by a, then t s optmal to pack b a unts of tem and no unts of tem,, tem n. Remark 3.: If ' S s any partal soluton of an optmal soluton, and the weght of ' S 0(mod a ), then ' S conssts entrely of unts of tem. 0

11 Next for the followng two lemmas, assume the contrary that S s an optmal soluton wth no tem ncluded. Also, recall that r A(mod a ). c c c3 Lemma 3.3 Let > (.e., > 3 ). If S s an optmal soluton wth no a a a 3 tem ncluded, then r 0, for =,, m. Proof. If r = 0,.e., A 0(mod a ), then t mples that the total weght of S s dsble by a. Thus strctly (as S can be replaced entrely by unts of tem, and the total alue of S wll ncrease S was a partal soluton of S). Ths olates the optmalty of the orgnal soluton and completes the proof. c c c3 Lemma 3.4 Let > (.e., > 3 ). If S s an optmal soluton wth no a a a 3 tem ncluded, then r r for k. k Proof. WLOG assume that k >. If r = r k, then Ak A 0(mod a ). Ths mples that the weght dfference between partal solutons partal soluton S k and S s dsble by a. Thus, the resultng S k \ S can be replaced by a sutable number of unts of tem (see Remark 3.), and the total alue of S wll ncrease strctly. Ths olates the optmalty of the orgnal soluton and completes the proof. Theorem 3.5 Let L max j { a j } =. Then there s a weght hii = ( a ) L such that f >, b> h II, then t s optmal to pack at least one unt of tem nto the knapsack. Proof. We hae > 3. Assume the contrary that there s an optmal soluton S wthout tem ncluded. There are two cases and seeral sub-cases. Case : Suppose that a > m, where m s the number of unts n S. We hae A ml. Thus, the remanng weght capacty b = b Am > ( a ) L ml L. Therefore, we may pack at least one more unt from tem,, tem n nto the knapsack. Thus, the orgnal soluton can not be optmal. Case : Now consder a m, there are two sub-cases: a < m and a = m. m

12 Sub-case : a < m. Recall that r A(mod a ), =,, m. There are m many r s. Furthermore, t follows from two precedng lemmas that these r s can not be zero and should all be dstnct. Thus, we conclude that there exst m dstnct non-zero resdues, where m> a. Snce there are at most a non-zero possbltes n a complete resdue system (mod a ) (Serre, 973), ths s, therefore, an absurd concluson. Sub-case : a = m. The remanng weght capacty b = b Am > ( a ) L ml= 0. We thus hae b (snce b > 0 and b s an nteger). Furthermore we hae two possbltes: b a and b < a. Sub-sub-case : b a. Then we would be able to add at least one unt of tem and thus, the orgnal soluton can not be optmal. a b a Sub-sub-case : b < a. Then 0< a b a, ths s the same as <. Thus, a b s a resdue of a complete resdue system (mod a ). On the other hand, we hae shown that there exst m dstnct non-zero resdues, here m= a whch s the maxmal non-zero possbltes n the complete resdue system (mod a ). Therefore, a b must be one of these r s. Thus, there exsts a k such that rk = a b, whch means r b a k + 0 (mod ). Ths says that the sum of k by a. So f we combne the partal soluton S s weght and the remanng weght s dsble S k and the remanng capacty b together, we could replace t entrely wth unts of tem. Hence the total alue of S wll ncrease strctly and thus, the orgnal soluton can not be optmal. The next corollary follows from Theorem 3.5. Corollary 3.6 Let L' max j { a j } =. Then there s a weght h II ' = ( a ) L' such that f >, b> h II ', then t s optmal to pack at least one unt of tem nto the knapsack. Proof. Follows at once from Theorem 3.5 snce L' L. The followng corollary s a known result. The reader s referred to problem 5 on p.47 of Garfnkel and Nemhauser (97).

13 Corollary 3.7 Let L' max j { a j } =. Then there s a weght h II '' = a ( ' ) L + such that f 3 >, b> h II '', then t s optmal to pack at least one unt of tem nto the knapsack. Proof. Follows at once from Theorem 3.5 snce a > a and L' + > L' L. Example 3.8: Our theorem s stronger than the corollares. Consder an example wth a = 0, a = and a3 =. Also, > 3. Then L= max j { aj} =, and L j { aj} ' = max = 0. Corollary 3.7 states that f b > 0 then t s optmal for us to pack at least one unt of tem nto the knapsack. Corollary 3.6 states that we may do so f b > 90. Howeer, Theorem 3.5 states that we should nclude tem to achee optmal wth just b > 8. Certanly t s adantageous to use our theorem. Also notce that nether corollary s applcable when b = 89. Usng Theorem 3.5, one mmedately gets an obous optmal soluton by loadng 8 unts of tem, 4 unts of tem and unt of tem 3. We may further show that hii = ( a ) L s a sharp bound n the sense that no smaller bound can be found under the condtons specfed n Theorem 3.5. Proposton 3.9 Gen assumptons as n Theorems 3.5. There are examples such that f b= h II, then t s not optmal to pack tem nto the knapsack. Proof. Consder tems wth a = a 3 = =, then L j { aj} = max =. If b= hii, then we hae b= a. Clearly tem can not be ncluded n any optmal soluton, and ths completes our proof. Agan, we performed computatonal experments n order to determne just how effecte our second turnpke theorem was. MATLAB 7.0 was used to randomly generate ntegral alues for c and a, =,,500 from a unform dstrbuton of ntegers from to,000 and, followng Gabrel and Mnoux (00), the knapsack capacty, b, was chosen to be an nteger randomly generated n the range a, a 3 = 3. After computng dozens 0,000 randomly = generated nstances, we ran nto 5 out of 0,000 cases where =,.e., the suffcent

14 condtons of Theorem 3.5 were not met and among all cases, b> h. Therefore, our theorem II b h could be appled n 99.95% of the cases, and the aerage reducton II = 98.0%. b 4. The Thrd Turnpke Theorem In Theorem., we examned the rato of bang-for-buck ratos. Explorng further a along the path, t may also be adantageous to examne the rato of weghts a. When a a <, we hae the followng new theorem. Theorem 4. Assume that > and a > a. Let the poste nteger q be unquely determned by the followng relaton q q q q <. + a Let k = q +. Then there s a weght h III = ka such that f a at least one unt of tem nto the knapsack. b h III, t s optmal to pack Proof. We hae > 3... Assume the contrary that there s an optmal soluton wth no tem ncluded and wth objecte functon alue z. Then, clearly z b. It follows from our condton that b ka 0. Now let y be the resdue of ( b ka )(mod a ) Then We clam that b= ka + pa + y for some non-negate nteger p. Thus, we hae z b = ( ka + pa + y) b = ( ka + pa + y) < ka + pa ( = kc + pc ) (4.) In other words, another packng wth k unts of tem and p unts of tem produces a hgher objecte functon alue than z. Ths would olate the optmalty of the orgnal soluton and complete the proof. 4

15 Now let us proe our clam. Notce that nequalty (4.) can be smplfed to ( ka + y) < ka (4.) Snce y s a resdue (mod a ), we know that y< a, so we may replace nequalty (4.) by the followng stronger nequalty ( ka + a ) < ka (4.3) We shall proe the aboe nequalty (4.3). It follows from our assumpton that q q+ < > = + q+ q q a a a k = q + q< k > a a q ka The aboe two relatons mply a > + ka > ( ka + a ) ka Ths establshes nequalty (4.3). Hence we may deduce nequaltes (4.) and (4.), and our proof s completed. Remark 4.: Recall that q can be determned n constant tme by Lemma.5 and Remark.6. a Now that k = q +, so k can also be obtaned n constant tme. a Remark 4.3: Snce a < a, t s clear that k q. Example 4.4: Theorem 4. s stronger than Theorem.. wheneer a > a and k < q. Consder an example wth = 00, = 9, for =,,99, and + a = 0, a = 0, a : poste ntegers, for = 3,,00. 5

16 9 9 Usng Lemma.5, r = = =, thus n = and r = Snce n =, we go one more step wth r = 9 =, thus n = 0 and r = Therefore, q= n + =, we hae h= I qa = 30. a 0 Let k = q + = + = a 0, then h III = ka = 0. Theorem. states that f b 30, then t s optmal to pack at least one unt of tem whereas Theorem 4. states that we may do so wth just b 0. Certanly t s adantageous to use Theorem 4. wheneer a > a and k < q. Specfcally, consder a knapsack problem wth seeral hundred tems and a carryng capacty of 00,000. Snce 00,000 = 833 x , usng Theorem 4. we may go ahead wth 833 unts of tem. The remanng capacty s thus reduced to a merely 40! Now t s obous that an optmal soluton s loadng 833 unts of tem and 4 unts of tem. Here we get an optmal soluton wthout nokng any algorthm. The followng proposton llustrates that hiii = ka s a sharp bound under condtons gen n Theorem 4.. Proposton 4.5 Gen assumptons as n Theorems 4.. There are examples such that f b= h III, then t s not optmal to pack tem nto the knapsack. Proof. Consder tems wth = q+, = q, where q s any poste nteger. We also q q assume that a > aq. Clearly, > and a > a. It s easy to erfy that <. q q + a aq Hence, we hae hiii = ka = a (here, k = q + = snce a a < ). If b= hiii = a, then tem can not be ncluded n any optmal soluton, and ths completes our proof. We also performed computatonal experments on our thrd turnpke theorem. MATLAB 7.0 was used to randomly generate ntegral alues of c and a for =,,500 from a unform 6

17 dstrbuton of ntegers from to,000 and, followng Gabrel and Mnoux (00), the knapsack capacty, b, was chosen to be an nteger randomly generated n the range a, a. = = After computng dozens 0,000 randomly generated nstances, we ran nto 8,667 out of 0,000 cases where ether = or a a,.e., the suffcent condtons of Theorem 4. were not met but among all cases, b> h. Therefore, although our theorem could only be appled n III hiii 3.33% of the cases, we obsered that = 65.89% n aerage. Ths mples that f the h I suffcent condtons of Theorem 4. are met, then usng Theorem 4. may achee further reducton than usng Theorem.. Out of those cases where the thrd turnpke theorem could be b h appled, the aerage reducton by Theorem 4. was III = 99.93% whle the aerage b b h reducton by Theorem. was I = 99.88%. b 5. Dscusson Ths paper deeloped sharp bounds under arous condtons on turnpke theorems for the unbounded knapsack problem. Turnpke theorems specfy when t s optmal to load at least one unt of the best tem (.e., the one wth the hghest bang-for-buck rato) and ther successe applcaton can drastcally reduce the sze of the knapsack problems. We also show that all three of our turnpke theorems specfy sharp bounds n the sense that no smaller bounds can be found under the assumed condtons and proe that two of our bounds can be obtaned n constant tme. Fnally, computatonal results on randomly generated problems clearly demonstrated the effecteness of our turnpke theorems. Indeed, the three turnpke theorems yelded aerage reductons n the knapsack capactes b h of 99.97%, 98.0% and 99.93%, respectely. b There are at least three extensons that we ntend to pursue. The frst noles consderng both the frst as well as the second best tems. The resultng turnpke theorems could then be based upon results from the famous Frobenus problem (Ramírez-Alfonsín, 005). The second extenson was suggested by one of the referees and noles conductng a comprehense computatonal study to see f the turnpke theorems actually reduce the total computng tme 7

18 when compared wth solng the knapsack problems wthout them. The thrd topc noles extendng the turnpke theorems to the multdmensonal knapsack problem. Acknowledgments: It s ndeed a pleasure to thank Kwe Tang and the two anonymous referees for ther numerous nsghtful comments, correctons and suggestons that sgnfcantly mproed the paper. References -Gabrel V, Mnoux M. A Scheme for Exact Separaton of Extended Coer Inequaltes and Applcaton to Multdmensonal Knapsack Problems. Operatons Research Letters 00; 30; Garey M, Johnson D. Computers and Intractablty: A Gude to the Theory of NP Completeness. W. H. Freeman and Co.: New York; Garfnkel R, Nemhauser G. Integer Programmng. John Wley & Sons: New York; 97. -Glmore P, Gomory R. The Theory and Computaton of Knapsack Functons. Operatons Research 966; 4; Kellerer H, Pferschy U, Psnger D. Knapsack Problems. Sprnger: Berln; Hardy G, Wrght E. An Introducton to the Theory of Numbers. Oxford Unersty Press, New York; Hu T. Integer Programmng and Network Flows. Addson-Wesley: Readng, MA; Martello S, Toth P. Knapsack Problems: Algorthms and Computer Implementaton. John Wley & Sons: New York; Morn T, Marsten R. An Algorthm for Nonlnear Knapsack Problems. Management Scence 976; ; Nemhauser G. Need and Potental for Real-Tme Mxed-Integer Programmng. Operatons Research Management Scence Today 007; 34; -. -Ramírez-Alfonsín J. The Dophantne Frobenus Problem. Oxford Unersty Press; Serre J-P. A Course n Arthmetc. Sprnger: Berln;

19 -Shapro J. Turnpke Theorem for Integer Programmng Problem. Operatons Research 970; 8; Zukerman M, Ja L, Neame T, Woegnger G. A Polynomal Solable Specal Case of the Unbounded Knapsack Problem. Operatons Research Letters 00; 9;