Relative Improvement by Alternative Solutions for Classes of Simple Shortest Path Problems with Uncertain Data Part II: Strings of Pearls G n,r with Biased Perturbations Jörg Sameith Graduiertenkolleg Approximation und algorithmische Verfahren Fakultät für Mathematik und Informatik Friedrich-Schiller-Universität Jena 0770 Jena Germany August 5, 005 Abstract For classes of simple shortest path problems we analyze the following situation: We are given a graph G = {V, E, w} and are allowed to provide two solutions A and B which are paths from a given starting node to a given target node. Then we learn about randomly perturbed edge lengths ŵ and are allowed to choose the solution from our pair {A, B} that has the better value with respect to these new lengths ŵ. We analyze how to choose two solutions A and B such that E[min e A ŵe, e B ŵe] is minimal. sameith@mathematik.uni-jena.de
Contents Introduction. The Models..................................... The Optimal Alternative 3. The k - Model................................... 3. The p - Model.................................... 3 3 Optimal Pairs vs. Optimal Alternatives 8 Failed Assumption 38 5 Conclusions 39 References 0
Introduction Introduction For classes of simple shortest path problems we analyze the following situation: We are given a graph G = {V, E, w} and are allowed to provide two solutions S 0 and S which are paths from a given starting node to a given target node. Then we learn about randomly perturbed edge lengths ŵ and are allowed to choose the solution from our pair {S 0, S } that has the better value with respect to these new lengths ŵ. We analyze how to choose two solutions S 0 and S such that E[min e S 0 ŵe, e S ŵe] is minimal. The report is organized as follows. In the remaining part of this section we introduce the considered models in detail: the graph model string of pearls G n,r and two different biased perturbation models k-model, p-model. Let S 0 be the optimal solution with respect to the original weights w. In Section we analyze only the optimal choice of an additional alternative solution S for the case that S 0 is fixed. In Section 3 we relax this condition and allow two arbitrary solutions. In Section we briefly comment on an assumption which turns out to be wrong that initially motivated us to study these models. We finish with conclusions in Section 5.. The Models Definition. string of pearls G n,r Consider a weighted directed graph G n,r = {V, E, w} with a set of nodes V and a set of edges E with V = {v 0, v,..., v n }, E = {e 0,i = v i, v i, e,i = v i, v i : i =,,..., n} and a weight function w : E R + that assigns a weight to each edge see Figure. The weights are we 0,i = and we,i = + r for i =,..., n. We interpret the edge-weights as lengths. We call this type of graph a string of pearls G n,r. Furthermore we call n the problem size and the edges {e,i : i =,,..., n} we call detours. A feasible solution S is a path from starting node v 0 to the final node v n. In every node v i 0 i < n it is to decide whether to use the short edge e 0,i+ around the top or the detour e,i+ down below. The length ws of a solution S is the sum of the lengths of its edges e S: ws = e S we.
. The Models v 0 v v v 3 v n v n + r + r + r + r Figure : String of Pearls G n,r We consider two different perturbation models: Definition. k-model Consider a string of pearls G n,r. Exactly k 0 k n randomly chosen edges get lengthperturbed. Let π be a random permutation of,,..., n, then ŵ : E R is: { + δ wei,j : if πi n + j k ŵ k e i,j := we i,j : otherwise Definition.3 p-model Consider a string of pearls G n,r. Every edge independently gets length-perturbed with the probability p 0 p. Then ŵ : E R is: { + δ wei,j : with probability p, ŵ p e i,j := we i,j : otherwise The parameter δ R we call the perturbation intensity. Remarks For the following cases the choice of solutions is obvious. The initial optimal solution the solution that uses no detour remains optimal with respect to ŵ.. p = 0 or k = 0: No edge-weight gets perturbed.. p = or k = n: All edge-weights get perturbed. 3. δ r: It is not advantageous to use a detour since a non-detour edge is never longer than a detour edge.
The Optimal Alternative The Optimal Alternative Let S 0 be the solution that uses no detour the optimal solution with respect to the original weights w. Which additional solution S we have to select such that the expected minimal value E[minŵS 0, ŵs ] is minimal? Since all detours of the graph G have the same length it does not matter which detours a solution uses. Only the number of detours is important. Let S be the set of available solutions: The length of path s i is: S = {s i : 0 i n} with: s i = e,, e,,..., e,i, e 0,i+, e 0,i+,..., e 0,n. i ws i = we,t + t= n t=i+ we 0,t.. The k - Model How many detours d opt = dn, r, δ, k the alternative solution S has to use if exactly k randomly chosen edges get length perturbed? We are able to calculate for any given n, r, δ and k the optimal number of detours that the additional solution S should use exactly. We generate all n k possible cases for ŵk with exactly k perturbations. For each case we calculate the minimal values of all possible solution pairs {{S 0, S } : S 0 = s 0 and S = s i with i {,,..., n}}. Finally we calculate the average values for all pairs and observe the optimal number of detours d opt. First we give some results that show d opt for different numbers of perturbations k and perturbation intensities δ see Tables -3. All results we show are for r = 0.. We tested other values too, but in view of the dependence on k and δ the results are quite similar. δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 3 3 3 3 3 3 k 5 3 3 3 3 7 3 3 3 3 3 Table : The optimal number of detours d opt of the alternative solution S [ k-model with fixed n =, r = 0. ] 3
. The k - Model δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 3 3 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 k 5 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 7 3 3 5 5 5 8 5 5 5 5 5 5 5 5 5 5 9 3 3 3 3 3 Table : The optimal number of detours d opt of the alternative solution S [ k-model with fixed n = 5, r = 0. ] δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 3 3 5 3 3 3 5 5 5 3 3 5 5 5 5 5 k 7 3 3 5 5 5 5 5 8 9 3 3 5 5 5 5 0 3 3 3 3 3 Table 3: The optimal number of detours d opt of the alternative solution S [ k-model with fixed n =, r = 0. ] We observe that the optimal number of detours d opt is increasing in δ. On the other hand d opt behaves unexpectedly irregular in k. For small values of δ the optimal number of detours is small and identical for all k. But with increasing δ some peaks occur. The optimal number of detours jumps up and down. At first sight this seems to be an odd-even effect. For odd values of k the optimal number of detours is unimodal in k. It first increases then decreases. For even values of k the alternative should use the full number of detours, whereas for k and k + the optimal number is significantly smaller. Definition. We say that d opt = dn, r, δ, k has a peak in k k N and 0 < k < n if for a given combination of the model parameters n N, r R and δ R it holds that: dn, r, δ, k < dn, r, δ, k > dn, r, δ, k +. The minimal difference of the optimal number of detours in k to k + and k : mindn, r, δ, k dn, r, δ, k, dn, r, δ, k dn, r, δ, k + we call the height of the peak.
. The k - Model We originally expected that the optimal number of detours is unimodal in k as it seems to hold if we focus on odd values for k only. The existence of a single peak does not prove that the assumption of unimodality is wrong. Thus it is much more interesting if there are multiple peaks with a gap in between. Definition. We say that d opt = dn, r, δ, k has a gap in k k N and 0 < k < n if for a given combination of the model parameters n N, r R and δ R it holds that: dn, r, δ, k > dn, r, δ, k < dn, r, δ, k +. The minimal difference of the optimal number of detours in k to k + and k : mindn, r, δ, k dn, r, δ, k, dn, r, δ, k dn, r, δ, k + we call the height of the gap. Next we show that in most cases the shown solutions are really the only optimal solutions. There do not exist other sets of optimal solutions for the different values of k such that dn, r, δ, k can be made smooth in k by the right choice of solutions. Table shows the three best results for each value of k. Let d denote the second best number of detours and d 3 the third best number of detours. Furthermore let E k i denote the expected minimal value E[minŵ k s 0, ŵ k s i ]. d opt E k d opt d E k d E k d opt d 3 E k d 3 E k d opt 5.90 3 0.000 0.00 5 5.389 3 0.0 0.03 3 3 5.9 0.00 0.0 5 5.857 3 0.0 0.0 k 5. 3 0.000 0.008 5.370 0.008 3 0.00 7.3 3 0.00 0.0 8.9 5 0.007 0.03 9 7.00 0.00 3 0.00 Table : The three best results for the number of detours to use [ k-model with n = 5, r = 0. and δ = 0.5 ] So far peaks occurred only in even values of k. Astonishingly this doesn t hold in general. For n = we noticed the first peaks in odd values of k see columns δ = 0.8, 0.9,.0 in Table 5. The most peaks still occur in even values of k, but not only. This is no isolated case. For larger problem sizes n we also found peaks in odd values of k. If we look for values of the perturbation intensity δ for different problem sizes n that produce really high peaks, values in the range of δ nr are conspicuous. Table shows the 5
. The k - Model δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 3.0 3 3 5 7 8 9 0 3 3 3 3 3 5 7 8 9 0 0 3 3 5 5 3 3 5 5 8 9 9 0 0 3 3 0 7 3 3 9 0 0 0 9 9 0 0 8 3 3 8 9 9 3 3 8 0 0 0 0 0 0 0 0 0 3 8 8 0 k 3 8 0 0 0 0 0 0 0 3 3 9 0 3 3 3 3 0 0 0 0 0 0 3 3 3 8 9 5 3 3 9 0 0 0 0 9 3 3 3 0 7 3 3 3 3 5 5 5 7 8 3 3 3 9 3 3 3 3 5 0 3 3 3 3 3 3 3 3 3 3 Table 5: The optimal number of detours d opt of the alternative solution S [ k-model with fixed n =, r = 0. ] optimal numbers of detours for this special case δ = nr for different problem sizes n. Since the number of possible perturbations n k grows heavily in n we are not able to calculate the optimal detours for all possible values of k within acceptable time. We limited our calculations to k. n 3 5 7 8 9 0 8 0 5 30 3 3 5 7 8 9 0 5 3 5 7 8 9 0 8 0 5 30 3 3 3 5 5 7 8 0 5 8 k 3 5 7 8 9 0 8 0 5 30 5 3 5 8 9 8 5 7 8 9 0 8 0 5 30 Table : The optimal number of detours d opt of the alternative solution S [ k-model with fixed r = 0. and δ = nr ] Table shows that with δ = nr the alternative should always use the full number of detours if k is even. For odd values of k the optimal number of detours seems to be a constant part of n. Table 7 shows the quotients d opt n. For k = the optimal number of detours d opt seems to be about n, for k = 3 about 3 5 n and for k = 5 about 3.
. The k - Model n 3 5 7 8 9 0 8 0 5 30 3 5 3 7 9 5 k 3 3 3 5 7 5 8 5 9 0 7 7 5 8 8 3 5 3 5 3 5 5 3 5 7 5 8 3 0 3 9 3 7 0 8 5 30 Table 7: Quotients d opt n [ k-model with fixed r = 0. and δ = nr ].. Theoretical Analysis of the Special Case δ = nr k = First we are interested in the expected minimal value E k= i = E[minŵ k= s 0, ŵ k= s i ] if exactly one edge gets perturbed. Let the set of edges E of the graph G = {V, E, w} be divided into these four parts: A = {e 0,, e 0,,..., e 0,i } C = {e 0,i, e 0,i+,..., e 0,n } B = {e,, e,,..., e,i } D = {e,i, e,i+,..., e,n } The solution s 0 uses all edges of A and C and solution s i uses all edges of B and C. Let e be the edge that gets perturbed. Four cases are possible: ŵ k= s 0 ŵ k= s i frequency e A : min n + nr, n + ir = n + ir i n e B : min n, n + ir + + rnr = n e C : min n + nr, n + ir + nr = n + nr n i n e D : min n, n + ir = n Now we are able to give the formula for the expected minimal value E k= i : E k= i = E[minŵ k= s 0, ŵ k= s i ] = i i n + ir + n n n + n i n = i n i n + ir + n + nr n n = i r inr + n r + n n + nr + n i n n 7
. The k - Model and check the assumption dn, r, δ = nr, k = n. If the assumption is true the following condition must hold for all 0 < i n with i n. Obviously the number of detours i the alternative solution s i uses has to be a discrete number. Nevertheless we consider i R for the analysis, but have to interpret the results with care. 0 < E k= i E k= n = i r inr + n r + n = i n r + n inr n = i + n inr n = i n r 8n n r n nr + n r + n We immediately see that the right side of the inequality never gets negative. Lemma.3 For the k-model with δ = nr and arbitrary r > 0 the alternative should always use about n detours if exactly k = edge gets perturbed. If n is even the alternative has to use exactly n detours. If n is odd the alternative has to use n or n+ detours which are equally good. k = The case k = we check in a similar way. Here already ten cases can occur. The first column gives the sets where the perturbations occur. ŵ k= s 0 ŵ k= s i frequency A, A : min n + nr, n + ir = n + ir i n B, B : min n, n + ir + + rnr = n C, C : min n + nr, n + ir + nr = n + nr n i n D, D : min n, n + ir = n A, B : min n + nr, n + ir + + rnr = n + nr i n in i A, C : min n + nr, n + ir + nr = n + ir + nr n A, D : min n + nr, n + ir = n + ir B, C : min n + nr, n + ir + + rnr + nr = n + nr B, D : min n, n + ir + + rnr = n C, D : min n + nr, n + ir + nr = n + nr n i n 8
. The k - Model So the expected minimal value for k = is: E k= i = E[minŵ k= s 0, ŵ k= s i ] = i n n + ir + n i n n + nr + i n n + nr + in i n n + nr + ir + n i n n + nr = ii n + ir nn + + ii n + nr + ir nn n in i n + nr nn + n i n + nr nn + i n + nr nn =... = 3ri3 + nr ri + n r + nri + n 3 r + n 3 n n r nn We assume that solution s n is optimal. So we check whether the difference of E k= i E k= n is greater than zero for all 0 < i < n. and 0 < E k= i E k= n = 3ri3 + nr ri + n r + nri + n 3 r + n 3 n n r nn n3 + 3n 3 r n r n nn = r 3i3 i + ni n i + ni + n 3 n nn = rn i3i + i 3ni n + n nn 0 < 3i + i 3ni n + n =: fi, n At first sight we can not examine wether fi, n is strictly positive. So we check for roots: 0 = 3i + i 3ni n + n i = n ± 3n + n + }{{} C for n 3, n N Since there are no real roots for n 3 we set i = and n = 3 and see that E k= i E k= n is greater than zero. So the above inequality has to be true for n 3. = r 30 Lemma. For the k-model with δ = nr, n 3 and arbitrary r > 0 the alternative should always use n detours if exactly k = edges get perturbed. 9
. The k - Model k = 3 Again we enumerate all possible cases for exactly k = 3 perturbations. ŵ k=3 s 0 ŵ k=3 s i frequency A, A, A : min n + 3nr, n + ir = n + ir 3 i n 3 B, B, B : min n, n + ir + 3 + rnr = n C, C, C : min n + 3nr, n + ir + 3nr = n + 3nr n i 3 n 3 D, D, D : min n, n + ir = n A, A, B : min n + nr, n + ir + + rnr = n + nr + m i i n 3 B, B, A : min n + nr, n + + rnr + ir = n + nr A, A, C : min n + 3nr, n + ir + nr = n + ir + nr n i i n 3 A, A, D : min n + nr, n + ir = n + ir B, B, C : min n + nr, n + + rnr + ir + nr = n + nr B, B, D : min n, n + + rnr + ir = n C, C, A : min n + 3nr, n + ir + nr = n + ir + nr i n 3 C, C, B : min n + nr, n + ir + + rnr + nr = n + nr D, D, A : min n + nr, n + ir = n + ir D, D, B : min n, n + ir + + rnr = n C, C, D : min n + nr, n + ir + nr = n + nr n i n i n 3 D, D, C : min n + nr, n + ir + nr = n + nr i A, B, C : min n + nr, n + ir + + rnr + nr = n + nr n i n 3 A, B, D : min n + nr, n + ir + + rnr = n + nr in i A, C, D : min n + nr, n + ir + nr = n + ir + nr n 3 B, C, D : min n + nr, n + ir + + rnr + nr = n + nr For the case that two perturbations are in A and one is in B it happens the first time that we are not able to decide whether to take s 0 or s. The remaining minimum term is m = minnr, ir + nr. Then the expected minimal value E k=3 i is: 0
. The k - Model E k=3 i = E[minŵ k=3 s 0, ŵ k=3 s i ] with i n i i n + ir + n + 3nr + i m + n + nr 3 3 }{{}}{{}}{{} =: g =: g =: g 3 i n i + n i n + nr + ir + i n + nr + ir }{{}}{{} =: g =: g 5 n i + n i n + 3nr + i n i n + 3nr }{{}}{{} =: g =: g 7 = n 3 = + in i n + nr + ir }{{} =: g 8 g nn n + g + g 3 + g + g 5 + g + g 7 + g 8 ii i g = n + ir = i ni + ri 3 ni ri = ni3 + ri ni ri 3 ni ri 3 + ni + ri = ri + n 3ri 3 + n + ri + ni n in i n i g = n + 3nr n i n i = n + 3n r ni 3nri n i = n 3 + 3n 3 r n i 3n ri n i 3n ri + ni + 3nri n 3n r + ni + 3nri = n i n 3 + 3n 3 r n i n ri + ni + 3nri n 3n r + ni + 3nri
. The k - Model = n + 3n r n 3 i n 3 ri + n i + 3n ri n 3 3n 3 r + n i + 3n ri + n 3 i 3n 3 ri + n i + n ri ni 3 3nri 3 + n i + 3n ri ni 3nri + n 3 n 3 r + 8n i + n ri ni nri + n + n r ni nri = n 3nri 3 + n + 9n r n 9nri + n 3 9n 3 r + n + 8n r n nri + n + 3n r n 3 9n 3 r + n + n r g 3 = i i m + n + nr = m i 3 + ni 3 + nri 3 m i ni nri g = = = n iii n + nr + ir n i + n ri + ni 3 r n i n ri nri ni 3 nri 3 ri + ni + nri + ri 3 = n ini + nri + i 3 r ni nri i r ri + r ni 3 + n + n r + ni + n n ri g 5 = in in i n + nr + ir = n i n i + n ri + nri ni nri ri 3 = n 3 i + n 3 ri + n ri n i n ri nri 3 n i n ri nri 3 + ni 3 + nri 3 + ri = n i n ri nri + ni + nri + ri 3 ri + n + ri 3 + n r 8n + nr + ni + n 3 + n 3 r n n ri
. The k - Model g = = = n in i n in + 3nr n ni + i n + 3n r ni 3nri n 3nr = n ni + i n i n + 3nr n 3 + 3n 3 r n i 3n ri n 3n r n 3 i n 3 ri + n i + n ri + n i + n ri + n i + 3n ri ni 3 3nri 3 ni 3nri = n 3nri 3 + n + 9n r n 3nri + n + 3n r n 3 n 3 ri + n 3 + 3n 3 r n 3n r g 7 = i n in + 3nr = n i + 3n ri ni 3 3nri 3 = n 3nri 3 + n + 3n ri g 8 = in i n + nr + ir = n ni + i ni + nri + ri = n 3 i + n 3 ri + rn i n i n ri nri 3 + ni 3 + nri 3 + ri = ri + ni 3 + rn n n ri + n 3 + n 3 ri E k=3 i = = g nn n + g + g 3 + g + g 5 + g + g 7 + g 8 =... nn n 7ri + 3m + 9r nri 3 + n r + r 8nr 3m i + 8n r n 3 r nri + n r 8n 3 r + n r + n n 3 + 8n and m = minnr, ir + nr Now we have the exact formula and can calculate for arbitrary model parameters i, n N, r R the expected minimal value without much effort. So we first check whether the observation that the optimal number of detours d opt 3 5 for r = 0. is true for arbitrary values 3
. The k - Model of r and n. We must not forget that the alternative can use only a discrete number of detours. But for a better understanding of the behavior of the function we still consider i R. For each combination of r and n we want to check we insert values i = 0.00, 0.00,..., n and look for the minimum in i. Table 8 shows the quotients i opt n. r 0.0 0.0 0.05 0.0 0.0 0.30 0.50.00.00 5.00 0.00 00.00 3 0.3 0.3 0. 0. 0.39 0.38 0.3 0.35 0.35 0.35 0.35 0.35 0.5 0.55 0.5 0.9 0.5 0.3 0.0 0.0 0.0 0.0 0.0 0.0 5 0.0 0.59 0.57 0.53 0.8 0.5 0. 0. 0. 0. 0. 0. 0. 0. 0.59 0.55 0.50 0.7 0.3 0. 0. 0. 0. 0. 7 0.3 0. 0.0 0.57 0.5 0.8 0. 0. 0. 0. 0. 0. 8 0. 0.3 0. 0.57 0.5 0.9 0.5 0.7 0.7 0.7 0.7 0.7 9 0. 0. 0. 0.58 0.53 0.50 0.5 0.8 0.8 0.8 0.8 0.8 0 0.5 0. 0. 0.58 0.53 0.50 0. 0.8 0.8 0.8 0.8 0.8 0 0.7 0. 0. 0. 0.5 0.5 0.7 0.5 0.5 0.5 0.5 0.5 n 30 0.7 0.7 0.5 0. 0.5 0.53 0.8 0.53 0.53 0.53 0.53 0.53 0 0.8 0.7 0.5 0. 0.57 0.53 0.53 0.53 0.53 0.53 0.53 0.53 50 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.53 0.5 0.5 0.5 0.5 0 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.5 0.5 0.5 0.5 0.5 70 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.5 0.5 0.5 0.5 0.5 80 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.5 0.5 0.5 0.5 0.5 90 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.5 0.5 0.5 0.5 0.5 00 0.8 0.7 0.5 0. 0.57 0.53 0.5 0.5 0.5 0.5 0.5 0.5 50 0.8 0.8 0. 0. 0.57 0.5 0.5 0.5 0.5 0.5 0.5 0.5 00 0.8 0.8 0. 0. 0.57 0.5 0.55 0.55 0.55 0.55 0.55 0.55 Table 8: i opt n [ k-model with fixed k = 3 and δ = nr ] Unfortunately we have to find out that i opt n depends on n and r. It varies between i opt iopt n = 0.35 for small n and large r and n = 0.8 for large n and small r. So we don t know the value for i such that E k=3 i gets minimal and have to calculate it first. We could not eliminate all minimum terms and have to distinguish two cases: m = minnr, ir + nr = { nr : n r i n case ir + nr : 0 < i < n r case
. The k - Model We define two functions f and f : f i, n, r : = E k=3 i for i n r = 7ri + nr + 9ri 3 + n r nr + ri nn n + n 3 r + 8n r nri + n r + 8n 8n 3 r n 3 + n r + n f i, n, r : = E k=3 i for i < n r = 0ri + 3nr nr + ri 3 + n r 3nr 8nr + ri nn n + n 3 r + 8n r nri + n r + 8n 8n 3 r n 3 + n r + n We want to show that i opt 7 7 0n. For the case that n r 0n we need to look at f only. The first partial derivative i f is: i f = r8i3 + 3n + 7i + 8n n + i + n 3 + 8n n nn n The increase rate of f for i = 7 0n and i = n is strictly positive, since i f > 0 di := 8i 3 + 3n + 7i + 8n n + i + n 3 + 8n n > 0 and d 7 90n3 n = 0 000 3087n3 + 33n + 33n3 9n + 8n 00 00 0 0 0 n3 + 8n n = 7 500 n3 + 83 00 n 5 n > 0 for n dn = 8n 3 3n 3 + 7n + 8n 3 n + n n 3 + 8n n = n 3 + 3n n > 0 for n. The second partial derivative of f in i is: i f = 8ri r3n 7i r + n 8n nn n 5
. The k - Model The solutions of i f i, n, r = 0 are: i = 9 8 + 3 n ± C for n {}}{ 393 + n 3n Thus f has no turning point. Thus we know that f is monotonically increasing at least for n, r > 0 and i 7 0 n. If 7 0 n < n r what is exactly if r < 3 0 we still have to look at f. Again we need two steps: a The increase of f for i = 7 0n and i = n is larger than zero: 8 i f = r0i3 + 9nr 7n + 8i + 8n nr 3n + i + n 3 + 8n n nn n i 0 < i f i = 7 0 n = rn + n 30 + n r 0nr 00n n 0 < n + n 30 + n r 0nr r > n + n 30 n + 0n n ii 0 < i f i = n = rn + 9n r nr n n n b f has no turning point: i f = r0i + 9nr 7n + 8i 8n 3nr + n + nn n 0 = r0i + 9nr 7n + 8i 8n 3nr + n + nn n i = 3 0 nr + 3 5 n 3 8n 0 ± r 9n r 57n + 0nr + 78n 5 0
. The k - Model Since it holds for r 3 0 that 8n r 9n r 57n + 0nr + 78n 5 7.9n 388.8n 57n + 33.n + 78n 5 = 957.5n + 0.n 5 < 0 n 3 both roots get complex. Thus f is monotonically increasing for n, 0 < r < 3 0 and i 7 0n, too. For δ = nr, n and r > 0 the optimal number of detours of the alternative solution is at most 7 0 n. This holds for n = 3 as well. We simply insert n = 3 into E k=3 i. Then the expected minimal values are: Since E k=3 = 3 + 39 0 r E k=3 = 3 + 37 0 r + 0 min3r, 3r + r E k=3 3 = 3 + 0 r E k=3 E k=3 = 0 min3r, 3r + r 0 r and E k=3 3 E k=3 = 3 0 r > 0 = r min3, 3r + > 0 0 the optimal number of detours is d opt = for all r > 0. Lemma.5 For the k-model with δ = nr, n 3 and r > 0 the optimal number of detours of the alternative solution is at most 7 0 n. From Lemma.3 -.5 it directly follows Proposition.. Proposition. The k-model with δ = nr, n 3 and r > 0 has a peak in k =. The height of the peak is at least 3 0 n. For k = the optimal number of detours is n n± for even values of n and for odd values of n. For k = the optimal number of detours is n. And for k = 3 the optimal number of detours is at most 7 0 n. 7
. The k - Model k = To prove the existence of a gap we need to look at k =. As the experiments showed it is to assume that the optimal number of detours is equal to n. The number of possible cases gets quite large now: ŵ k= s 0 ŵ k= s i frequency A, A, A, A : min n + nr, n + ir = n + ir i n B, B, B, B : min n, n + ir + + rnr = n C, C, C, C : min n + nr, n + ir + nr = n + nr n i n D, D, D, D : min n, n + ir = n A, A, A, B : min n + 3nr, n + ir + + rnr = n + nr + m i 3i n B, B, B, A : min n + nr, n + ir + 3 + rnr = n + nr A, A, A, C : min n + nr, n + ir + nr = n + ir + nr 3n i i n A, A, A, D : min n + 3nr, n + ir = n + ir B, B, B, C : min n + nr, n + ir + nr + 3 + rnr = n + nr B, B, B, D : min n, n + ir + 3 + rnr = n C, C, C, A : min n + nr, n + ir + 3nr = n + ir + 3nr 3 i n C, C, C, B : min n + 3nr, n + ir + 3nr + + rnr = n + 3nr D, D, D, A : min n + nr, n + ir = n + ir D, D, D, B : min n, n + ir + + rnr = n C, C, C, D : min n + 3nr, n + ir + 3nr = n + 3nr n i 3 n i n D, D, D, C : min n + nr, n + ir + nr = n + nr A, A, B, B : min n + nr, n + ir + + rnr = n + nr i i n A, A, C, C : min n + nr, n + ir + nr = n + ir + nr i n i n A, A, D, D : min n + nr, n + ir = n + ir B, B, C, C : min n + nr, n + ir + nr + + rnr = n + nr B, B, D, D : min n, n + ir + + rnr = n C, C, D, D : min n + nr, n + ir + nr = n + nr n i n i n 8
. The k - Model ŵ k= s 0 ŵ k= s i frequency A, A, C, D : min n + 3nr, n + ir + nr = n + ir + nr n i i n B, B, C, D : min n + nr, n + ir + nr + + rnr = n + nr C, C, A, B : min n + 3nr, n + ir + nr + + rnr = n + 3nr n i i n D, D, A, B : min n + nr, n + ir + + rnr = n + nr A, A, B, C : min n + 3nr, n + ir + nr + + rnr = n + nr + m i in i n A, A, B, D : min n + nr, n + ir + + rnr = n + nr + m B, B, A, C : min n + nr, n + ir + nr + + rnr = n + nr B, B, A, D : min n + nr, n + ir + + rnr = n + nr C, C, A, D : min n + 3nr, n + ir + nr = n + ir + nr in i n C, C, B, D : min n + nr, n + ir + nr + + rnr = n + nr D, D, A, C : min n + nr, n + ir + nr = n + ir + nr D, D, B, C : min n + nr, n + ir + nr + + rnr = n + nr A, B, C, D : min n + nr, n + ir + nr + + rnr = n + nr i n i n It remain the two minimum terms: m = minnr, ir + nr and m = minnr, ir + nr The expected minimal value E k= i is: E k= i = E[minŵ k= s 0, ŵ k= s i ] = i n n + ir + n i n n + nr + i 3 i n + nr + m n + + i 3 n i n n + nr + ir + n i 3 n n i n + nr + n n i 3 n i n + nr + ir i i i n i n + nr + n + nr + ir n 9
+ n i n i n n + nr + i n i n n + nr + ir + n i n. The k - Model i n + nr + i in i n n + nr + m + n i in i n n + nr + ir + i n i n n + nr =... = 5ri 5 + r m + m + 7nri nn n n 3 + m m + nm 33r 0n r + 3nri 3 + nm r n r + 80n 3 r + 7nr + 8m i + 88n r + 9n 3 r 3n r + nri + n 3 8n + n 5 n n r 9n r + 3n 5 r + 88n 3 r As for the case of k = 3 perturbations we first check the assumption d opt = n experimentally. For each combination of r and n to check we insert values i = 0.00, 0.00,..., n and look for the minimum in i. Table 9 shows the quotients i opt n. We have to recognize that the assumption d opt = n is not true for all values of r. There seems to be a r n such that the assumption is only true for r < r n. Figure shows this r n numerically calculated for a large number of problem sizes n. At this point we decided to give up. To prove the observations for the case k = in the way we proved the observations for the case k = 3 would fill a lot of pages. The expenditure is out of all proportion to the result. Unfortunately this way we are not able to give a proposition that ensures the existence of a gap. But with the given formula for E k= i we feel save enough to state our observation as an assertion. Conjecture.7 For a given problem size n it exists a r n > 0 such that the k-model with δ = nr, n 3 and 0 < r < r n has a gap in k = 3. 0
. The k - Model r 0.0 0.0 0.05 0.0 0.0 0.30 0.50.00.00 5.00 0.0 00.0 3.00.00.00.00 0.8 0.7 0. 0. 0.5 0.5 0.5 0.5.00.00.00.00.00 0.3 0.3 0.9 0.9 0.9 0.9 0.9 5.00.00.00.00.00 0.38 0.3 0.3 0.3 0.3 0.3 0.3.00.00.00.00.00 0. 0.35 0.3 0.3 0.3 0.3 0.3 7.00.00.00.00.00.00 0.37 0.3 0.35 0.35 0.35 0.35 8.00.00.00.00.00.00 0.38 0.37 0.3 0.3 0.3 0.3 9.00.00.00.00.00.00 0.39 0.38 0.37 0.37 0.37 0.37 0.00.00.00.00.00.00 0.39 0.39 0.37 0.37 0.37 0.37 0.00.00.00.00.00.00 0. 0. 0.0 0.0 0.0 0.0 n 30.00.00.00.00.00.00 0.3 0.3 0. 0. 0. 0. 0.00.00.00.00.00.00 0.3 0. 0. 0. 0. 0. 50.00.00.00.00.00.00 0. 0. 0. 0. 0. 0. 0.00.00.00.00.00.00 0. 0.5 0. 0. 0. 0. 70.00.00.00.00.00.00 0.88 0.5 0. 0. 0. 0. 80.00.00.00.00.00.00 0.88 0.5 0. 0. 0. 0. 90.00.00.00.00.00.00 0.88 0.5 0. 0. 0. 0. 00.00.00.00.00.00.00 0.88 0.5 0. 0. 0. 0. 50.00.00.00.00.00.00 0.88 0.5 0. 0. 0. 0. 00.00.00.00.00.00.00 0.88 0. 0. 0. 0. 0. Table 9: i opt n [ k-model with fixed k = and δ = nr ] 0,5 0, 0,3 0, 0, 0 5 0 35 50 5 80 95 0 5 0 55 70 85 00 n Figure : r n [ k-model with fixed k = and δ = nr ]
. The k - Model.. Special Case n = 3 and δ = 3r Since we did not prove the existence of a gap for arbitrary n 3 we quickly want to show it at least for n = 3. If exactly k = edges get perturbed the expected minimal values are: E k= = 3 + 8 5 r E k= = 3 + 7 5 r + 5 min3r, 3r + r E k= 3 = 3 + 7 5 r + 3 5 minr, 3r + 3r To prove that d opt = 3 it must hold: E k= > E k= 3 and E k= > E k= 3 : E k= E k= 3 = r 0 3 min, 3r + 3 > 0 5 For r this is obviously wrong. So we assume r. With r < 9 E k= E k= 3 = r 0 33r + 3 > 0 5 r < 9 we get for Ek= > E k= 3 : E k= E k= 3 = r + min3, 3r + 3 min, 3r + 3 5 = r r + 3r + 33r + 3 = + 3r > 0 5 5 For r < 9 the optimal number of detours is d opt = 3. Proposition.8 The k-model with δ = 3r, n = 3 and 0 < r < 9 has a gap in k = 3. For k = and k = the optimal number of detours is 3. For k = 3 the optimal number of detours is.
. The p - Model. The p - Model How many detours d opt = dn, r, δ, p the alternative solution S has to use if every edge of the string of pearls gets independently length perturbed with probability p? In contrast to the k-model we are not able to calculate the expected minimal value numerically. For a given combination of the model parameters we have to simulate using a pseudo random number generator a large number of runs T and calculate the average value to get an approximation for the expected minimal value. We have to do this for all possible numbers of detours and then check which number is best. Dependence of d opt on the perturbation intensity δ First we have a look at the dependence on δ. Tables 0 and show simulation results for the optimal numbers of detours d opt for a fixed length of the detours r = 0. and problem size n = 5 and n = 0 respectively. δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 0. 3 3 5 5 5 5 5 5 5 0. 3 3 5 5 5 5 5 5 5 0.3 3 3 5 5 5 5 5 5 5 5 0. 3 5 5 5 5 5 5 5 p 0.5 3 5 5 5 5 5 5 5 0. 3 3 5 5 5 5 5 0.7 3 3 5 5 5 5 5 0.8 3 3 3 5 5 5 0.9 3 3 3 3 3 Table 0: The optimal number of detours d opt of the alternative solution S [ p-model with fixed n = 5, r = 0.; T = 0 runs ] δ 0. 0.3 0. 0.5 0. 0.7 0.8 0.9.0....8.0 0. 3 3 5 5 7 8 0 0 0 0. 3 3 5 5 8 0 0 0 0 0.3 3 3 9 0 0 0 0 0 0 0 0. 3 9 0 0 0 0 0 0 0 p 0.5 3 8 9 0 0 0 0 0 0 0. 3 3 8 8 9 0 0 0 0 0 0.7 3 3 0 0 0 0 0 0.8 3 3 3 3 5 5 5 0.9 3 3 3 3 3 Table : The optimal number of detours d opt of the alternative solution S [ p-model with fixed n = 0, r = 0.; T = 0 runs ] As to be expected d opt increases in δ. For small values of δ the optimal number of detours is small and equal for all tested probabilities p. For intermediate values of p p 0.3 the optimal number of detours increases fastest in δ such that it becomes unimodal in p. Thus the optimal number of detours for these intermediate values of p reach the maximal value 3
. The p - Model d opt = n first. Then for larger values of δ the optimal number of detours d opt gets equal to n for the small values of p, too. Astonishingly the entries of both tables are nearly identical for values of δ 0.7. Dependence of d opt on the detour length r For the sake of completeness we have a look at some results for the dependence on r. Tables and 3 show simulation results for the optimal numbers of detours d opt for a fixed perturbation intensity δ = 3.0. Obviously d opt decreases in r. Again it is conspicuous that the entries of both tables are nearly identical for values of r 0.3. r 0. 0.5 0.0 0.5 0.30 0.35 0.0 0.5 0.50 0.0 0.70 0.80 0.90 0. 5 5 5 5 5 3 3 0. 5 5 5 5 5 3 3 3 0.3 5 5 5 5 3 3 0. 5 5 5 5 3 3 p 0.5 5 5 5 3 0. 5 5 3 0.7 5 5 3 0.8 5 3 0.9 5 3 Table : The optimal number of detours d opt of the alternative solution S [ p-model with fixed n = 5, δ = 3.0; T = 0 runs ] r 0. 0.5 0.0 0.5 0.30 0.35 0.0 0.5 0.50 0.0 0.70 0.80 0.90 0. 0 0 8 5 3 3 0. 0 0 0 5 3 3 3 0.3 0 0 0 0 3 3 0. 0 0 0 8 3 p 0.5 0 0 9 3 0. 0 0 8 3 0.7 0 0 3 0.8 3 0.9 5 3 Table 3: The optimal number of detours d opt of the alternative solution S [ p-model with fixed n = 0, δ = 3.0; T = 0 runs ] Dependence of d opt on the problem size n Finally we have a look at the dependence on n. Table shows what happens for the different values of p if we increase n and let r and δ fixed. Then Table 5 shows what happens for the different values of r if we increase n and let p and δ fixed. Finally Table shows what happens for the different values of δ if we increase n and let r = 0. and p = 0. fixed.
. The p - Model n 3 5 7 8 9 0 3 5 7 8 9 0 0. 3 5 5 5 5 5 5 5 5 5 5 5 5 5 0. 3 5 3 0.3 3 5 0 0. 3 5 8 9 0 p 0.5 3 5 5 5 8 9 0 0 0 0 0 0 0 0 0 0 0. 3 5 5 5 8 9 9 9 9 9 9 9 0 9 9 9 9 0.7 3 0.8 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 0.9 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 Table : The optimal number of detours d opt of the alternative solution S [ p-model with fixed r = 0., δ =.0; T = 0 runs ] n 3 5 7 8 9 0 3 5 7 8 9 0 0.0 3 5 7 8 9 0 3 5 7 8 9 0 0.5 3 5 7 7 8 8 3 5 0.0 3 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 0.5 3 0.30 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 0.35 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 r 0.0 0.5 0.50 0.55 0.0 0.5 0.70 Table 5: The optimal number of detours d opt of the alternative solution S [ p-model with fixed p = 0., δ =.0; T = 0 runs ] n 3 5 7 8 9 0 3 5 7 8 9 0 0. 0. 0. 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 0.8 3.0 3 5 5 5 5 5 5 5 5 5 5 5. 3 5 7 7 7 7 7 7 7 7 7 7 7 7 7 7. 3 5 7 8 8 8 8 8 8 8 8 8 8 8 9 8 δ. 3 5 7 8 9 0 0 0 0 9 0 0 9 0 9 0.8 3 5 7 8 9 0 0 0.0 3 5 7 8 9 0 3 3 3 3 0. 3 5 7 8 9 0 3 5 5 5 5 5. 3 5 7 8 9 0 3 5 7 7 5 0. 3 5 7 8 9 0 3 5 5 7 8 8 0.8 3 5 7 8 9 0 3 5 7 8 9 0 3.0 3 5 7 8 9 0 3 5 7 8 9 0 Table : The optimal number of detours d opt of the alternative solution S [ p-model with fixed p = 0., r = 0.; T = 0 runs ] 5
. The p - Model The results are quite smooth. There are no large peaks or gaps. But there are some exceptions, too. For example the row p = 0. in Table contains an anomaly. With increasing n the optimal number increases up to then for n = jumps up to, jumps back to at n =, jumps up to 3 at n =, jumps back.... Also some other entries differ from the pattern, but usually only by. These are maybe imprecisions caused by the simulation. We observe that with some exceptions for increasing problem size n the optimal number of detours d opt = dn, r, δ, p nearly always increases up to a number d = dr, δ, p and then remains constant for larger values of n. Such that: { n : n < d d opt n, r, δ, p = r, δ, p, d r, δ, p : otherwise We observe the following dependencies of d r, δ, p: d is unimodal in p see Table d is decreasing in r see Table 5 d is increasing in δ see Table theoretical analysis of the special case n = 3 In contrast to the k-model, we assume that the optimal number of detours is unimodal in p. For fixed n 3, r > 0 and δ > r and probabilities 0 < p < p < p 3 it should hold one of the cases: d opt p < d opt p < d opt p 3 d opt p < d opt p > d opt p 3 d opt p > d opt p > d opt p 3. So there exist no parameters n, r, δ, p, p, p 3 such that d opt p > d opt p < d opt p 3. Since there is no way to build up a formula for arbitrary n, r, δ we try the special case n = 3. Let E p i denote the expected minimal value E[minŵ ps 0, ŵ p s i ] of the solution pair. The probability that exactly k perturbations occur is: p k p n k. By enumeration of all possible cases of perturbations each time checking the minimal value of ŵ p s 0 and ŵ p s i for i =,, 3 we observed the following formulas for the expected minimal values E p, Ep and E p 3 : E p = p 3 + p p 5 8 + δ + r + p p 5 + δ + r + p 3 p 3 0 + δ + r + p p 5 + δ + r + p 5 p 8 + δ + r + p 3 + 3δ
. The p - Model E p = p 3 + p p 5 8 + δ + minδ, r + p p 5 + 9δ + r + minδ, r + p 3 p 3 0 + 0δ + r + minδ, r + minδ, r + δ + rδ + p p 5 + 0δ + r + minδ, r + δ + rδ + p 5 p 8 + δ + minδ, r + δ + rδ + p 3 + 3δ E p 3 = p 3 + p p 5 8 + 3 minδ, 3r + p p 5 + 9δ + 3 minδ, 3r + p 3 p 3 0 + 9δ + 3r + 9 minδ, 3r + δ + rδ + p p 5 + δ + 3 min3δ, 3r + δ + rδ + p 5 p 8 + δ + 3 min3δ, 3r + δ + rδ + p 3 + 3δ If we want to eliminate all minimum terms we have to distinguish cases and the resulting polynomials are still of degree. We found no way to prove our assumption. case n = 3 and δ = nr If we additionally fix the perturbation intensity to δ = nr, where we found the highest peaks and gaps for the k-model, the polynomials get much more handy. There are only two remaining minimum terms. The expected minimal values are E p = 3 + 7pr + p r E p = 3 + 7pr + p r + p 3 rm p rm E p 3 = 3 + r + 9rm p + 3r 8rm p 5 + 5r + 9rm p + 30p 3 r 9p r + 9pr with m = min, 3r and m = min, r. We immediately see that E p < Ep for all 0 < p < and r > 0, since: E p Ep = p3 rm p rm = p 3 pm > 0 If it holds that E p < Ep 3 for all 0 < p < and r > 0 the existence of gaps is out of question. 7
3 Optimal Pairs vs. Optimal Alternatives 0 < E p 3 Ep = r + 3rm p + 3r rm p 5 + 5r + 3rm p + 30p 3 r p r + pr = pprp 3p m p 3 + 3p 3 m + p 9p + 0 < p 3p m p 3 + 3p 3 m + p 9p + = 3p 3 pm + p p 3 + p 9p + 0 < p p 3 + p 9p + =: gp Since g p = 8p 3 7p + p 9 g p = 0 p = { and g p = p p + g = ± I C gp is minimal for p = and since g = it holds that gp E p < Ep 3 for all 0 < p < and r > 0 d opt p = for all 0 < p < and r > 0 for 0 < p <. Proposition.9 The p-model with r > 0 and 0 < p < has neither a peak nor a gap for the special case n = 3, δ = 3r. 3 Optimal Pairs vs. Optimal Alternatives The original question was how to choose two solutions S 0 and S such that the expected minimal value E[minŵS 0, ŵs ] gets minimal. Is it generally optimal to fix one of the solutions of the pair to the initial best solution that uses no detour? Since it makes no sense that both solutions use the same detour we redefine the set S of solutions we look at: S = {s i,j : i, j 0; i + j n} with s i,j = e 0,,..., e 0,i, e,i+,..., e,i+j, e 0,i+j+,..., e 0,n 8
3 Optimal Pairs vs. Optimal Alternatives The length of the path s i,j is: i ws i,j = we 0,t + t= i+j t=i+ we,t + n t=i+j+ we 0,t The first i edges of a solution are non-detour edges, the next j edges are detours and the remaining n i j edges are non-detour edges again. Since E[minŵs 0,i, ŵs i,j ] = E[minŵs 0,j, ŵs j,i ] we assume that the second solution uses at least as much detours as the first solution from the pair. So the set of solution pairs P to analyze is: P = {p i,j = {s 0,i, s i,j } : 0 i n, i j n i} with wp i,j = minws 0,i, ws i,j Now we are able to put the question in concrete terms: How to choose i and j such that the expected minimal value E[ŵp i,j ] gets minimal? For the k - model we are able to calculate the expected minimal value of a solution pair exactly as described in subsection.. For the p - model we have to do simulations again as described in subsection.. We tested both problems with similar parameter sets: k - model with exact calculation k =,,..., n n = 3,,..., 0 r = 0.05, 0.0,..., δ = r + 0., r + 0.,..., n + rn 0. 0. p - model with T = 0 simulation runs p = 0., 0.,..., 0.9 n = 3,,..., 0 r = 0.05, 0.0,..., δ = r + 0., r + 0.,..., n + rn 0. 0. We made one main observation: For all settings the optimal solution with respect to w the solution that uses no detour at all was part of the optimal pairs. Thus it seems to hold for both models: min E[ŵp 0,l] 0 l n min E[ŵp i,j ] i,j n i+j n In detail the results are identical to those of subsection where we analyzed only the choice of an optimal alternative solution. So there is no need to show them again explicitly. To give a theoretical prove for the correctness of this observation for arbitrary problem parameters would be a quite hard task. But we give a prove at least for n = and n = 3. 9
3 Optimal Pairs vs. Optimal Alternatives We start with n =. The first column of Table 7 shows all possible cases for perturbations. We symbolize the graph by a matrix. The entries of the matrix show which edges get perturbed denoted by X and which edges not denoted by O. There are only three solution pairs to analyze. If the first solution uses no detour the second solution can use either one or two detours. And if the first solution uses already one detour the second has also to use one detour. ŵp 0, ŵp 0, ŵp, OO OO + r XO OO + δ + minδ, r + r OX OO + r + minδ, r + r OO XO + r OO OX + r XO XO + δ + δ + r + δ XO OX + δ + δ + r XX OO + r + δ + r + r + δ OO XX + r + δ + rδ OX OX + δ + δ + r + δ OX XO + r + δ + r XX XO + r + δ + δ + minδ, r + rδ + r + δ XX OX + δ + δ + minδ, r + rδ + r + δ OX XX + δ + δ + r + δ + rδ XO XX + δ + δ + r + δ + rδ XX XX + δ + δ + r + δ + rδ Table 7: Enumeration of all possible cases of perturbations for n = 30
3 Optimal Pairs vs. Optimal Alternatives Then the expected minimal values for the solution pairs of the k-model are: E k= 0, = 8 + r + δ E k= 0, = + r + δ E k=3 0, = 8 + r + 5δ E k= 0, = 8 + minδ, r E k= 0, = + r + δ E k=3 0, = 8 + δ + minδ, r + rδ E k=, = 8 + r E k=, = + r + δ + rδ E k=3, = 8 + r + δ + rδ We immediately see that E k 0, Ek, for k =,, 3. Lemma 3. k-model For a string of pearls G n=,r with detours r > 0, perturbation intensity δ > r and 0 < k < perturbations there is always an optimal solution pair that contains the solution s 0,0. For some parameter combinations it exist other optimal solution pairs that do not contain the solution s 0,0. It holds: min E[ŵ kp 0,l ] 0 l min E[ŵ k p i,j ] i,j i+j The expected minimal values for the solution pairs of the p-model are: E p 0, = p + p p 3 8 + r + δ + p p + r + δ + p 3 p8 + r + 5δ + p + δ E p 0, = p + p p 3 8 + minδ, r + p p + r + δ + p 3 p8 + δ + minδ, r + rδ + p + δ E p, = p + r + p p 3 8 + r + p p + r + δ + rδ + p 3 p8 + r + δ + rδ + p + r + δ + rδ 3
3 Optimal Pairs vs. Optimal Alternatives Again every summand of E p 0, is at most equal to the corresponding summand of Ep, and some summands are strictly smaller. Thus the question is immediately answered for the p-model, too. Lemma 3. p-model For a string of pearls G n=,r with detours r > 0, perturbation intensity δ > r and a perturbation probability 0 < p < the optimal solution pair always contains the solution s 0,0. It holds: min E[ŵ pp 0,l ] < 0 l min E[ŵ p p i,j ] i,j i+j We do the same procedure for n = 3. We have to look at five solution pairs: p 0,, p 0,, p 0,3 and p,, p, and possible cases of perturbations. We don t show every single case, but give the formulas for k = 0,,..., n. For each k we have to show that the expected value of one of p 0,, p 0,, p 0,3 is at least as good as the expected value of the pair p,. The same for the pair p,. k = 0: E k=0 0, = E k=0 0, = E k=0 0,3 = 3 < 3 + r = E k=0, = E k=0, k = : a E k= 0, = 8 + r + δ E k= 0, = 8 + δ + minδ, r E k= 0,3 = 8 + 3 minδ, 3r vs. E k= 0,3 E k=, 8 + 3 minδ, 3r 8 + r + δ E k=, = 8 + r + δ E k=, = 8 + 9r 3 minδ, 3r r + δ for δ 3r : 3δ r + δ δ 3r for δ 3r : 9r r + δ δ 3r b E k= 0,3 E k=, 8 + 3 minδ, 3r 8 + 9r 9r 9r 3
3 Optimal Pairs vs. Optimal Alternatives k = : E k= 0, = 5 + r + δ 5 E k= 0, = 5 + r + 9δ + minδ, r 5 E k= 0,3 = 5 + 9δ + 3 minδ, 3r 5 vs. E k=, = 5 + 5r + 9δ + rδ 5 E k=, = 5 + 8r + 9δ + r min, δ + rδ 5 a E k= 0,3 E k=, 5 + 9δ + 3 minδ, 3r 5 + 5r + 9δ + rδ 5 5 9r 5r + rδ b E k= 0,3 E k=, 5 + 9δ + 3 minδ, 3r 5 + 8r + 9δ + r min, d + rδ 5 5 9r 8r + r min, δ + rδ k = 3: E k=3 0, = 0 + r + δ 0 E k=3 0, = 0 + r + δ + minδ, r + minδ, r + rδ 0 E k=3 0,3 = 0 + 3r + 8δ + 9 minδ, 3r + rδ 0 vs. E k=3, = 0 + 0r + δ + rδ 0 E k=3, = 0 + 8r + 8δ + minδ, r + rδ + 7rδ 0 a E k=3 0, E k=3, 0 0 + r + δ + minδ, r + minδ, r + rδ 0 + 0r + δ + rδ 0 minδ, r + minδ, r + rδ r + rδ 8r + rδ r + rδ 33
3 Optimal Pairs vs. Optimal Alternatives b E k=3 0,3 E k=3, 0 + 3r + 8δ + 9 minδ, 3r + rδ 0 + 8r + 8δ + minδ, r + rδ + 7rδ 0 0 9 minδ, 3r + rδ 5r + minδ, r + rδ + 7rδ for δ r + rδ : 9δ 5r + 7rδ + δ 7δ 7r + 7rδ 5r + 7rδ for δ r + rδ : 7r + 9rδ 7r + 9rδ k = : E k= 0, = 5 + r + δ 5 E k= 0, = 5 + r + δ + minδ, r + rδ 5 E k= 0,3 = 5 + δ + 3 minδ, 3r + rδ 5 vs. E k=, = 5 + 5r + δ + rδ 5 E k=, = 5 + 8r + δ + 9rδ 5 a b E k= 0, E k=, 5 5 + r + δ + minδ, r + rδ 5 + 5r + δ + rδ 5 minδ, r + rδ 3r + rδ 8r + rδ 3r + rδ E k= 0, E k=, 5 5 + r + δ + minδ, r + rδ 5 + 8r + δ + 9rδ 5 minδ, r + rδ r + 9rδ 8r + rδ r + 9rδ 3
3 Optimal Pairs vs. Optimal Alternatives k = 5: E k=5 0, = 8 + r + δ E k=5 0, = 8 + 3δ + minδ, r + rδ E k=5 0,3 = 8 + δ + 3 minδ, 3r + rδ vs. E k=5, = 8 + r + 3δ + rδ E k=5, = 0 + 8r + δ + minδ, r + rδ + 5rδ a E k=5 0,3 E k=5, 8 + δ + 3 minδ, 3r + rδ 8 + r + 3δ + rδ 3 minδ, 3r + rδ r + δ + rδ for δ 3r + rδ : 3δ r + rδ + δ δ 3r + rδ for δ 3r + rδ : 9r + rδ r + rδ + δ 3r + rδ δ b E k=5 0,3 E k=5, 8 + δ + 3 minδ, 3r + rδ 0 + 8r + δ + minδ, r + rδ + 5rδ 3 minδ, 3r + rδ 8r + minδ, r + rδ + 5rδ for δ r + rδ : 3δ 8r + δ + 5rδ δ 8r + 5rδ for δ r + rδ : 9r + rδ 8r + δ + 5rδ r + rδ δ k = : E k= 0, = E k= 0, = E k= 0,3 = 3 + 3δ < 3 + r + 3δ + rδ = E k=, = E k=, 35
3 Optimal Pairs vs. Optimal Alternatives Lemma 3.3 k-model For a string of pearls G n=3,r with detours r > 0, perturbation intensity δ > r and 0 < k < perturbations there is always an optimal solution pair that contains the solution s 0,0. For some parameter combinations there exist other optimal solution pairs that do not contain the solution s 0,0. It holds: min E[ŵ kp 0,l ] 0 l 3 min E[ŵ k p i,j ] i,j 3 i+j 3 With the formulas for the k-model we are ready to give the formulas for the p-model: E p 0, = p 3 + p p 5 8 + r + δ + p p 5 + δ + r + p 3 p 3 0 + δ + r + p p 5 + δ + r + p 5 p8 + δ + r + p 3 + 3δ E p 0, = p 3 + p p 5 8 + δ + minδ, r + p p 5 + 9δ + r + minδ, r + p 3 p 3 0 + δ + r + minδ, r + minδ, r + rδ + p p 5 + δ + r + minδ, r + rδ + p 5 p8 + 3δ + minδ, r + rδ + p 3 + 3δ E p 0,3 = p 3 + p p 5 8 + 3 minδ, 3r + p p 5 + 9δ + 3 minδ, 3r + p 3 p 3 0 + 8δ + 3r + 9 minδ, 3r + rδ + p p 5 + δ + 3 minδ, 3r + rδ + p 5 p8 + δ + 3 minδ, 3r + rδ + p 3 + 3δ E p, = p 3 + r + p p 5 8 + r + δ + p p 5 + 5r + 9δ + rδ + p 3 p 3 0 + 0r + δ + rδ + p p 5 + 5r + δ + rδ + p 5 p8 + r + 3δ + rδ + p 3 + r + 3δ + rδ E p, = p 3 + r + p p 5 8 + 9r + p p 5 + 9δ + 8r + r min, δ + rδ + p 3 p 3 0 + 8r + 8δ + minδ, r + rδ + 7rδ + p p 5 + δ + 8r + 9rδ + p 5 p8 + 8r + δ + 5rδ + minδ, r + rδ + p 3 + r + 3δ + rδ 3
3 Optimal Pairs vs. Optimal Alternatives We show that d := E p, Ep 0,3 > 0 for arbitrary r > 0, δ > r and 0 < p < : d = p r {> 0 + p p 5 r + δ 3 minδ, 3r { 0 + p p 5r + rδ 3 minδ, 3r { p p r + rδ + p 3 p 3 7r + δ + rδ 9 minδ, 3r + rδ { p 3 p 3 r rδ + p p 5r + rδ 3 minδ, 3r + rδ { p p r + 3rδ + p 5 pr + δ + rδ 3 minδ, 3r + rδ { 0 + p r + rδ {> 0 > p p r + rδ + p 3 p 3 r rδ + p p r + 3rδ for p < p: d > p 3 p 3 r + rδ + r rδ > 0 for p p: d > p 3 p 3 r + 3rδ + r rδ > 0 We do the same for d := E p, Ep 0,3 > 0: d = p r {> 0 + p p 5 9r 3 minδ, 3r { 0 + p p 8r + r min, δ + rδ 3 minδ, 3r {> 0 + p 3 p 3 5r + minδ, r + rδ + 7rδ 9 minδ, 3r + rδ { 0 + p p 8r + 9rδ 3 minδ, 3r + rδ {> 0 + p 5 p8r + 5rδ + minδ, r + rδ 3 minδ, 3r + rδ { 0 + p r + rδ {> 0 Lemma 3. p-model For a string of pearls G n=3,r with detours r > 0, perturbation intensity δ > r and a perturbation probability 0 < p < the optimal solution pair always contains the solution s 0,0. It holds: min E[ŵ pp 0,l ] < 0 l 3 min E[ŵ p p i,j ] i,j 3 i+j 3 37
Failed Assumption Failed Assumption One of our initial interests for this abstract string of pearls model was to prove theoretically one of the experimentally observed results that are presented in [Sam 005a]. There a similar but more complex scenario is analyzed where the chance to give a prove is quite small. Translated to the string of pearls model we assumed to be able to prove the following assumption: Consider a string of pearls G n,r = {V, E, w}. Let s i denote the solution that uses i detours and let E i denote the expected minimal value E[minŵs 0, ŵs i ] with respect to ŵ. Furthermore let n 3, r > 0, δ > r and either 0 < p < p-model or 0 < k < n k-model. Then there do not exist parameters a, b, c N with 0 a < b < c n such that E a < E b > E c. We had to find out that this assumption does not generally hold for the p-model and the k-model. For n = 3 we found counter examples with the help of the theoretical formulas for the expected minimal values shown in Subsections. and.. For larger problem sizes n we found counter examples experimentally. We could give a list of parameter combinations where the assumption fails. We are content with showing one counter-example for the p- model and a set of counter-examples for the k-model: p-model: n = 3, r = 0., δ = 0. and p = 0. E p < Ep > Ep 3 : E p 0 = 3.000 E p = 3.000 E p = 3.88 E p 3 = 3.30 k-model: n = 3, 0 < r < 5 9, δ = 3r and k = Ek < Ek > Ek 3 : E k 0 = 3 + 90r 5 E k = 3 + 8r 5 E k = 3 + 8r 5 + 5 min0, 3r r E k 3 = 3 + 90r 5 + 3 5 min0, 3r 3r If we change the string of pearls graph such that the detours are iteratively increasing see Figure 3 the assumption seems to hold. At least we found no counter examples. But with this modification the problems get more complicated again. 38
5 Conclusions v 0 v v v 3 v n v n + r + r + 3r + nr Figure 3: Extended String of Pearls with Iteratively Increasing Detours 5 Conclusions We analyzed strings of pearls G n,r see Definition. with biased perturbations see Definition.: k-model and Definition.3: p-model. In Section we analyzed the optimal choice of an alternative solution S additionally to the solution S 0 that uses no detour, such that E[minŵS 0, ŵs ] is minimal. Since all detours of the graph have the same length it does not matter which detours S uses. Only the number d of detours is important. The models turned out to be awkward to manage theoretically. Thus the results mainly are observations by computer calculations or computer simulations. For certain parameter subsets we were able to prove the results. For the k-model we observed that the optimal number of detours d opt decreases in the detour length r, increases in the perturbation intensity δ, behaves unexpectedly irregular in the number of perturbations k. For certain combinations of n, r, δ it jumps up and down if k increases see Pages.., Proposition. and Conjecture.7. For the p-model we observed that for increasing problem size n the optimal number of detours d opt = dn, r, δ, p increases up to a number d = dr, δ, p and then remains constant, such that: { n : n < d d opt n, r, δ, p = r, δ, p, d r, δ, p : otherwise. We observe the following dependencies of d r, δ, p: d first increases and then decreases in the perturbation probability p, d decreases in the detour length r, d increases in perturbation intensity δ. 39
REFERENCES Furthermore we conjecture that for both models it is optimal to fix one of the solutions to the solution that uses no detour. If the alternative is correctly chosen there exist no other pairs of solutions that are better proved for n 3. References [Sam 005a] J. Sameith. 005 Penalty Methods Generating Alternative Solutions for Discrete Optimization Problems with Uncertain Data. submitted as doctoral dissertation, Faculty of Mathematics and Computer Science, Friedrich-Schiller- University Jena. [Sam 005b] J. Sameith. 005 Relative Improvement by Alternative Solutions for Classes of Simple Shortest Path Problems with Uncertain Data Part I: Strings of Pearls G n with Unbiased Perturbations. Technical report. Available at http://www.minet.uni-jena.de/math-net/reports/. 0