· relative improvement by alternative solutions for classes of simple shortest path problems with...

Relative Improvement by Alternative Solutions

for Classes of Simple Shortest Path Problems

with Uncertain Data

—

Part II: Strings of Pearls Gn,r with Biased Perturbations

l l l l l l l ls s s s s s s

¸ ¸ ¸ ¸ ¸ ¸ ¸

Jorg Sameith∗

Graduiertenkolleg “Approximation und algorithmische Verfahren”Fakultat fur Mathematik und Informatik

Friedrich-Schiller-Universitat Jena07740 Jena – Germany

August 15, 2005

AbstractFor classes of simple shortest path problems we analyze the following situation: We aregiven a graph G = {V, E,w} and are allowed to provide two solutions A and B which arepaths from a given starting node to a given target node. Then we learn about randomlyperturbed edge lengths w and are allowed to choose the solution from our pair {A,B}that has the better value with respect to these new lengths w. We analyze how to choosetwo solutions A and B such that

E[min(∑

e∈A

w(e),∑

e∈B

w(e))]

is minimal.

∗[email protected]

Contents

1 Introduction 11.1 The Models . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

2 The Optimal Alternative 32.1 The k - Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32.2 The p - Model . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 23

3 Optimal Pairs vs. Optimal Alternatives 28

4 Failed Assumption 38

5 Conclusions 39

References 40

1 Introduction

1 Introduction

For classes of simple shortest path problems we analyze the following situation: We are givena graph G = {V, E, w} and are allowed to provide two solutions S0 and S1 which are pathsfrom a given starting node to a given target node. Then we learn about randomly perturbededge lengths w and are allowed to choose the solution from our pair {S0, S1} that has thebetter value with respect to these new lengths w. We analyze how to choose two solutionsS0 and S1 such that

E[min(∑

e∈S0

w(e),∑

e∈S1

w(e))]

is minimal.

The report is organized as follows. In the remaining part of this section we introducethe considered models in detail: the graph model (string of pearls Gn,r) and two differentbiased perturbation models (k-model, p-model). Let S0 be the optimal solution with respectto the original weights w. In Section 2 we analyze only the optimal choice of an additionalalternative solution S1 for the case that S0 is fixed. In Section 3 we relax this condition andallow two arbitrary solutions. In Section 4 we briefly comment on an assumption (whichturns out to be wrong) that initially motivated us to study these models. We finish withconclusions in Section 5.

1.1 The Models

Definition 1.1 (string of pearls Gn,r)Consider a weighted directed graph Gn,r = {V, E, w} with a set of nodes V and a set ofedges E with

V = {v0, v1, . . . , vn},E = {e0,i = (vi−1, vi) , e1,i = (vi−1, vi) : i = 1, 2, . . . , n}

and a weight function w : E → R+ that assigns a weight to each edge (see Figure 1). Theweights are w(e0,i) = 1 and w(e1,i) = 1 + r for i = 1, . . . , n. We interpret the edge-weightsas lengths.

We call this type of graph a string of pearls Gn,r. Furthermore we call n the problemsize and the edges {e1,i : i = 1, 2, . . . , n} we call detours.

A feasible solution S is a path from starting node v0 to the final node vn. In every nodevi (0 ≤ i < n) it is to decide whether to use the short edge e0,i+1 around the top or thedetour e1,i+1 down below. The length w(S) of a solution S is the sum of the lengths of itsedges e ∈ S:

w(S) =∑

e∈S

w(e) .

1

1.1 The Models

µ´¶³

µ´¶³

µ´¶³

µ´¶³

· · · µ´¶³

µ´¶³R R R

¸ ¸ ¸

R

¸

v0 v1 v2 v3 vn−1 vn

1 1 1

1 + r 1 + r 1 + r

1

1 + r

Figure 1: String of Pearls Gn,r

We consider two different perturbation models:

Definition 1.2 (k-model)Consider a string of pearls Gn,r. Exactly k (0 ≤ k ≤ 2n) randomly chosen edges get length-perturbed. Let π be a random permutation of (1, 2, . . . , 2n), then w : E → R is:

wk(ei,j) :={

(1 + δ) · w(ei,j) : if π(i · n + j) ≤ kw(ei,j) : otherwise

Definition 1.3 (p-model)Consider a string of pearls Gn,r. Every edge independently gets length-perturbed with theprobability p (0 ≤ p ≤ 1). Then w : E → R is:

wp(ei,j) :={

(1 + δ) · w(ei,j) : with probability p,w(ei,j) : otherwise

The parameter δ ∈ R we call the perturbation intensity.

RemarksFor the following cases the choice of solutions is obvious. The initial optimal solution – thesolution that uses no detour – remains optimal with respect to w.

1. p = 0 or k = 0: No edge-weight gets perturbed.

2. p = 1 or k = 2n: All edge-weights get perturbed.

3. δ ≤ r: It is not advantageous to use a detour since a non-detour edge is never longerthan a detour edge.

2

2 The Optimal Alternative

2 The Optimal Alternative

Let S0 be the solution that uses no detour – the optimal solution with respect to the originalweights w. Which additional solution S1 we have to select such that the expected minimalvalue

E[min(w(S0), w(S1))] is minimal?

Since all detours of the graph G have the same length it does not matter which detoursa solution uses. Only the number of detours is important. Let S be the set of availablesolutions:

S = {si : 0 ≤ i ≤ n}with: si = (e1,1, e1,2, . . . , e1,i, e0,i+1, e0,i+2, . . . , e0,n) .

The length of path si is:

w(si) =i∑

t=1

w(e1,t) +n∑

t=i+1

w(e0,t) .

2.1 The k - Model

How many detours dopt = d(n, r, δ, k) the alternative solution S1 has to use if exactly k ran-domly chosen edges get length perturbed?

We are able to calculate for any given n, r, δ and k the optimal number of detours thatthe additional solution S1 should use exactly. We generate all

(2nk

)possible cases for wk

with exactly k perturbations. For each case we calculate the minimal values of all possiblesolution pairs {{S0, S1} : S0 = s0 and S1 = si with i ∈ {1, 2, . . . , n}}. Finally we calculatethe average values for all pairs and observe the optimal number of detours dopt.

First we give some results that show dopt for different numbers of perturbations k andperturbation intensities δ (see Tables 1-3). All results we show are for r = 0.1. We testedother values too, but in view of the dependence on k and δ the results are quite similar.

δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0

1 1 1 2 2 3 3 4 4 4 4 4 4 4 42 1 1 4 4 4 4 4 4 4 4 4 4 4 43 1 1 2 3 3 3 4 4 4 4 4 4 4 4

k 4 1 1 4 4 4 4 4 4 4 4 4 4 4 45 1 1 2 2 3 3 3 3 4 4 4 4 4 46 1 1 4 4 4 4 4 4 4 4 4 4 4 47 1 1 2 2 2 2 2 3 3 3 3 3 4 4

Table 1: The optimal number of detours dopt of the alternative solution S1 — [ k-model with fixedn = 4, r = 0.1 ]

3

2.1 The k - Model

δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0

1 1 1 2 2 3 3 4 4 5 5 5 5 5 52 1 1 2 5 5 5 5 5 5 5 5 5 5 53 1 1 2 3 3 4 4 4 5 5 5 5 5 54 1 1 2 5 5 5 5 5 5 5 5 5 5 5

k 5 1 1 2 2 3 4 4 4 4 5 5 5 5 56 1 1 2 5 5 5 5 5 5 5 5 5 5 57 1 1 2 2 2 3 3 4 4 4 4 5 5 58 1 1 2 2 5 5 5 5 5 5 5 5 5 59 1 1 2 2 2 2 2 3 3 3 3 3 4 4


δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0

1 1 1 2 2 3 3 4 4 5 6 6 6 6 62 1 1 2 6 6 6 6 6 6 6 6 6 6 63 1 1 2 3 3 4 4 5 5 6 6 6 6 64 1 1 2 6 6 6 6 6 6 6 6 6 6 65 1 1 2 3 3 4 5 5 5 5 5 6 6 6

k 6 1 1 2 6 6 6 6 6 6 6 6 6 6 67 1 1 2 2 3 3 4 5 5 5 5 5 6 68 1 1 2 2 6 6 6 6 6 6 6 6 6 69 1 1 2 2 2 3 3 4 4 4 5 5 5 5

10 1 1 2 2 2 2 6 6 6 6 6 6 6 611 1 1 2 2 2 2 2 3 3 3 3 3 4 4


We observe that the optimal number of detours dopt is increasing in δ. On the otherhand dopt behaves unexpectedly irregular in k. For small values of δ the optimal numberof detours is small and identical for all k. But with increasing δ some peaks occur. Theoptimal number of detours jumps up and down. At first sight this seems to be an odd-eveneffect. For odd values of k the optimal number of detours is unimodal in k. It first increasesthen decreases. For even values of k the alternative should use the full number of detours,whereas for k − 1 and k + 1 the optimal number is significantly smaller.

Definition 2.1We say that dopt = d(n, r, δ, k) has a peak in k (k ∈ N and 0 < k < 2n) if for a givencombination of the model parameters n ∈ N, r ∈ R and δ ∈ R it holds that:

d(n, r, δ, k − 1) < d(n, r, δ, k) > d(n, r, δ, k + 1).

The minimal difference of the optimal number of detours in k to k + 1 and k − 1:

min(d(n, r, δ, k)− d(n, r, δ, k − 1), d(n, r, δ, k)− d(n, r, δ, k + 1))

we call the height of the peak.

4

2.1 The k - Model

We originally expected that the optimal number of detours is unimodal in k as it seemsto hold if we focus on odd values for k only. The existence of a single peak does not provethat the assumption of unimodality is wrong. Thus it is much more interesting if there aremultiple peaks with a gap in between.

Definition 2.2We say that dopt = d(n, r, δ, k) has a gap in k (k ∈ N and 0 < k < 2n) if for a givencombination of the model parameters n ∈ N, r ∈ R and δ ∈ R it holds that:

d(n, r, δ, k − 1) > d(n, r, δ, k) < d(n, r, δ, k + 1).

The minimal difference of the optimal number of detours in k to k + 1 and k − 1:

min(d(n, r, δ, k)− d(n, r, δ, k − 1), d(n, r, δ, k)− d(n, r, δ, k + 1))

we call the height of the gap.

Next we show that in most cases the shown solutions are really the only optimal solu-tions. There do not exist other sets of optimal solutions for the different values of k suchthat d(n, r, δ, k) can be made smooth in k by the right choice of solutions. Table 4 shows thethree best results for each value of k. Let d2 denote the second best number of detours andd3 the third best number of detours. Furthermore let Ek

i denote the expected minimal valueE[min(wk(s0), wk(si))].

dopt Ekdopt

d2 Ekd2− Ek

doptd3 Ek

d3− Ek

dopt

1 2 5.190 3 0.000 1 0.0202 5 5.389 3 0.011 2 0.0133 3 5.629 2 0.002 4 0.0114 5 5.857 3 0.014 2 0.014

k 5 2 6.121 3 0.000 4 0.0086 5 6.370 2 0.008 3 0.0107 2 6.643 3 0.006 1 0.0148 2 6.916 5 0.007 1 0.0139 2 7.200 1 0.010 3 0.020

Table 4: The three best results for the number of detours to use — [ k-model with n = 5, r = 0.1 andδ = 0.5 ]

So far peaks occurred only in even values of k. Astonishingly this doesn’t hold in general.For n = 11 we noticed the first peaks in odd values of k (see columns δ = 0.8, 0.9, 1.0 inTable 5). The most peaks still occur in even values of k, but not only. This is no isolatedcase. 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 thatproduce really high peaks, values in the range of δ ≈ nr are conspicuous. Table 6 shows the

5

2.1 The k - Model

δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0 3.0

1 1 1 2 2 3 3 4 4 5 6 7 8 9 10 112 1 1 2 3 3 4 4 11 11 11 11 11 11 11 113 1 1 2 3 3 4 4 5 6 7 8 9 10 10 114 1 1 2 3 3 4 5 11 11 11 11 11 11 11 115 1 1 2 3 3 4 5 5 6 8 9 9 10 10 116 1 1 2 3 3 4 10 11 11 11 11 11 11 11 117 1 1 2 3 3 4 9 10 10 10 9 9 10 10 118 1 1 2 3 3 8 9 11 11 11 11 11 11 11 119 1 1 2 3 3 8 11 10 10 10 10 10 10 10 10

10 1 1 2 2 3 8 8 10 11 11 11 11 11 11 11k 11 1 1 2 2 3 4 8 11 10 10 10 10 10 10 10

12 1 1 2 2 3 3 4 9 10 11 11 11 11 11 1113 1 1 2 2 3 3 3 11 11 10 10 10 10 10 1014 1 1 2 2 3 3 3 8 9 11 11 11 11 11 1115 1 1 2 2 2 3 3 4 4 9 10 10 10 10 916 1 1 2 2 2 3 3 3 4 10 11 11 11 11 1117 1 1 2 2 2 3 3 3 3 4 5 5 5 6 718 1 1 2 2 2 2 3 3 3 4 4 4 11 11 1119 1 1 2 2 2 2 3 3 3 3 4 4 4 4 520 1 1 2 2 2 2 2 3 3 3 3 3 4 4 421 1 1 2 2 2 2 2 3 3 3 3 3 4 4 4


optimal numbers of detours for this special case δ = nr for different problem sizes n. Sincethe number of possible perturbations

(2nk

)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 cal-culations to k ≤ 6.

n3 4 5 6 7 8 9 10 12 14 16 18 20 25 30

1 1 2 2 3 3 4 4 5 6 7 8 9 10 12 152 3 4 5 6 7 8 9 10 12 14 16 18 20 25 303 1 2 3 3 4 5 5 6 7 8 10 11 12 15 18

k 4 3 4 5 6 7 8 9 10 12 14 16 18 20 25 305 1 2 2 3 4 5 6 6 8 9 11 12 14 18 216 4 5 6 7 8 9 10 12 14 16 18 20 25 30

Table 6: The optimal number of detours dopt of the alternative solution S1 — [ k-model with fixed r = 0.1and δ = nr ]

Table 6 shows that with δ = nr the alternative should always use the full number ofdetours if k is even. For odd values of k the optimal number of detours seems to be aconstant part of n. Table 7 shows the quotients dopt

n . For k = 1 the optimal number ofdetours dopt seems to be about 1

2n, for k = 3 about 35n and for k = 5 about 2

3 .

6

2.1 The k - Model

n3 4 5 6 7 8 9 10 12 14 16 18 20 25 30

1 13

12

25

12

37

12

49

12

12

12

12

12

12

1225

12

k 3 13

12

35

12

47

58

59

610

711

47

58

1118

35

35

35

5 13

12

25

12

47

58

23

610

23

914

1116

23

710

1825

2130

Table 7: Quotientsdopt

n— [ k-model with fixed r = 0.1 and δ = nr ]

2.1.1 Theoretical Analysis of the Special Case δ = nr

k = 1

First we are interested in the expected minimal value Ek=1i = E[min(wk=1(s0), wk=1(si))]

if exactly one edge gets perturbed. Let the set of edges E of the graph G = {V, E, w} bedivided into these four parts:

A = {e0,1, e0,2, . . . , e0,i} C = {e0,i, e0,i+1, . . . , e0,n}B = {e1,1, e1,2, . . . , e1,i} D = {e1,i, e1,i+1, . . . , e1,n}

The solution s0 uses all edges of A and C and solution si uses all edges of B and C. Lete∗ be the edge that gets perturbed. Four cases are possible:

wk=1(s0) wk=1(si) frequency

e∗ ∈ A : min( n + nr , n + ir ) = n + ir i2n

e∗ ∈ B : min( n , n + ir + (1 + r)nr ) = n ↖

e∗ ∈ C : min( n + nr , n + ir + nr ) = n + nr n−i2n

e∗ ∈ D : min( n , n + ir ) = n ↖

Now we are able to give the formula for the expected minimal value Ek=1i :

Ek=1i = E[min(wk=1(s0), wk=1(si))]

=i

2n· (n + ir) +

i

2n· (n) +

n− i

2n· (n + nr) +

n− i

2n· (n)

=i

2n· (2n + ir) +

n− i

2n· (2n + nr)

=i2r − inr + n2(r + 2)

2n

7

2.1 The k - Model

and check the assumption d(n, r, δ = nr, k = 1) ≈ n2 . If the assumption is true the following

condition must hold for all 0 < i ≤ n with i 6= n2 . Obviously the number of detours i the

alternative solution si uses has to be a discrete number. Nevertheless we consider i ∈ R forthe analysis, but have to interpret the results with care.

0 < Ek=1i − Ek=1

n2

=i2r − inr + n2(r + 2)

2n−

n2

2r − n2 nr + n2(r + 2)

2n

=(i2 − n2

4 )r + (n2 − i)nr

2n

=(i2 + n2

4 − in)r2n

=(2i− n)2r

8n

We immediately see that the right side of the inequality never gets negative.

Lemma 2.3For the k-model with δ = nr and arbitrary r > 0 the alternative should always use about n

2detours if exactly k = 1 edge gets perturbed. If n is even the alternative has to use exactlyn2 detours. If n is odd the alternative has to use n−1

2 or n+12 detours which are equally good.

k = 2

The case k = 2 we check in a similar way. Here already ten cases can occur. The first columngives the sets where the perturbations occur.


A,A : min( n + 2nr , n + ir ) = n + ir(i2)

(2n2 )

B, B : min( n , n + ir + 2(1 + r)nr ) = n ↖

C, C : min( n + 2nr , n + ir + 2nr ) = n + 2nr(n−i

2 )(2n

2 )D, D : min( n , n + ir ) = n ↖

A,B : min( n + nr , n + ir + (1 + r)nr ) = n + nr i2

(2n2 )

A,C : min( n + 2nr , n + ir + nr ) = n + ir + nr i(n−i)

(2n2 )

A,D : min( n + nr , n + ir ) = n + ir ↖B, C : min( n + nr , n + ir + (1 + r)nr + nr ) = n + nr ↖B,D : min( n , n + ir + (1 + r)nr ) = n ↖

C, D : min( n + nr , n + ir + nr ) = n + nr (n−i)2

(2n2 )

8

2.1 The k - Model

So the expected minimal value for k = 2 is:


=

(i2

)(2n2

) · (2n + ir) +

(n−i2

)(2n2

) · (2n + 2nr) +i2(2n2

) · (n + nr)

+i(n− i)(

2n2

) · (4n + 2nr + 2ir) +(n− i)2(

2n2

) · (n + nr)

=i(i− 1)(2n + ir)

2n(2n− 1)+

(n− i)(n− i− 1)(2n + 2nr)2n(2n− 1)

+i2(n + nr)n(2n− 1)

+i(i− 1)(4n + 2nr + 2ir)

n(2n− 1)+

(n− i)2(n + nr)n(2n− 1)

= . . . =−3ri3 + (6nr − r)i2 + (−4n2r + 2nr)i + (4n3r + 4n3 − 2n2 − 2n2r)

2n(2n− 1)

We assume that solution sn is optimal. So we check whether the difference of Ek=2i and

Ek=2n is greater than zero for all 0 < i < n.

0 < Ek=2i − Ek=2

n =−3ri3 + (6nr − r)i2 + (−4n2r + 2nr)i + (4n3r + 4n3 − 2n2 − 2n2r)

2n(2n− 1)

− 4n3 + 3n3r − n2r − 2n2

2n(2n− 1)

=r(−3i3 − i2 + 6ni2 − 4n2i + 2ni + n3 − n2)

2n(2n− 1)

=r(n− i)(3i2 + i− 3ni− n + n2)

2n(2n− 1)⇔ 0 < 3i2 + i− 3ni− n + n2 =: f(i, n)

At first sight we can not examine wether f(i, n) is strictly positive. So we check for roots:

0 = 3i2 + i− 3ni− n + n2

⇔ i =n

2− 1

6±√−3n2 + 6n + 1

6︸︷︷︸∈ C for n ≥ 3, n ∈ N

Since there are no real roots for n ≥ 3 we set i = 1 and n = 3 and see that Ek=2i −Ek=2

n = r30

is greater than zero. So the above inequality has to be true for n ≥ 3.

Lemma 2.4For the k-model with δ = nr, n ≥ 3 and arbitrary r > 0 the alternative should always use ndetours if exactly k = 2 edges get perturbed.

9

2.1 The k - Model

k = 3

Again we enumerate all possible cases for exactly k = 3 perturbations.


A,A, A : min( n + 3nr , n + ir ) = n + ir(i3)

(2n3 )

B, B,B : min( n , n + ir + 3(1 + r)nr ) = n ↖

C, C, C : min( n + 3nr , n + ir + 3nr ) = n + 3nr(n−i

3 )(2n

3 )D,D, D : min( n , n + ir ) = n ↖

A,A, B : min( n + 2nr , n + ir + (1 + r)nr ) = n + nr + m1(i2)i

(2n3 )

B, B,A : min( n + nr , n + 2(1 + r)nr + ir ) = n + nr ↖

A,A, C : min( n + 3nr , n + ir + nr ) = n + ir + nr(i2)(n−i)

(2n3 )

A,A, D : min( n + 2nr , n + ir ) = n + ir ↖B, B, C : min( n + nr , n + 2(1 + r)nr + ir + nr ) = n + nr ↖B,B, D : min( n , n + 2(1 + r)nr + ir ) = n ↖

C,C, A : min( n + 3nr , n + ir + 2nr ) = n + ir + 2nr(n−i

2 )i

(2n3 )

C,C, B : min( n + 2nr , n + ir + (1 + r)nr + 2nr ) = n + 2nr ↖D, D, A : min( n + nr , n + ir ) = n + ir ↖D, D, B : min( n , n + ir + (1 + r)nr ) = n ↖

C,C, D : min( n + 2nr , n + ir + 2nr ) = n + 2nr(n−i

2 )(n−i)

(2n3 )

D, D,C : min( n + nr , n + ir + nr ) = n + nr ↖

A,B, C : min( n + 2nr , n + ir + (1 + r)nr + nr ) = n + 2nr i2(n−i)

(2n3 )

A,B, D : min( n + nr , n + ir + (1 + r)nr ) = n + nr ↖

A,C, D : min( n + 2nr , n + ir + nr ) = n + ir + nr i(n−i)2

(2n3 )

B,C, D : min( n + nr , n + ir + (1 + r)nr + nr ) = n + nr ↖

For the case that two perturbations are in A and one is in B it happens the first timethat we are not able to decide whether to take s0 or s1. The remaining minimum term ism1 = min(nr, ir + nr2). Then the expected minimal value Ek=3

i is:

10

2.1 The k - Model


=1(2n3

)((

i

3

)· (2n + ir)

︸︷︷︸=: g1

+(

n− i

3

)· (2n + 3nr)

︸︷︷︸=: g2

+(

i

2

)i · (m1 + 2n + 2nr)

︸︷︷︸=: g3

+(

i

2

)(n− i) · (4n + 2nr + 2ir)

︸︷︷︸=: g4

+(

n− i

2

)i · (4n + 4nr + 2ir)

︸︷︷︸=: g5

+(

n− i

2

)(n− i) · (2n + 3nr)

︸︷︷︸=: g6

+(i2(n− i)

)· (2n + 3nr)

︸︷︷︸=: g7

+(i(n− i)2

)· (2n + 2nr + ir)

︸︷︷︸=: g8

)

=6

2n(2n− 1)(2n− 2)·(g1 + g2 + g3 + g4 + g5 + g6 + g7 + g8

)

with

g1 =i(i− 1)(i− 2)

6(2n + ir) =

(i− 2)(2ni2 + ri3 − 2ni− ri2)6

=2ni3 + ri4 − 2ni2 − ri3 − 4ni2 − 2ri3 + 4ni + 2ri2

6

=16

(ri4 + (2n− 3r)i3 + (−6n + 2r)i2 + (4n)i

)

g2 =(n− i)(n− i− 1)(n− i− 2)

6(2n + 3nr)

=(n− i− 1)(n− i− 2)

6(2n2 + 3n2r − 2ni− 3nri)

=(n− i− 2)

6(2n3 + 3n3r − 2n2i− 3n2ri− 2n2i− 3n2ri + 2ni2

+ 3nri2 − 2n2 − 3n2r + 2ni + 3nri)

=(n− i− 2)

6(2n3 + 3n3r − 4n2i− 6n2ri + 2ni2 + 3nri2 − 2n2 − 3n2r + 2ni + 3nri)

11

2.1 The k - Model

=16

(2n4 + 3n4r − 4n3i− 6n3ri + 2n2i2 + 3n2ri2 − 2n3 − 3n3r + 2n2i + 3n2ri)

+ (−2n3i− 3n3ri + 4n2i2 + 6n2ri2 − 2ni3 − 3nri3 + 2n2i + 3n2ri− 2ni2 − 3nri2)

+ (−4n3 − 6n3r + 8n2i + 12n2ri− 4ni2 − 6nri2 + 4n2 + 6n2r − 4ni− 6nri))

=16

((−2n− 3nr)i3 + (6n2 + 9n2r − 6n− 9nr)i2

+ (−6n3 − 9n3r + 12n2 + 18n2r − 4n− 6nr)i

+ (2n4 + 3n4r − 6n3 − 9n3r + 4n2 + 6n2r))

g3 =i2(i− 1)(m1 + 2n + 2nr)

2=

12

(m1i

3 + 2ni3 + 2nri3 −m1i2 − 2ni2 − 2nri2

)

g4 =(n− i)i(i− 1)(4n + 2nr + 2ir)

2=

(n− i)(4ni2 + 2nri2 + 2i3r − 4ni− 2nri− 2i2r)2

=12

(4n2i2 + 2n2ri2 + 2ni3r − 4n2i− 2n2ri− 2nri2

− 4ni3 − 2nri3 − 2ri4 + 4ni2 + 2nri2 + 2ri3)

=12

(− 2ri4 + (2r − 4n)i3 + (4n2 + 2n2r + 4n)i2 + (−4n2 − 2n2r)i

)

g5 =i(n− i)(n− i− 1)(4n + 4nr + 2ir)

2

=n− i− 1

2

(4n2i + 4n2ri + 2nri2 − 4ni2 − 4nri2 − 2ri3

)

=12

(4n3i + 4n3ri + 2n2ri2 − 4n2i2 − 4n2ri2 − 2nri3

− 4n2i2 − 4n2ri2 − 2nri3 + 4ni3 + 4nri3 + 2ri4

− 4n2i− 4n2ri− 2nri2 + 4ni2 + 4nri2 + 2ri3)

=12

(2ri4 + (4n + 2r)i3 + (−6n2r − 8n2 + 2nr + 4n)i2 + (4n3 + 4n3r − 4n2 − 4n2r)i

)

12

2.1 The k - Model

g6 =(n− i)(n− i− 1)(n− i)(2n + 3nr)

2=

(n− 2ni + i2)(n− i− 1)(2n + 3nr)2

=n− 2ni + i2

2

(2n2 + 3n2r − 2ni− 3nri− 2n− 3nr

)

=12

(2n3 + 3n3r − 2n2i− 3n2ri− 2n2 − 3n2r − 4n3i− 6n3ri + 4n2i2 + 6n2ri2

+ 4n2i + 6n2ri + 2n2i2 + 3n2ri2 − 2ni3 − 3nri3 − 2ni2 − 3nri2)

=12

((−2n− 3nr)i3 + (6n2 + 9n2r − 2n− 3nr)i2

+ (2n2 + 3n2r − 4n3 − 6n3r)i + (2n3 + 3n3r − 2n2 − 3n2r))

g7 = i2(n− i)(2n + 3nr) = 2n2i2 + 3n2ri2 − 2ni3 − 3nri3 = (−2n− 3nr)i3 + (2n2 + 3n2r)i2

g8 = i(n− i)2(2n + 2nr + ir) = (n2 − 2ni + i2)(2ni + 2nri + ri2)

= 2n3i + 2n3ri + rn2i2 − 4n2i2 − 4n2ri2 − 2nri3 + 2ni3 + 2nri3 + ri4

= ri4 + (2n)i3 + (rn2 − 4n2 − 4n2r)i2 + (2n3 + 2n3r)i

Ek=3i =

62n(2n− 1)(2n− 2)

·(g1 + g2 + g3 + g4 + g5 + g6 + g7 + g8

)= . . .

=1

2n(2n− 1)(2n− 2)·(

7ri4 + (3m1 + 9r − 24nr)i3 + (24n2r + 2r − 18nr − 3m1)i2

+ (18n2r − 12n3r − 6nr)i + (12n4r − 18n3r + 6n2r + 4n2 − 12n3 + 8n4)

)

and

m1 = min(nr, ir + nr2)

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 ob-servation that the optimal number of detours dopt ≈ 3

5 for r = 0.1 is true for arbitrary values

13

2.1 The k - Model

of r and n. We must not forget that the alternative can use only a discrete number of de-tours. 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 = 0.001, 0.002, . . . , nand look for the minimum in i. Table 8 shows the quotients iopt

n .

r0.01 0.02 0.05 0.10 0.20 0.30 0.50 1.00 2.00 5.00 10.00 100.00

3 0.43 0.43 0.42 0.41 0.39 0.38 0.36 0.35 0.35 0.35 0.35 0.354 0.56 0.55 0.52 0.49 0.45 0.43 0.40 0.40 0.40 0.40 0.40 0.405 0.60 0.59 0.57 0.53 0.48 0.45 0.42 0.42 0.42 0.42 0.42 0.426 0.62 0.61 0.59 0.55 0.50 0.47 0.43 0.44 0.44 0.44 0.44 0.447 0.63 0.62 0.60 0.57 0.51 0.48 0.44 0.46 0.46 0.46 0.46 0.468 0.64 0.63 0.61 0.57 0.52 0.49 0.45 0.47 0.47 0.47 0.47 0.479 0.64 0.64 0.62 0.58 0.53 0.50 0.45 0.48 0.48 0.48 0.48 0.48

10 0.65 0.64 0.62 0.58 0.53 0.50 0.46 0.48 0.48 0.48 0.48 0.4820 0.67 0.66 0.64 0.61 0.56 0.52 0.47 0.51 0.51 0.51 0.51 0.51

n 30 0.67 0.67 0.65 0.61 0.56 0.53 0.48 0.53 0.53 0.53 0.53 0.5340 0.68 0.67 0.65 0.62 0.57 0.53 0.53 0.53 0.53 0.53 0.53 0.5350 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.53 0.54 0.54 0.54 0.5460 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.54 0.54 0.54 0.54 0.5470 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.54 0.54 0.54 0.54 0.5480 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.54 0.54 0.54 0.54 0.5490 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.54 0.54 0.54 0.54 0.54

100 0.68 0.67 0.65 0.62 0.57 0.53 0.54 0.54 0.54 0.54 0.54 0.54150 0.68 0.68 0.66 0.62 0.57 0.54 0.54 0.54 0.54 0.54 0.54 0.54200 0.68 0.68 0.66 0.62 0.57 0.54 0.55 0.55 0.55 0.55 0.55 0.55

Table 8:iopt

n— [ k-model with fixed k = 3 and δ = nr ]

Unfortunately we have to find out that iopt

n depends on n and r. It varies betweeniopt

n = 0.35 for small n and large r and iopt

n = 0.68 for large n and small r. So we don’tknow the value for i such that Ek=3

i gets minimal and have to calculate it first. We couldnot eliminate all minimum terms and have to distinguish two cases:


={

nr : n(1− r) ≤ i ≤ n ← case (1)ir + nr2 : 0 < i < n(1− r) ← case (2)

14

2.1 The k - Model

We define two functions f1 and f2:

f1(i, n, r) : = Ek=3i for i ≥ n(1− r)

=1

2n(2n− 1)(2n− 2)

(7ri4 + (−21nr + 9r)i3 + (24n2r − 21nr + 2r)i2

+ (−12n3r + 18n2r − 6nr)i + 12n4r + 8n4 − 18n3r − 12n3 + 6n2r + 4n2)

f2(i, n, r) : = Ek=3i for i < n(1− r)

=1

2n(2n− 1)(2n− 2)

(10ri4 + (3nr2 − 24nr + 6r)i3 + (24n2r − 3nr2 − 18nr + 2r)i2

+ (−12n3r + 18n2r − 6nr)i + 12n4r + 8n4 − 18n3r − 12n3 + 6n2r + 4n2)

We want to show that iopt ≤ 710n. For the case that n(1 − r) ≤ 7

10n we need to look atf1 only. The first partial derivative ∂

∂if1 is:

∂

∂if1 =

r(28i3 + (−63n + 27)i2 + (48n2 − 42n + 4)i + (−12n3 + 18n2 − 6n))2n(2n− 1)(2n− 2)

The increase rate of f1 for i = 710n and i = n is strictly positive, since

∂

∂if1 > 0 ⇔ d(i) := 28i3 + (−63n + 27)i2 + (48n2 − 42n + 4)i + (−12n3 + 18n2 − 6n) > 0

and

d(710

n) =9604n3

1000− 3087n3

100+

1323n2

100+

336n3

10− 294n2

10+

28n

10− 12n3 + 18n2 − 6n

=167500

n3 +183100

n2 − 165

n > 0 for n ≥ 2

d(n) = 28n3 − 63n3 + 27n2 + 48n3 − 42n2 + 4n− 12n3 + 18n2 − 6n

= n3 + 3n2 − 2n > 0 for n ≥ 1.

The second partial derivative of f1 in i is:

∂2

∂i2f1 =

84ri2 − 2r(63n− 27)i− r(−4 + 42n− 48n2)2n(2n− 1)(2n− 2)

15

2.1 The k - Model

The solutions of ∂2

∂i2f1(i, n, r) = 0 are:

i = − 928

+34n±

∈ C for n ≥ 4︷︸︸︷√393 + 126n− 63n2

84

Thus f1 has no turning point. Thus we know that f1 is monotonically increasing at least forn ≥ 4, r > 0 and i ≥ 7

10n.

If 710n < n(1− r) what is exactly if r < 3

10 we still have to look at f2. Again we need twosteps:

(a) The increase of f2 for i = 710n and i = n is larger than zero:

∂

∂if2 =

r(40i3 + (9nr − 72n + 18)i2 + (48n2 − 6nr − 36n + 4)i + (−12n3 + 18n2 − 6n))2n(2n− 1)(2n− 2)

(i) 0 <∂

∂if2(i =

710

n) =r(4n2 + 162n− 320 + 441n2r − 420nr)

200(2n− 1)(2n− 2)

⇔ 0 < 4n2 + 162n− 320 + 441n2r − 420nr

⇔ r >4n2 + 162n− 320−441n2 + 420n

⇔ n ≥ 2

(ii) 0 <∂

∂if2(i = n) =

r(4n2 + 9n2r − 2− 6nr)2(2n− 1)(2n− 2)

⇔ n ≥ 2

(b) f2 has no turning point:

∂2

∂i2f2 =

r(60i2 + (9nr − 72n + 18)i− 18n− 3nr + 24n2 + 2)n(2n− 1)(2n− 2)

0 =r(60i2 + (9nr − 72n + 18)i− 18n− 3nr + 24n2 + 2)

n(2n− 1)(2n− 2)

⇔ i = − 340

nr +35n− 3

20±√

81n2r2 − 1296n2r − 576n2 + 1044nr + 1728n− 156120

16

2.1 The k - Model

Since it holds for r ≤ 310 that

81n2r2 − 1296n2r − 576n2 + 1044nr + 1728n− 156

≤ 7.29n2 − 388.8n2 − 576n2 + 313.2n + 1728n− 156

= −957.51n2 + 2041.2n− 156 < 0 ⇐ n ≥ 3

both roots get complex.Thus f2 is monotonically increasing for n ≥ 4, 0 < r < 3

10 and i ≥ 710n, too.

⇒ For δ = nr, n ≥ 4 and r > 0 the optimal number of detours of the alternative solutionis at most d 7

10ne.

This holds for n = 3 as well. We simply insert n = 3 into Ek=3i . Then the expected

minimal values are:

Ek=31 = 3 +

3910

r

Ek=32 = 3 +

3710

r +110

min(3r, 3r2 + 2r)

Ek=33 = 3 +

4210

r

Since

Ek=32 − Ek=3

1 =110

min(3r, 3r2 + 2r)− 210

r

=r

10(min(3, 3r + 2)− 2) > 0

and Ek=33 − Ek=3

1 =310

r > 0

the optimal number of detours is dopt = 1 for all r > 0.

Lemma 2.5For the k-model with δ = nr, n ≥ 3 and r > 0 the optimal number of detours of thealternative solution is at most d 7

10ne.From Lemma 2.3 - 2.5 it directly follows Proposition 2.6.

Proposition 2.6The k-model with δ = nr, n ≥ 3 and r > 0 has a peak in k = 2. The height of the peak is atleast b 3

10nc. For k = 1 the optimal number of detours is n2 for even values of n and n±1

2 forodd values of n. For k = 2 the optimal number of detours is n. And for k = 3 the optimalnumber of detours is at most d 7

10ne.

17

2.1 The k - Model

k = 4

To prove the existence of a gap we need to look at k = 4. As the experiments showed it isto assume that the optimal number of detours is equal to n. The number of possible casesgets quite large now:


A,A, A,A : min( n + 4nr , n + ir ) = n + ir(i4)

(2n4 )

B,B, B, B : min( n , n + ir + 4(1 + r)nr ) = n ↖

C, C,C, C : min( n + 4nr , n + ir + 4nr ) = n + 4nr(n−i

4 )(2n

4 )D, D,D, D : min( n , n + ir ) = n ↖

A,A, A,B : min( n + 3nr , n + ir + (1 + r)nr ) = n + nr + m2(i3)i

(2n4 )

B,B, B, A : min( n + nr , n + ir + 3(1 + r)nr ) = n + nr ↖

A,A, A, C : min( n + 4nr , n + ir + nr ) = n + ir + nr(i3)(n−i)

(2n4 )

A,A,A, D : min( n + 3nr , n + ir ) = n + ir ↖B, B,B,C : min( n + nr , n + ir + nr + 3(1 + r)nr ) = n + nr ↖B, B, B, D : min( n , n + ir + 3(1 + r)nr ) = n ↖

C, C, C,A : min( n + 4nr , n + ir + 3nr ) = n + ir + 3nr(n−i

3 )i

(2n4 )

C, C, C,B : min( n + 3nr , n + ir + 3nr + (1 + r)nr ) = n + 3nr ↖D, D,D, A : min( n + nr , n + ir ) = n + ir ↖D, D,D, B : min( n , n + ir + (1 + r)nr ) = n ↖

C, C, C,D : min( n + 3nr , n + ir + 3nr ) = n + 3nr(n−i

3 )(n−i)

(2n4 )

D,D, D,C : min( n + nr , n + ir + nr ) = n + nr ↖

A,A, B, B : min( n + 2nr , n + ir + 2(1 + r)nr ) = n + 2nr(i2)(

i2)

(2n4 )

A, A,C, C : min( n + 4nr , n + ir + 2nr ) = n + ir + 2nr(i2)(

n−i2 )

(2n4 )

A,A, D, D : min( n + 2nr , n + ir ) = n + ir ↖B, B, C, C : min( n + 2nr , n + ir + 2nr + 2(1 + r)nr ) = n + 2nr ↖B, B,D, D : min( n , n + ir + 2(1 + r)nr ) = n ↖

C, C, D, D : min( n + 2nr , n + ir + 2nr ) = n + 2nr(n−i

2 )(n−i2 )

(2n4 )

18

2.1 The k - Model


A, A,C, D : min( n + 3nr , n + ir + nr ) = n + ir + nr(i2)(n−i)2

(2n4 )

B, B, C,D : min( n + nr , n + ir + nr + 2(1 + r)nr ) = n + nr ↖

C, C, A,B : min( n + 3nr , n + ir + 2nr + (1 + r)nr ) = n + 3nr(n−i

2 )i2

(2n4 )

D, D, A,B : min( n + nr , n + ir + (1 + r)nr ) = n + nr ↖

A,A, B, C : min( n + 3nr , n + ir + nr + (1 + r)nr ) = n + 2nr + m1(i2)i(n−i)

(2n4 )

A, A,B, D : min( n + 2nr , n + ir + (1 + r)nr ) = n + nr + m1 ↖B,B,A, C : min( n + 2nr , n + ir + nr + 2(1 + r)nr ) = n + 2nr ↖B, B, A,D : min( n + nr , n + ir + 2(1 + r)nr ) = n + nr ↖

C, C,A, D : min( n + 3nr , n + ir + 2nr ) = n + ir + 2nr(n−i

2 )i(n−i)

(2n4 )

C, C,B,D : min( n + 2nr , n + ir + 2nr + (1 + r)nr ) = n + 2nr ↖D, D, A, C : min( n + 2nr , n + ir + nr ) = n + ir + nr ↖D, D, B, C : min( n + nr , n + ir + nr + (1 + r)nr ) = n + nr ↖

A, B,C, D : min( n + 2nr , n + ir + nr + (1 + r)nr ) = n + 2nr i2(n−i)2

(2n4 )

It remain the two minimum terms:


and m2 = min(2nr, ir + nr2)

The expected minimal value Ek=4i is:


=

(i4

)(2n4

) · (2n + ir) +

(n−i4

)(2n4

) · (2n + 4nr) +

(i3

)i(

2n4

) · (2n + 2nr + m2)

+

(i3

)(n− i)(2n4

) · (4n + 2nr + 2ir) +

(n−i3

)i(

2n4

) · (4n + 6nr + 2ir)

+

(n−i3

)(n− i)(2n4

) · (2n + 4nr) +

(i2

)(i2

)(2n4

) · (n + 2nr) +

(i2

)(n−i2

)(2n4

) · (4n + 4nr + 2ir)

19

2.1 The k - Model

+

(n−i2

)(n−i2

)(2n4

) · (n + 2nr) +

(i2

)(n− i)2(

2n4

) · (2n + 2nr + ir) +

(n−i2

)i2(

2n4

) · (2n + 4nr)

+

(i2

)i(n− i)(

2n4

) · (4n + 6nr + 2m1) +

(n−i2

)i(n− i)(2n4

) · (4n + 6nr + 2ir)

+i2(n− i)2(

2n4

) · (n + 2nr)

= . . . =1

2n(2n− 1)(2n− 2)(2n− 3)(− 15ri5 + (−42r − 24m1 + 4m2 + 76nr)i4

+ (24m1 − 12m2 + 24nm1 − 33r − 120n2r + 132nr)i3

+ (−24nm1 − 6r − 144n2r + 80n3r + 72nr + 8m2)i2

+ (−88n2r + 96n3r − 32n4r + 24nr)i

+ (44n3 − 48n4 + 16n5 − 12n2 − 24n2r − 96n4r + 32n5r + 88n3r))

As for the case of k = 3 perturbations we first check the assumption dopt = n experimen-tally. For each combination of r and n to check we insert values i = 0.001, 0.002, . . . , n andlook for the minimum in i. Table 9 shows the quotients iopt

n . We have to recognize that theassumption dopt = n is not true for all values of r. There seems to be a r∗(n) such that theassumption is only true for r < r∗(n). Figure 2 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 = 4 in theway we proved the observations for the case k = 3 would fill a lot of pages. The expenditureis out of all proportion to the result.

Unfortunately this way we are not able to give a proposition that ensures the existenceof a gap. But with the given formula for Ek=4

i we feel save enough to state our observationas an assertion.

Conjecture 2.7For a given problem size n it exists a r∗(n) > 0 such that the k-model with δ = nr, n ≥ 3and 0 < r < r∗(n) has a gap in k = 3.

20

2.1 The k - Model

r0.01 0.02 0.05 0.10 0.20 0.30 0.50 1.00 2.00 5.00 10.0 100.0

3 1.00 1.00 1.00 1.00 0.28 0.27 0.26 0.24 0.25 0.25 0.25 0.254 1.00 1.00 1.00 1.00 1.00 0.34 0.31 0.29 0.29 0.29 0.29 0.295 1.00 1.00 1.00 1.00 1.00 0.38 0.34 0.32 0.32 0.32 0.32 0.326 1.00 1.00 1.00 1.00 1.00 0.41 0.35 0.34 0.34 0.34 0.34 0.347 1.00 1.00 1.00 1.00 1.00 1.00 0.37 0.36 0.35 0.35 0.35 0.358 1.00 1.00 1.00 1.00 1.00 1.00 0.38 0.37 0.36 0.36 0.36 0.369 1.00 1.00 1.00 1.00 1.00 1.00 0.39 0.38 0.37 0.37 0.37 0.37

10 1.00 1.00 1.00 1.00 1.00 1.00 0.39 0.39 0.37 0.37 0.37 0.3720 1.00 1.00 1.00 1.00 1.00 1.00 0.42 0.42 0.40 0.40 0.40 0.40

n 30 1.00 1.00 1.00 1.00 1.00 1.00 0.43 0.43 0.41 0.41 0.41 0.4140 1.00 1.00 1.00 1.00 1.00 1.00 0.43 0.44 0.41 0.41 0.41 0.4150 1.00 1.00 1.00 1.00 1.00 1.00 0.44 0.44 0.42 0.42 0.42 0.4260 1.00 1.00 1.00 1.00 1.00 1.00 0.44 0.45 0.42 0.42 0.42 0.4270 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.45 0.42 0.42 0.42 0.4280 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.45 0.42 0.42 0.42 0.4290 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.45 0.42 0.42 0.42 0.42

100 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.45 0.42 0.42 0.42 0.42150 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.45 0.42 0.42 0.42 0.42200 1.00 1.00 1.00 1.00 1.00 1.00 0.88 0.46 0.42 0.42 0.42 0.42

Table 9:iopt

n— [ k-model with fixed k = 4 and δ = nr ]

0

0,1

0,2

0,3

0,4

0,5

5 20 35 50 65 80 95 110 125 140 155 170 185 200

n

Figure 2: r∗(n) — [ k-model with fixed k = 4 and δ = nr ]

21

2.1 The k - Model

2.1.2 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 itat least for n = 3.

If exactly k = 4 edges get perturbed the expected minimal values are:

Ek=41 = 3 +

8215

r

Ek=42 = 3 +

7415

r +415

min(3r, 3r2 + 2r)

Ek=43 = 3 +

7215

r +315

min(6r, 3r2 + 3r)

To prove that dopt = 3 it must hold: Ek=41 > Ek=4

3 and Ek=42 > Ek=4

3 :(1)

Ek=41 − Ek=4

3 =r

15(10− 3min(6, 3r + 3)) > 0

For r ≥ 1 this is obviously wrong. So we assume r ≤ 1.

⇒ Ek=41 − Ek=4

3 =r

15(10− 3(3r + 3)) > 0

⇔ r <19

(2) With r < 19 we get for Ek=4

2 > Ek=43 :

Ek=42 − Ek=4

3 =r

15(2 + 4min(3, 3r + 2)− 3min(6, 3r + 3))

=r

15(2 + 4(3r + 2)− 3(3r + 3)) =

r

15(1 + 3r)) > 0

⇒ For r < 19 the optimal number of detours is dopt = 3.

Proposition 2.8The k-model with δ = 3r, n = 3 and 0 < r < 1

9 has a gap in k = 3. For k = 2 and k = 4 theoptimal number of detours is 3. For k = 3 the optimal number of detours is 1.

22

2.2 The p - Model

2.2 The p - Model

How many detours dopt = d(n, r, δ, p) the alternative solution S1 has to use if every edge ofthe 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 nu-merically. For a given combination of the model parameters we have to simulate – usinga pseudo random number generator – a large number of runs T and calculate the averagevalue to get an approximation for the expected minimal value. We have to do this for allpossible numbers of detours and then check which number is best.

Dependence of dopt on the perturbation intensity δ

First we have a look at the dependence on δ. Tables 10 and 11 show simulation results forthe optimal numbers of detours dopt for a fixed length of the detours r = 0.1 and problemsize n = 5 and n = 10 respectively.

δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0

0.1 1 1 2 3 3 4 4 5 5 5 5 5 5 50.2 1 1 2 3 3 4 4 5 5 5 5 5 5 50.3 1 1 2 3 3 4 5 5 5 5 5 5 5 50.4 1 1 2 2 3 4 4 5 5 5 5 5 5 5

p 0.5 1 1 2 2 3 4 4 5 5 5 5 5 5 50.6 1 1 2 2 3 3 4 4 4 5 5 5 5 50.7 1 1 2 2 2 3 3 4 4 5 5 5 5 50.8 1 1 2 2 2 2 3 3 3 4 4 5 5 50.9 1 1 2 2 2 2 3 3 3 3 3 4 4 4

Table 10: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedn = 5, r = 0.1; T = 106 runs ]

δ0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0 1.2 1.4 1.6 1.8 2.0

0.1 1 1 2 3 3 4 4 5 5 7 8 10 10 100.2 1 1 2 3 3 4 5 5 6 8 10 10 10 100.3 1 1 2 3 3 4 9 10 10 10 10 10 10 100.4 1 1 2 2 3 4 9 10 10 10 10 10 10 10

p 0.5 1 1 2 2 3 4 8 9 10 10 10 10 10 100.6 1 1 2 2 3 3 8 8 9 10 10 10 10 100.7 1 1 2 2 2 3 3 4 4 10 10 10 10 100.8 1 1 2 2 2 3 3 3 3 4 4 5 5 50.9 1 1 2 2 2 2 3 3 3 3 3 4 4 4

Table 11: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedn = 10, r = 0.1; T = 106 runs ]

As to be expected dopt increases in δ. For small values of δ the optimal number of detoursis small and equal for all tested probabilities p. For intermediate values of p (p ≈ 0.3) theoptimal number of detours increases fastest in δ such that it becomes unimodal in p. Thusthe optimal number of detours for these intermediate values of p reach the maximal value

23

2.2 The p - Model

dopt = n first. Then for larger values of δ the optimal number of detours dopt gets equal ton for the small values of p, too. Astonishingly the entries of both tables are nearly identicalfor values of δ ≤ 0.7.

Dependence of dopt on the detour length r

For the sake of completeness we have a look at some results for the dependence on r. Tables12 and 13 show simulation results for the optimal numbers of detours dopt for a fixed per-turbation intensity δ = 3.0. Obviously dopt decreases in r. Again it is conspicuous that theentries of both tables are nearly identical for values of r ≥ 0.3.

r0.1 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.60 0.70 0.80 0.90

0.1 5 5 5 5 5 4 4 3 3 2 2 2 20.2 5 5 5 5 5 4 3 3 3 2 2 2 20.3 5 5 5 5 4 3 3 2 2 2 2 2 10.4 5 5 5 5 3 3 2 2 2 2 1 1 1

p 0.5 5 5 5 4 3 2 2 2 1 1 1 1 10.6 5 5 4 3 2 2 2 1 1 1 1 1 10.7 5 5 3 2 2 2 1 1 1 1 1 1 10.8 5 4 3 2 2 1 1 1 1 1 1 1 10.9 5 3 2 2 2 1 1 1 1 1 1 1 1

Table 12: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedn = 5, δ = 3.0; T = 106 runs ]

r0.1 0.15 0.20 0.25 0.30 0.35 0.40 0.45 0.50 0.60 0.70 0.80 0.90

0.1 10 10 8 6 5 4 4 3 3 2 2 2 20.2 10 10 10 6 5 4 3 3 3 2 2 2 20.3 10 10 10 10 4 3 3 2 2 2 2 2 10.4 10 10 10 8 4 3 2 2 2 2 1 1 1

p 0.5 10 10 9 4 3 2 2 2 2 1 1 1 10.6 10 10 8 3 2 2 2 1 1 1 1 1 10.7 10 10 3 2 2 2 1 1 1 1 1 1 10.8 6 4 3 2 2 2 1 1 1 1 1 1 10.9 5 3 2 2 2 1 1 1 1 1 1 1 1

Table 13: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedn = 10, δ = 3.0; T = 106 runs ]

Dependence of dopt on the problem size n

Finally we have a look at the dependence on n. Table 14 shows what happens for the dif-ferent values of p if we increase n and let r and δ fixed. Then Table 15 shows what happensfor the different values of r if we increase n and let p and δ fixed. Finally Table 16 showswhat happens for the different values of δ if we increase n and let r = 0.1 and p = 0.1 fixed.

24

2.2 The p - Model

n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.1 2 3 4 5 5 5 5 5 6 5 6 5 5 6 5 5 5 5 50.2 2 3 4 5 6 6 6 6 6 6 12 12 6 6 13 6 6 12 120.3 2 3 4 5 6 6 6 6 10 11 12 12 12 12 12 12 12 12 110.4 2 3 4 5 6 6 8 9 10 11 11 11 11 11 11 11 11 11 11

p 0.5 2 3 4 5 5 5 8 9 10 10 10 10 10 10 10 10 10 10 110.6 2 3 4 5 5 5 8 9 9 9 9 9 9 9 10 9 9 9 90.7 2 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 40.8 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 30.9 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

Table 14: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedr = 0.1, δ = 1.0; T = 106 runs ]

n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.10 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 200.15 2 3 4 5 6 7 7 8 8 11 12 13 14 14 14 14 16 15 140.20 2 3 4 5 5 5 5 5 5 5 5 5 5 5 5 5 5 5 50.25 2 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 40.30 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 30.35 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3

r 0.40 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.45 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.50 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.55 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.60 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.65 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10.70 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

Table 15: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedp = 0.2, δ = 2.0; T = 106 runs ]

n2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

0.2 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 10.4 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 20.6 2 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 3 30.8 2 3 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 4 41.0 2 3 4 5 5 6 6 5 5 5 6 5 5 6 5 5 6 5 51.2 2 3 4 5 6 7 7 7 7 7 7 7 7 7 7 7 7 7 71.4 2 3 4 5 6 7 8 8 8 8 8 8 8 8 8 8 8 9 8

δ 1.6 2 3 4 5 6 7 8 9 10 10 10 10 9 10 10 9 10 9 101.8 2 3 4 5 6 7 8 9 10 11 11 12 12 11 12 11 12 10 202.0 2 3 4 5 6 7 8 9 10 11 12 13 12 13 12 12 13 13 202.2 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 15 14 15 152.4 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 17 15 202.6 2 3 4 5 6 7 8 9 10 11 12 13 14 15 15 17 18 18 202.8 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 203.0 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Table 16: The optimal number of detours dopt of the alternative solution S1 — [ p-model with fixedp = 0.1, r = 0.1; T = 106 runs ]

25

2.2 The p - Model

The results are quite smooth. There are no large peaks or gaps. But there are someexceptions, too. For example the row p = 0.2 in Table 14 contains an anomaly. With in-creasing n the optimal number increases up to 6 then for n = 12 jumps up to 12, jumpsback to 6 at n = 14, jumps up to 13 at n = 16, jumps back . . . . Also some other entriesdiffer from the pattern, but usually only by 1. These are maybe imprecisions caused by thesimulation.

We observe that – with some exceptions – for increasing problem size n the optimalnumber of detours dopt = d(n, r, δ, p) nearly always increases up to a number d∗ = d(r, δ, p)and then remains constant for larger values of n. Such that:

dopt(n, r, δ, p) ={

n : n < d∗(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 14)

• d∗ is decreasing in r (see Table 15)

• d∗ is increasing in δ (see Table 16)

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 < p1 < p2 < p3 ≤ 1 it should hold oneof the cases:

• dopt(p1) < dopt(p2) < dopt(p3)

• dopt(p1) < dopt(p2) > dopt(p3)

• dopt(p1) > dopt(p2) > dopt(p3).

So there exist no parameters n, r, δ, p1, p2, p3 such that dopt(p1) > dopt(p2) < dopt(p3).

Since there is no way to build up a formula for arbitrary n, r, δ we try the special casen = 3. Let Ep

i denote the expected minimal value E[min(wp(s0), wp(si))] of the solution pair.The probability that exactly k perturbations occur is: pk(1−p)n−k. By enumeration of all 64possible cases of perturbations – each time checking the minimal value of wp(s0) and wp(si)for i = 1, 2, 3 – we observed the following formulas for the expected minimal values Ep

1,Ep2

and Ep3:

Ep1 = (1− p)6(3) + p1(1− p)5(18 + 2δ + r)

+ p2(1− p)4(45 + 11δ + 4r)

+ p3(1− p)3(60 + 24δ + 6r)

+ p4(1− p)2(45 + 26δ + 4r)

+ p5(1− p)1(18 + 14δ + r) + p6(3 + 3δ)

26

2.2 The p - Model

Ep2 = (1− p)6(3) + p1(1− p)5(18 + δ + 2 min(δ, 2r))

+ p2(1− p)4(45 + 9δ + 2r + 4 min(δ, 2r))

+ p3(1− p)3(60 + 20δ + 4r + 2min(δ, 2r) + 2min(2δ, 2r + δ + rδ))

+ p4(1− p)2(45 + 20δ + 2r + 4min(2δ, 2r + δ + rδ))

+ p5(1− p)1(18 + 11δ + 2 min(2δ, 2r + δ + rδ)) + p6(3 + 3δ)

Ep3 = (1− p)6(3) + p1(1− p)5(18 + 3 min(δ, 3r))

+ p2(1− p)4(45 + 9δ + 3min(2δ, 3r))

+ p3(1− p)3(60 + 9δ + 3r + 9 min(2δ, 3r + δ + rδ))

+ p4(1− p)2(45 + 21δ + 3 min(3δ, 3r + δ + rδ))

+ p5(1− p)1(18 + 6δ + 3min(3δ, 3r + 2δ + 2rδ)) + p6(3 + 3δ)

If we want to eliminate all minimum terms we have to distinguish 24 cases and the resultingpolynomials are still of degree 6. 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 highestpeaks and gaps for the k-model, the polynomials get much more handy. There are only tworemaining minimum terms. The expected minimal values are

Ep1 = 3 + 7pr + 2p2r

Ep2 = 3 + 7pr + 2p2r + 2p3rm1 − 2p4rm1

Ep3 = 3 + (−12r + 9rm2)p6 + (36r − 18rm2)p5 + (−45r + 9rm2)p4 + 30p3r − 9p2r + 9pr

with m1 = min(1, 3r) and m2 = min(1, r).

We immediately see that Ep1 < Ep

2 for all 0 0, since:

Ep2 − Ep

1 = 2p3rm1 − 2p4rm1 = 2p3(1− p)m1 > 0

If it holds that Ep1 < Ep

3 for all 0 0 the existence of gaps is out of question.

27

3 Optimal Pairs vs. Optimal Alternatives

0 < Ep3 − Ep

1

= (−12r + 3rm2)p6 + (36r − 6rm2)p5 + (−45r + 3rm2)p4 + 30p3r − 11p2r + 2pr

= (1− p)pr(12p4 − 3p4m2 − 24p3 + 3p3m2 + 21p2 − 9p + 2)

⇔ 0 < 12p4 − 3p4m2 − 24p3 + 3p3m2 + 21p2 − 9p + 2

= 3p3(1− p)m2 + 12p4 − 24p3 + 21p2 − 9p + 2

⇐ 0 < 12p4 − 24p3 + 21p2 − 9p + 2 =: g(p)

Since

g′(p) = 48p3 − 72p2 + 42p− 9

g′(p) = 0 ⇔ p =

{1212 ± 1

4I√

12 ∈ C

and g′′(p) = 144p2 − 144p + 42

g′′(12) = 6

g(p) is minimal for p = 12 and since g(1

2) = 12 it holds that g(p) ≥ 1

2 for 0 0

⇒ dopt(p) = 1 for all 0 0

Proposition 2.9The p-model with r > 0 and 0 < p < 1 has neither a peak nor a gap for the special casen = 3, δ = 3r.


The original question was how to choose two solutions S0 and S1 such that the expectedminimal value E[min(w(S0), w(S1))] gets minimal. Is it generally optimal to fix one of thesolutions 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 ofsolutions we look at:

S = {si,j : i, j ≥ 0; i + j ≤ n}with si,j = (e0,1, . . . , e0,i, e1,i+1, . . . , e1,i+j , e0,i+j+1, . . . , e0,n)

28


The length of the path si,j is:

w(si,j) =i∑

t=1

w(e0,t) +i+j∑

t=i+1

w(e1,t) +n∑

t=i+j+1

w(e0,t)

The first i edges of a solution are non-detour edges, the next j edges are detours and theremaining n− i− j edges are non-detour edges again. Since

E[min(w(s0,i), w(si,j)] = E[min(w(s0,j), w(sj,i)]

we assume that the second solution uses at least as much detours as the first solution fromthe pair. So the set of solution pairs P to analyze is:

P = {pi,j = {s0,i, si,j} : 0 ≤ i ≤ bn2c, i ≤ j ≤ n− i}

with w(pi,j) = min(w(s0,i), w(si,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[w(pi,j)] gets minimal?

For the k - model we are able to calculate the expected minimal value of a solution pairexactly – as described in subsection 2.1. For the p - model we have to do simulations again– as described in subsection 2.2. We tested both problems with similar parameter sets:

k - model with exact calculation

k = 1, 2, . . . , 2n− 1n = 3, 4, . . . , 10r = 0.05, 0.10, . . . , 2

δ = r + 0.1, r + 0.2, . . . ,⌊n + rn

0.1

⌋0.1

p - model with T = 106 simulation runs

p = 0.1, 0.2, . . . , 0.9n = 3, 4, . . . , 10r = 0.05, 0.10, . . . , 2

δ = r + 0.1, r + 0.2, . . . ,⌊n + rn

0.1

⌋0.1

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 tohold for both models:

min0≤l≤n

E[w(p0,l)] ≤ min1≤i,j≤ni+j≤n

E[w(pi,j)]

In detail the results are identical to those of subsection 2 where we analyzed only thechoice 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 problemparameters would be a quite hard task. But we give a prove at least for n = 2 and n = 3.

29


We start with n = 2. The first column of Table 17 shows all possible cases for pertur-bations. We symbolize the graph by a 2× 2 matrix. The entries of the matrix show whichedges get perturbed (denoted by ’X’) and which edges not (denoted by ’O’). There are onlythree solution pairs to analyze. If the first solution uses no detour the second solution canuse either one or two detours. And if the first solution uses already one detour the secondhas also to use one detour.

w(p0,1) w(p0,2) w(p1,1)

OOOO 2 2 2 + r

XOOO 2 + δ 2 + min(δ, 2r) 2 + rOXOO 2 + r 2 + min(δ, 2r) 2 + rOOXO 2 2 2 + rOOOX 2 2 2 + r

XOXO 2 + δ 2 + δ 2 + r + δXOOX 2 + δ 2 + δ 2 + rXXOO 2 + r + δ 2 + 2r 2 + r + δOOXX 2 2 2 + r + δ + rδOXOX 2 + δ 2 + δ 2 + r + δOXXO 2 + r 2 + δ 2 + r

XXXO 2 + r + δ 2 + δ + min(δ, 2r + rδ) 2 + r + δXXOX 2 + 2δ 2 + δ + min(δ, 2r + rδ) 2 + r + δOXXX 2 + δ 2 + δ 2 + r + δ + rδXOXX 2 + δ 2 + δ 2 + r + δ + rδ

XXXX 2 + 2δ 2 + 2δ 2 + r + 2δ + rδ

Table 17: Enumeration of all possible cases of perturbations for n = 2

30

Then the expected minimal values for the solution pairs of the k-model are:

Ek=10,1 =

14(8 + r + δ)

Ek=20,1 =

16(12 + 2r + 4δ)

Ek=30,1 =

14(8 + r + 5δ)

Ek=10,2 =

14(8 + 2min(δ, 2r))

Ek=20,2 =

16(12 + 2r + 4δ)

Ek=30,2 =

14(8 + 4δ + 2 min(δ, 2r + rδ))

Ek=11,1 =

14(8 + 4r)

Ek=21,1 =

16(12 + 6r + 4δ + rδ)

Ek=31,1 =

14(8 + 4r + 4δ + 2rδ)

We immediately see that Ek0,2 ≤ Ek

1,1 for k = 1, 2, 3. ⇒

Lemma 3.1 (k-model)For a string of pearls Gn=2,r with detours r > 0, perturbation intensity δ > r and 0 < k < 4perturbations there is always an optimal solution pair that contains the solution s0,0. Forsome parameter combinations it exist other optimal solution pairs that do not contain thesolution s0,0. It holds:

min0≤l≤2

E[wk(p0,l)] ≤ min1≤i,j≤2i+j≤2

E[wk(pi,j)]

The expected minimal values for the solution pairs of the p-model are:

Ep0,1 = (1− p)4(2) + p(1− p)3(8 + r + δ) + p2(1− p)2(12 + 2r + 4δ)

+ p3(1− p)(8 + r + 5δ) + p4(2 + 2δ)

Ep0,2 = (1− p)4(2) + p(1− p)3(8 + 2min(δ, 2r)) + p2(1− p)2(12 + 2r + 4δ)

+ p3(1− p)(8 + 4δ + 2 min(δ, 2r + rδ)) + p4(2 + 2δ)

Ep1,1 = (1− p)4(2 + r) + p(1− p)3(8 + 4r) + p2(1− p)2(12 + 6r + 4δ + rδ)

+ p3(1− p)(8 + 4r + 4δ + 2rδ) + p4(2 + r + 2δ + rδ)

31

Again every summand of Ep0,2 is at most equal to the corresponding summand of Ep

1,1 andsome summands are strictly smaller. Thus the question is immediately answered for thep-model, too.

Lemma 3.2 (p-model)For a string of pearls Gn=2,r with detours r > 0, perturbation intensity δ > r and a pertur-bation probability 0 < p < 1 the optimal solution pair always contains the solution s0,0. Itholds:

min0≤l≤2

E[wp(p0,l)] < min1≤i,j≤2i+j≤2

E[wp(pi,j)]

We do the same procedure for n = 3. We have to look at five solution pairs: p0,1, p0,2,p0,3 and p1,1, p1,2 and 64 possible cases of perturbations. We don’t show every single case,but give the formulas for k = 0, 1, . . . , 2n. For each k we have to show that the expectedvalue of one of p0,1, p0,2, p0,3 is at least as good as the expected value of the pair p1,1. Thesame for the pair p1,2.

k = 0:Ek=0

0,1 = Ek=00,2 = Ek=0

0,3 = 3 < 3 + r = Ek=01,1 = Ek=0

1,2

k = 1:

Ek=10,1 =

16(18 + r + 2δ)

Ek=10,2 =

16(18 + δ + 2min(δ, 2r))

Ek=10,3 =

16(18 + 3 min(δ, 3r))

vs.Ek=1

1,1 =16(18 + 6r + δ)

Ek=11,2 =

16(18 + 9r)

a)

Ek=10,3 ≤ Ek=1

1,1

⇔ 16(18 + 3 min(δ, 3r)) ≤ 1

6(18 + 6r + δ)

⇔ 3min(δ, 3r) ≤ 6r + δ

⇐ (for δ ≤ 3r) : 3δ ≤ 6r + δ ↔ δ ≤ 3r

⇐ (for δ ≥ 3r) : 9r ≤ 6r + δ ↔ δ ≥ 3r

b)

Ek=10,3 ≤ Ek=1

1,2

⇔ 16(18 + 3 min(δ, 3r)) ≤ 1

6(18 + 9r)

⇔ 9r ≤ 9r

32


k = 2:

Ek=20,1 =

115

(45 + 4r + 11δ)

Ek=20,2 =

115

(45 + 2r + 9δ + 4 min(δ, 2r))

Ek=20,3 =

115

(45 + 9δ + 3min(2δ, 3r))

vs. Ek=21,1 =

115

(45 + 15r + 9δ + rδ)

Ek=21,2 =

115

(45 + 18r + 9δ + r min(1, δ) + 2rδ)

a)

Ek=20,3 ≤ Ek=2

1,1

⇔ 115

(45 + 9δ + 3 min(2δ, 3r)) ≤ 115

(45 + 15r + 9δ + rδ)

⇐ 9r ≤ 15r + rδ

b)

Ek=20,3 ≤ Ek=2

1,2

⇔ 115

(45 + 9δ + 3 min(2δ, 3r)) ≤ 115

(45 + 18r + 9δ + r min(1, d) + 2rδ)

⇐ 9r ≤ 18r + r min(1, δ) + 2rδ

k = 3:

Ek=30,1 =

120

(60 + 6r + 24δ)

Ek=30,2 =

120

(60 + 4r + 22δ + 2min(δ, 2r) + 2 min(δ, 2r + rδ))

Ek=30,3 =

120

(60 + 3r + 18δ + 9min(δ, 3r + rδ))

vs. Ek=31,1 =

120

(60 + 20r + 22δ + 4rδ)

Ek=31,2 =

120

(60 + 28r + 18δ + 2 min(δ, r + rδ) + 7rδ)

a)

Ek=30,2 ≤ Ek=3

1,1

⇔ 120

(60 + 4r + 22δ + 2 min(δ, 2r) + 2min(δ, 2r + rδ)) ≤ 120

(60 + 20r + 22δ + 4rδ)

⇔ 2min(δ, 2r) + 2min(δ, 2r + rδ) ≤ 16r + 4rδ

⇐ 8r + 2rδ ≤ 16r + 4rδ

33


b)

Ek=30,3 ≤ Ek=3

1,2

⇔ 120

(60 + 3r + 18δ + 9 min(δ, 3r + rδ)) ≤ 120

(60 + 28r + 18δ + 2 min(δ, r + rδ) + 7rδ)

⇔ 9min(δ, 3r + rδ)) ≤ 25r + 2min(δ, r + rδ) + 7rδ

⇐ (for δ ≤ r + rδ) : 9δ ≤ 25r + 7rδ + 2δ ↔ 7δ ≤ 7r + 7rδ ≤ 25r + 7rδ

⇐ (for δ ≥ r + rδ) : 27r + 9rδ ≤ 27r + 9rδ

k = 4:

Ek=40,1 =

115

(45 + 4r + 26δ)

Ek=40,2 =

115

(45 + 2r + 24δ + 4min(δ, 2r + rδ))

Ek=40,3 =

115

(45 + 24δ + 3 min(2δ, 3r + rδ))

vs. Ek=41,1 =

115

(45 + 15r + 24δ + 6rδ)

Ek=41,2 =

115

(45 + 18r + 24δ + 9rδ)

a)

Ek=40,2 ≤ Ek=4

1,1

⇔ 115

(45 + 2r + 24δ + 4 min(δ, 2r + rδ)) ≤ 115

(45 + 15r + 24δ + 6rδ)

⇔ 4min(δ, 2r + rδ) ≤ 13r + 6rδ

⇐ 8r + 4rδ ≤ 13r + 6rδ

b)

Ek=40,2 ≤ Ek=4

1,2

⇔ 115

(45 + 2r + 24δ + 4 min(δ, 2r + rδ)) ≤ 115

(45 + 18r + 24δ + 9rδ)

⇔ 4min(δ, 2r + rδ) ≤ 16r + 9rδ

⇐ 8r + 4rδ ≤ 16r + 9rδ

34

k = 5:

Ek=50,1 =

16(18 + r + 14δ)

Ek=50,2 =

16(18 + 13δ + 2 min(δ, 2r + rδ))

Ek=50,3 =

16(18 + 12δ + 3 min(δ, 3r + 2rδ))

vs. Ek=51,1 =

16(18 + 6r + 13δ + 4rδ)

Ek=51,2 =

16(60 + 8r + 12δ + min(δ, r + rδ) + 5rδ)

a)

Ek=50,3 ≤ Ek=5

1,1

⇔ 16(18 + 12δ + 3min(δ, 3r + 2rδ)) ≤ 1

6(18 + 6r + 13δ + 4rδ)

⇔ 3min(δ, 3r + 2rδ) ≤ 6r + δ + 4rδ

⇐ (for δ ≤ 3r + 2rδ) : 3δ ≤ 6r + 4rδ + δ ↔ δ ≤ 3r + 2rδ

⇐ (for δ ≥ 3r + 2rδ) : 9r + 6rδ ≤ 6r + 4rδ + δ ↔ 3r + 2rδ ≤ δ

b)

Ek=50,3 ≤ Ek=5

1,2

⇔ 16(18 + 12δ + 3min(δ, 3r + 2rδ)) ≤ 1

6(60 + 8r + 12δ + min(δ, r + rδ) + 5rδ)

⇔ 3min(δ, 3r + 2rδ) ≤ 8r + min(δ, r + rδ) + 5rδ

⇐ (for δ ≤ r + rδ) : 3δ ≤ 8r + δ + 5rδ ↔ 2δ ≤ 8r + 5rδ

⇐ (for δ ≥ r + rδ) : 9r + 6rδ ≤ 8r + δ + 5rδ ↔ r + rδ ≤ δ

k = 6:

Ek=60,1 = Ek=6

0,2 = Ek=60,3 = 3 + 3δ < 3 + r + 3δ + rδ = Ek=6

1,1 = Ek=61,2

35

Lemma 3.3 (k-model)For a string of pearls Gn=3,r with detours r > 0, perturbation intensity δ > r and 0 < k < 6perturbations there is always an optimal solution pair that contains the solution s0,0. Forsome parameter combinations there exist other optimal solution pairs that do not containthe solution s0,0. It holds:

min0≤l≤3

E[wk(p0,l)] ≤ min1≤i,j≤3i+j≤3

E[wk(pi,j)]

With the formulas for the k-model we are ready to give the formulas for the p-model:

Ep0,1 = (1− p)6(3) + p(1− p)5(18 + r + 2δ) + p2(1− p)4(45 + 11δ + 4r)

+ p3(1− p)3(60 + 24δ + 6r) + p4(1− p)2(45 + 26δ + 4r)

+ p5(1− p)(18 + 14δ + r) + p6(3 + 3δ)

Ep0,2 = (1− p)6(3) + p(1− p)5(18 + δ + 2 min(δ, 2r))

+ p2(1− p)4(45 + 9δ + 2r + 4min(δ, 2r))

+ p3(1− p)3(60 + 22δ + 4r + 2 min(δ, 2r) + 2 min(δ, 2r + rδ))

+ p4(1− p)2(45 + 24δ + 2r + 4 min(δ, 2r + rδ))

+ p5(1− p)(18 + 13δ + 2 min(δ, 2r + rδ)) + p6(3 + 3δ)

Ep0,3 = (1− p)6(3) + p(1− p)5(18 + 3 min(δ, 3r))

+ p2(1− p)4(45 + 9δ + 3 min(2δ, 3r))

+ p3(1− p)3(60 + 18δ + 3r + 9 min(δ, 3r + rδ))

+ p4(1− p)2(45 + 24δ + 3 min(2δ, 3r + rδ))

+ p5(1− p)(18 + 12δ + 3 min(δ, 3r + 2rδ)) + p6(3 + 3δ)

Ep1,1 = (1− p)6(3 + r) + p(1− p)5(18 + 6r + δ) + p2(1− p)4(45 + 15r + 9δ + rδ)

+ p3(1− p)3(60 + 20r + 22δ + 4rδ) + p4(1− p)2(45 + 15r + 24δ + 6rδ)

+ p5(1− p)(18 + 6r + 13δ + 4rδ) + p6(3 + r + 3δ + rδ)

Ep1,2 = (1− p)6(3 + r) + p(1− p)5(18 + 9r)

+ p2(1− p)4(45 + 9δ + 18r + r min(1, δ) + 2rδ)

+ p3(1− p)3(60 + 28r + 18δ + 2 min(δ, r + rδ) + 7rδ)

+ p4(1− p)2(45 + 24δ + 18r + 9rδ)

+ p5(1− p)(18 + 8r + 12δ + 5rδ + min(δ, r + rδ)) + p6(3 + r + 3δ + rδ)

36

We show that d1 := Ep1,1 − Ep

0,3 > 0 for arbitrary r > 0, δ > r and 0 0

+ p(1− p)5(6r + δ − 3 min(δ, 3r)) {≥ 0

+ p2(1− p)4(15r + rδ − 3min(2δ, 3r)) {≥ p2(1− p)4(6r + rδ)

+ p3(1− p)3(17r + 4δ + 4rδ − 9min(δ, 3r + rδ)) {≥ p3(1− p)3(2r − rδ)

+ p4(1− p)2(15r + 6rδ − 3min(2δ, 3r + rδ)) {≥ p4(1− p)2(6r + 3rδ)

+ p5(1− p)(6r + δ + 4rδ − 3min(δ, 3r + 2rδ)) {≥ 0

+ p6(r + rδ) {> 0

> p2(1− p)4(6r + rδ)

+ p3(1− p)3(2r − rδ)

+ p4(1− p)2(6r + 3rδ)

⇒ for p < (1− p): d1 > p3(1− p)3(6r + rδ + 2r − rδ) > 0

⇒ for p ≥ (1− p): d1 > p3(1− p)3(6r + 3rδ + 2r − rδ) > 0

We do the same for d2 := Ep1,2 − Ep

0,3 > 0:

d2 = (1− p)6(r) {> 0

+ p1(1− p)5(9r − 3min(δ, 3r)) {≥ 0

+ p2(1− p)4(18r + r min(1, δ) + 2rδ − 3 min(2δ, 3r)) {> 0

+ p3(1− p)3(25r + 2 min(δ, r + rδ) + 7rδ − 9min(δ, 3r + rδ)) {≥ 0

+ p4(1− p)2(18r + 9rδ − 3 min(2δ, 3r + rδ)) {> 0

+ p5(1− p)(8r + 5rδ + min(δ, r + rδ)− 3min(δ, 3r + 2rδ)) {≥ 0

+ p6(r + rδ) {> 0

Lemma 3.4 (p-model)For a string of pearls Gn=3,r with detours r > 0, perturbation intensity δ > r and a pertur-bation probability 0 < p < 1 the optimal solution pair always contains the solution s0,0. Itholds:

min0≤l≤3

E[wp(p0,l)] < min1≤i,j≤3i+j≤3

E[wp(pi,j)]

37

4 Failed Assumption

4 Failed Assumption

One of our initial interests for this abstract string of pearls model was to prove theoreti-cally one of the experimentally observed results that are presented in [Sam 2005a]. Therea similar but more complex scenario is analyzed where the chance to give a prove is quitesmall. Translated to the string of pearls model we assumed to be able to prove the followingassumption:

Consider a string of pearls Gn,r = {V, E,w}. Let si denote the solution that uses i de-tours and let Ei denote the expected minimal value E[min(w(s0), w(si))] with respect to w.Furthermore let n ≥ 3, r > 0, δ > r and either 0 Ec.

We had to find out that this assumption does not generally hold for the p-model and thek-model. For n = 3 we found counter examples with the help of the theoretical formulas forthe expected minimal values shown in Subsections 2.1 and 2.2. For larger problem sizes nwe found counter examples experimentally. We could give a list of parameter combinationswhere 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 = 0.1, δ = 0.2 and p = 0.2 ⇒ Ep1 < Ep

2 > Ep3:

Ep0 = 3.12000Ep

1 = 3.10400Ep

2 = 3.11488Ep

3 = 3.11360

k-model: n = 3, 0 < r < 59 , δ = 3r and k = 4 ⇒ Ek

1 < Ek2 > Ek

3:

Ek0 = 3 +

90r15

Ek1 = 3 +

82r15

Ek2 = 3 +

86r15

+415

min(0, 3r2 − r)

Ek3 = 3 +

90r15

+315

min(0, 3r2 − 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. Butwith this modification the problems get more complicated again.

38

5 Conclusions

½¼

¾»

½¼

¾»

½¼

¾»

½¼

¾»· · ·

½¼

¾»

½¼

¾»R R R R

µ µ ¸ ¸

v0 v1 v2 v3 vn−1 vn

1 1 1 1

1 + r 1 + 2r 1 + 3r 1 + nr

Figure 3: Extended String of Pearls with Iteratively Increasing Detours

5 Conclusions

We analyzed strings of pearls Gn,r (see Definition 1.1) with biased perturbations (see Defini-tion 1.2: k-model and Definition 1.3: p-model). In Section 2 we analyzed the optimal choiceof an alternative solution S1 additionally to the solution S0 that uses no detour, such thatE[min(w(S0), w(S1))] is minimal. Since all detours of the graph have the same length it doesnot matter which detours S1 uses. Only the number d of detours is important.

The models turned out to be awkward to manage theoretically. Thus the results mainlyare observations by computer calculations or computer simulations. For certain parametersubsets we were able to prove the results.

For the k-model we observed that the optimal number of detours dopt

• decreases in the detour length r,

• increases in the perturbation intensity δ,

• behaves unexpectedly irregular in the number of perturbations k. For certain combi-nations of n, r, δ it jumps up and down if k increases (see Pages 4.., Proposition 2.6and Conjecture 2.7).

For the p-model we observed that for increasing problem size n the optimal number ofdetours dopt = d(n, r, δ, p) increases up to a number d∗ = d(r, δ, p) and then remains constant,such that:

dopt(n, r, δ, p) ={

n : n < d∗(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 solutionsto the solution that uses no detour. If the alternative is correctly chosen there exist no otherpairs of solutions that are better (proved for n ≤ 3).

References

[Sam 2005a] J. Sameith. (2005) Penalty Methods — Generating Alternative Solutions forDiscrete Optimization Problems with Uncertain Data. submitted as doctoraldissertation, Faculty of Mathematics and Computer Science, Friedrich-Schiller-University Jena.

[Sam 2005b] J. Sameith. (2005) Relative Improvement by Alternative Solutions for Classesof Simple Shortest Path Problems with Uncertain Data — Part I: Strings ofPearls Gn with Unbiased Perturbations. Technical report. Available athttp://www.minet.uni-jena.de/math-net/reports/ .

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