Competitive Analysis for Dynamic Multiperiod. Uncapacitated Routing Problems Enrico Angelelli and M. Grazia Speranza, страница 5

There are two natural algorithms to consider and analyze as a first step towards constructing an effective algorithm for DMPRP.Namely,thealgorithmwhichservesallcustomersin C1|2 inthefirsttimeperiod,whichwewillcallIMMEDIATE, and the algorithm that postpones all customers in C1|2 to the second time period, which we will call DELAY.

Ouranalysisofthesetwosimplealgorithmswillshowthat√ their competitive ratio is quite far from the lower bound        2 shown in Theorem 1.

Algorithm 1 (IMMEDIATE). Visit all customers in C1 and C1|2 in the first time period and visit all customers in C2 in the second time period.

Theorem 2. The competitive ratio of Algorithm IMMEDIATE for instances with customer locations on the nonnegative real line is 2.

Proof. It is easy to see that z∗(I) ≥ Lall ≥ max(L1,1|2,L2) and z(IMMEDIATE,I) = L1,1|2 +L2. Therefore,

2Lall

               rIMMEDIATE               = 2

Lall

for all instances I. Thus rIMMEDIATE ≤ 2.

Toseethattheboundistight,consideranyinstanceI where C1 = ∅ and the sets C1|2 and C2 are identical and non-empty.

Then,

z(IMMEDIATE,I) = 2L1|2 = 2.  z∗(I)     L1|2

Algorithm 2 (DELAY). Visit all customers in C1 in the first time period and visit all customers in C1|2 and C2 in the second time period.

Theorem 3. The competitive ratio of Algorithm DELAY for instances with customer locations on the non-negative real

line is 2.

Proof. Since z∗(I) ≥ Lall ≥ max(L1,L1|2,2) and z(DELAY,I) = L1 + L1|2,2, we obtain

rDELAY(I) ≤ L1 + L1|2,2 ≤ 2Lall ≤ 2

                                               Lall                  Lall

for all instances I. Thus rDELAY ≤ 2.

To see that the bound is tight, consider any instance I where the sets C1 and C1|2 are identical and nonempty and C2 = ∅. Then,

                         z(DELAY,I)        L1 + L1

                                z∗(I)          =       L1          = 2.                       ■

We now consider a parameterized algorithm that applies either IMMEDIATE or DELAY, based on the information available at the beginning of the first period.

Algorithm 3 (SMART(p)).             If L1 > 0 and L1,1|2 ≤ pL1,

then apply IMMEDIATE, else apply DELAY.

Algorithm SMART(p) visits all customers in C1|2 in the first time period if C1 is nonempty and L1,1|2 ≤ pL1, i.e. if the customers in C1|2 are not “too far” from the customers in C1, while it postpones all customers in C1|2 to the second time period otherwise. Throughout the paper we will assume p > 1.

Next, we will show that the proposed algorithm, for an appropriate parameter choice, is optimal in the sense that no algorithm exists with a better competitive ratio.

Theorem4.√      ThecompetitiveratioofAlgorithmSMART(1+

2) for instances with customer locations on the non-

negative real line is    2. The algorithm is optimal.

Proof. The cases L1 = 0 and L1|2 ≤ L1 are trivial. In the first case Algorithm DELAY is applied and the offline optimal solution is found. In the second case Algorithm IMMEDIATE is applied and the off-line optimal solution is

found. In both cases the ratio rSMART(I) equals 1. Now, let us assume 0 < L1 < L1|2 and observe that L1,1|2 = L1|2.

When L1 < L2, the optimal strategy is to serve the customers in C1|2 in the second time period, resulting in a total cost z∗(I) = L1 ++√L1|2,2.

If L1,1|2 > (1 2)L1, then Algorithm DELAY is applied and it finds the optimal solution: rSMART(I) = 1.

If L1,1|2 ≤ (1 + √2)L1, then Algorithm IMMEDIATE is applied. We derive

                           z(IMMEDIATE,I)       L1,1|2      L2

rSMART(I) =                                          =         + +

                                      z∗(I)                   L1      L1|2,2

,

where the latter inequality is due to the fact that the ratio

LL11|2++xx is a decreasing function of x and L1|2,2 L1|2. Thus,

2L1

        rSMART(I) ≤            |2       ≤     2+(1 ++√2√)L1       = √2.

                              L1 + L1|2          L1       (1         2)L1

The latter equality is due to the fact that the ratio        2x                is an 1+