Relative Error with respect to the Tight Lower Bound 52

The relative error is used to measure the extent to which the cost given by the heuristic algorithm deviates from the optimal value, which is di-mensionless, making it convenient to compare. Analogically, the relative error with respect to the tight lower bound is introduced to measure the extent to which the cost given by the heuristic algorithm deviates from the corresponding tight lower bound in this study.

CHAPTER 4. MULTI-PORT CASE 53

Lemma 4.1. For a given problem instance, the relative error is less than or equal

to the relative error with respect to the tight lower bound for positive minimization problem.

Proof. Let I be a given problem instance, H (I) be the value obtained by the heuristic algorithm, M (I) be the theoretical minimum value, T LB (I) be the tight lower bound on M (I), and suppose H (I), M (I) and T LB (I) are all positive values. Let RE (I) be the relative error. Given the prob-lem instance I, RE (I) is defined as RE (I) := ^{H(I)−M (I)}_{M (I)} = _{M (I)}^H(I) − 1 and RE-T LB (I) := H(I)−T LB(I)

T LB(I) = _{T LB(I)}^H(I) − 1. Due to M (I) > T LB (I) > 0 and H (I) > 0, so RE (I) 6 RE-T LB (I).

Lemma 4.1 provides another possible way to evaluate the performance of a heuristic algorithm. In the situation when the theoretical minimum value is not easy or even impossible to calculate but a tight lower bound is determined, we are still able to draw the conclusion that the heuristic algorithm performs effectively as long as the calculated relative error with respect to the tight lower bound is sufficiently small, say, less than 5 per cent. It is because the underlying true relative error must be identical or even smaller according to Lemma 4.1.

CHAPTER 4. MULTI-PORT CASE 54

4.2 Algorithm

Since it is extremely difficult to derive optimal policy in the multi-port case, a polynomial time algorithm is developed to find an approximate repositioning policy.

Suppose we have already calculated A^k_n and S_n^k for all 1 6 n 6 N and 1 6 k 6 K beforehand. We treat this multi-stage decision problem as a sequence of several single-stage decision problems, and make a decision at the beginning of each decision period sequentially when i^k_n for every port is realized. Without loss of generality, assume that we are in the n-th period, based on A^k_n, S_n^k
for port k, we define three state sets as follows:

Ω^a_n:=k|i^k_n < A^k_n

Ω^b_n:=k|A^k_n 6 i^kn 6 Sn^k

Ω^c_n :=k|i^k_n> S_n^k

It is clear that in every period, each port must fall into only one of the three sets, and all ports therefore are classified into three different groups.

According to the optimal policy established in Theorem 3.1, in each period, importing empty containers for k ∈ Ω^a_n and exporting empty containers for k ∈ Ω^c_n can reduce the total operating cost, while either importing or exporting empty containers for k ∈ Ω^b_nwill add to the total operating cost.

For any port k, importing one empty container will change the total

oper-CHAPTER 4. MULTI-PORT CASE 55

ating cost by

∆^k_n i^k_n
= G^k_n i^k_n+ 1
− G^k_n i^k_n
+ c^k_i (4.12)

For any port k, exporting one empty container will change the total oper-ating cost by convex and the above inequalities hold strictly.

According to the optimal policy established in Theorem 3.1, it would never be optimal, in any circumstance, to export empty containers for k ∈ Ω^a_nor import empty containers for k ∈ Ω^c_n. Therefore, we re-define Delta used in Algorithm 4.1 as follows:

CHAPTER 4. MULTI-PORT CASE 56

∆^k_n i^k_n
=











G^k_n i^k_n− 1
− G^k_n i^k_n
+ c^k_e, i^k_n > A^k_n a huge number, i^k_n 6 A^kn

. (4.18)

Algorithm 4.1 is proposed to reposition the empty containers between multi-ports. We have calculated the A^k_n and S_n^k for the whole planning horizon, i.e., 1 6 n 6 N and all the ports involved, i.e., 1 6 k 6 K before-hand by Algorithm 3.1. Since Algorithm 4.1 is applicable to any period, we omit the subscript n for brevity.

Algorithm 4.1. Reposition

1. For each k within 1 6 k 6 K:

(a) If i^k < A^k, add k to state set Ω^a. (b) If A^k 6 i^k6 S^k, add k to state set Ω^b.

(c) If i^k > S^k, add k to state set Ω^c.

2. (a) If Ω^a= ∅and Ω^c= ∅, stop.

(b) If Ω^a6= ∅ and Ω^c6= ∅, go to step 3.

(c) If Ω^a= ∅and Ω^c6= ∅,

If Ω^b 6= ∅, go to step 4; else stop.

CHAPTER 4. MULTI-PORT CASE 57

(d) If Ω^a6= ∅ and Ω^c= ∅,

If Ω^b 6= ∅, go to step 5; else stop.

3. (a) Find the p with the minimum ∆^k i^k
from k ∈ Ω^a. (b) Fine the q with the minimum ∆^k i^k
from k ∈ Ω^c.

(c) Reset i^p = i^p + 1. (d) Reset i^q= i^q− 1.

(e) If i^p = A^p,

• Delete p from Ω^a;

• Add p to Ω^b. (f) If i^q = S^q,

• Delete q from Ω^c;

• Add q to Ω^b. (g) Go to step 2.

4. (a) Find the p with the minimum ∆^k i^k
from k ∈ Ω^c. (b) Find the q with the minimum ∆^k i^k
from k ∈ Ω^b.

(c) If ∆^p(i^p) + ∆^q(i^q) > 0, stop; else

• Reset i^p = i^p− 1.

• Reset i^q = i^q+ 1.

CHAPTER 4. MULTI-PORT CASE 58

• If i^p = S^p, delete p from Ω^c.

• If i^q = S^q, delete q from Ω^b.

• Go to step 2.

5. (a) Find the p with the minimum ∆^k i^k
from k ∈ Ω^a. (b) Fine the q with the minimum ∆^k i^k
from k ∈ Ω^b.

(c) If ∆^p(i^p) + ∆^q(i^q) > 0, stop; else

• Reset i^p = i^p+ 1.

• Reset i^q = i^q− 1.

• If i^p = A^p, delete p from Ω^a.

• If i^q = A^q, delete q from Ω^b.

• Go to step 2.

Theorem 4.4. Algorithm 4.1 has time complexity O (K).

Proof. The result is self-evident.

Theorem 4.4 demonstrates that Algorithm 4.1 is a polynomial time algo-rithm, which performs very efficiently due to its polynomial running time.

Theorem 4.5. The overall time complexity for solving the whole empty container repositioning problem is O (KM N R).

CHAPTER 4. MULTI-PORT CASE 59

Proof. Since Algorithm 4.1 is executed at the beginning of each decision period when the inventory levels for all the ports are realized, the time complexity corresponding to the whole planning horizon containing N consecutive decision periods is O (KN ). Moreover, Algorithm 3.1 is exe-cuted K times beforehand to calculate the A^k_nand S_n^kfor all the ports over the whole planning horizon. Thus, the corresponding time complexity is O (KM N R). Because max {O (KN ) , O (KM N R)} = O (KM N R), the overall time complexity to solve the whole problem is O (KM N R).

Theorem 4.5 demonstrates that the whole empty container repositioning problem can be solved within a polynomial running time, which reflects the high efficiency of the proposed Algorithms 3.1 and 4.1.