**3. Single-agent** *k***-CTP with uncertain edges**

In this section, we analyze the single-agent problem, namely, the S-*k*-CTP-U. We present a lower bound to this problem and prove its tightness by introducing a simple strategy. To suggest a lower bound on the competitive ratio of deterministic strategies, we need to analyze the performance of all of deterministic strategies on a special instance. Below, we propose our lower bound on the S-*k*-CTP-U, by analyzing an instance of O-D edge-disjoint graphs. Note that an O-D edge-disjoint graph is an undirected graph *G* with a given source node O and a destination node D, such that any two distinct O-D paths in *G* are edge-disjoint, that is, they do not have a common edge.

Theorem 1.1 For the S-*k*-CTP-U, there is no deterministic online strategy with competitive ratio less than min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> .

• Case 2. *<sup>d</sup>*<sup>1</sup>

complete.

*p*1

ratio equals *<sup>d</sup>*<sup>1</sup>

**3.1 Pessimistic strategy**

*D*1, otherwise go to step 2.

conditions.

• Condition 1. Check if

Otherwise, check condition 2.

**251**

of the problem.

be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*d*<sup>1</sup>

*DOI: http://dx.doi.org/10.5772/intechopen.88741*

*p*1

*p*1

explore more than one uncertain edge at each iteration.

>2*k* � 1. We also consider this case for *α* ¼ 0 and *α* ¼ 1 separately.

(*i* ¼ 0*,* 1*,* …*, k* � 1). The minimum competitive

, when *i* ¼ 0, which is greater than the proposed lower bound

◦ *<sup>α</sup>* <sup>¼</sup> 1. In this case the competitive ratio of the corresponding strategies can

*New Variations of the Online* k*-Canadian Traveler Problem: Uncertain Costs at Known Locations*

◦ *<sup>α</sup>* <sup>¼</sup> 0. In this case the minimum competitive ratio of the corresponding strategies equals 2*k* � 1, which matches the lower bound of the problem.

Since we proved that the competitive ratios of all of the deterministic strategies for this special instance are greater than or equal to min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � �, the proof is

Now, we introduce a new deterministic strategy which meets the presented lower bound. We call this strategy the *pessimistic strategy* since the agent avoids to

• **Initialization**. Put *i* ¼ 0, where *i* denotes the iteration number. At each iteration the agent starts her travel from O and explores the cost of one

when the agent reaches D, the strategy ends. Note that each iteration

*ci*. Let *S* denote the set of the uncertain edges in the graph.

2 P*<sup>i</sup>*�<sup>1</sup> *<sup>j</sup>*¼<sup>1</sup> *pj* � � <sup>þ</sup> *pi* <sup>þ</sup> *ci*

*pi*þ<sup>1</sup>

and there exists no uncertain edge in the selected path, and proceed to reach D.

uncertain edge or will reach D without visiting any unvisited uncertain edge. If the agent reaches D, then the strategy ends. Otherwise, she backtracks to O or reaches D without visiting any other unvisited uncertain edge. In the latter case

corresponds to one of the stages of the problem, because at each iteration the cost of one of the uncertain edges is learned. That is, the first iteration corresponds to stage 1 of the problem. Also note that *pi* is nondecreasing in *i*, where *pi* is the cost of the optimistic shortest O-D path at the beginning of the *i*th iteration. Let *ci* denote the cost of the uncertain edge which is learned at the *<sup>i</sup>*th iteration. Note that *pi*þ<sup>1</sup> is computable immediately after the agent observes

• **Step 1**. Compute *<sup>d</sup>*1, *<sup>p</sup>*1, and min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � �. If the minimum is *<sup>d</sup>*1*=p*1, take

• **Step 2**. If (*i* ¼ *k* � 1), then go to step 3; otherwise, put *i* ¼ *i* þ 1 and find *πi*. If it does not contain uncertain edges, the agent takes it to reach D. Otherwise, take *π<sup>i</sup>* to reach the *i*th visited uncertain edge, observe *ci*, set the value of the newly visited uncertain edge equal to *ci*, and remove it from *S*. That is, it is not considered as an uncertain edge hereafter. Next, check the following

< 2*k* � 1 (1)

**Proof**. Consider the special graph in **Figure 1**. For each of deterministic strategies, we consider the instance when the cost of all of the first *k* � 1 visited uncertain edges equals *M* and the cost of the last visited uncertain edge equals 0. Hence, the cost of the offline shortest path equals *p*1. For a strategy, we call this instance the *adverse instance*. In the special graph in **Figure 1**, any deterministic strategy corresponds to a permutation which specifies in which order the uncertain paths and *D*<sup>1</sup> (not necessarily all of them) are going to be selected. For each of these strategies, consider the adverse instance. We define *α* as a binary coefficient which equals 1, if the agent takes *D*1, and equals 0, if the agent does not take *D*<sup>1</sup> in the strategy. Suppose that the agent has taken *i* number of uncertain paths before taking *D*<sup>1</sup> when *α* equals 1. In this case, the competitive ratio of deterministic strategies on the

special graph shown in **Figure 1** can be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*α*ð Þþ *<sup>d</sup>*<sup>1</sup> ð Þ <sup>1</sup>�*<sup>α</sup> <sup>p</sup>*<sup>1</sup> *p*1

(*i* ¼ 0*;* 1*;* 2*,* …*, k* � 1). Note that in the adverse instance, the agent has to incur a cost equal to 2*p*<sup>1</sup> in her first *k* � 1 trials at the uncertain paths, since she has to come back to O after finding the uncertain edges blocked. However, since the cost of the *k*th visited uncertain edge equals 0, the agent incurs *p*<sup>1</sup> in her *k*th trial at the uncertain paths and reaches D. Now, we present our proof by considering two cases.

	- *<sup>α</sup>* <sup>¼</sup> 1. In this case the competitive ratio of the corresponding strategies can be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*d*<sup>1</sup> *p*1 (*i* ¼ 0*,* 1*,* …*, k* � 1). The minimum competitive ratio equals *<sup>d</sup>*<sup>1</sup> *p*1 , when *i* ¼ 0, which matches the proposed lower bound of the problem.
	- *<sup>α</sup>* <sup>¼</sup> 0. In this case the minimum competitive ratio of the corresponding strategies equals 2*k* � 1, which is greater than or equal to the lower bound of the problem.

**Figure 1.** *A special graph.*

*New Variations of the Online* k*-Canadian Traveler Problem: Uncertain Costs at Known Locations DOI: http://dx.doi.org/10.5772/intechopen.88741*

	- *<sup>α</sup>* <sup>¼</sup> 1. In this case the competitive ratio of the corresponding strategies can be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*d*<sup>1</sup> *p*1 (*i* ¼ 0*,* 1*,* …*, k* � 1). The minimum competitive ratio equals *<sup>d</sup>*<sup>1</sup> *p*1 , when *i* ¼ 0, which is greater than the proposed lower bound of the problem.
	- *<sup>α</sup>* <sup>¼</sup> 0. In this case the minimum competitive ratio of the corresponding strategies equals 2*k* � 1, which matches the lower bound of the problem.

Since we proved that the competitive ratios of all of the deterministic strategies for this special instance are greater than or equal to min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � �, the proof is complete.

Now, we introduce a new deterministic strategy which meets the presented lower bound. We call this strategy the *pessimistic strategy* since the agent avoids to explore more than one uncertain edge at each iteration.

### **3.1 Pessimistic strategy**

Theorem 1.1 For the S-*k*-CTP-U, there is no deterministic online strategy with

**Proof**. Consider the special graph in **Figure 1**. For each of deterministic strategies, we consider the instance when the cost of all of the first *k* � 1 visited uncertain edges equals *M* and the cost of the last visited uncertain edge equals 0. Hence, the cost of the offline shortest path equals *p*1. For a strategy, we call this instance the *adverse instance*. In the special graph in **Figure 1**, any deterministic strategy corresponds to a permutation which specifies in which order the uncertain paths and *D*<sup>1</sup> (not necessarily all of them) are going to be selected. For each of these strategies, consider the adverse instance. We define *α* as a binary coefficient which equals 1, if the agent takes *D*1, and equals 0, if the agent does not take *D*<sup>1</sup> in the strategy. Suppose that the agent has taken *i* number of uncertain paths before taking *D*<sup>1</sup> when *α* equals 1. In this case, the competitive ratio of deterministic strategies on the

*p*1

(*i* ¼ 0*,* 1*,* …*, k* � 1). The minimum competitive

, when *i* ¼ 0, which matches the proposed lower bound of the

special graph shown in **Figure 1** can be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*α*ð Þþ *<sup>d</sup>*<sup>1</sup> ð Þ <sup>1</sup>�*<sup>α</sup> <sup>p</sup>*<sup>1</sup>

paths and reaches D. Now, we present our proof by considering two cases.

(*i* ¼ 0*;* 1*;* 2*,* …*, k* � 1). Note that in the adverse instance, the agent has to incur a cost equal to 2*p*<sup>1</sup> in her first *k* � 1 trials at the uncertain paths, since she has to come back to O after finding the uncertain edges blocked. However, since the cost of the *k*th visited uncertain edge equals 0, the agent incurs *p*<sup>1</sup> in her *k*th trial at the uncertain

≤ 2*k* � 1. We consider this case for *α* ¼ 0 and *α* ¼ 1 separately.

◦ *<sup>α</sup>* <sup>¼</sup> 1. In this case the competitive ratio of the corresponding strategies can

◦ *<sup>α</sup>* <sup>¼</sup> 0. In this case the minimum competitive ratio of the corresponding strategies equals 2*k* � 1, which is greater than or equal to the lower bound of

competitive ratio less than min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> .

*Probability, Combinatorics and Control*

• Case 1. *<sup>d</sup>*<sup>1</sup>

**Figure 1.** *A special graph.*

**250**

*p*1

ratio equals *<sup>d</sup>*<sup>1</sup>

the problem.

problem.

be formulated as <sup>2</sup>*i p*ð Þ<sup>1</sup> <sup>þ</sup>*d*<sup>1</sup>

*p*1

*p*1


$$\frac{2\left(\sum\_{j=1}^{i-1} p\_j\right) + p\_i + c\_i}{p\_{i+1}} < 2k - 1\tag{1}$$

and there exists no uncertain edge in the selected path, and proceed to reach D. Otherwise, check condition 2.

• Condition 2. Note that immediately after the agent observes *ci*, *Di*þ<sup>1</sup> and *di*þ<sup>1</sup> are computable. Check if

$$\frac{2\left(\sum\_{i=1}^{i} p\_i\right) + d\_{i+1}}{p\_{i+1}} < 2k - 1,\tag{2}$$

Since, both *πk*þ<sup>1</sup> and *πk*�<sup>1</sup> (i.e., *πk*þ<sup>1</sup> ≥ *πk*�1) contain the ð Þ *k* � 1 th visited uncertain edge, *pk*�<sup>1</sup> <sup>þ</sup> *ck*�<sup>1</sup> is less than or equal to *pk*þ1. Hence, we can replace *pk*�<sup>1</sup> <sup>þ</sup> *ck*�<sup>1</sup> by *pk*þ<sup>1</sup> in the numerator above. We can also replace *pi* values for ð Þ *i* ¼ 1*;* 2*;* …*; k* � 2 by *pk* in the numerator since *pi* is nondecreasing in *i*. We obtain

*New Variations of the Online* k*-Canadian Traveler Problem: Uncertain Costs at Known Locations*

Now, we replace *pi* values for ð Þ *i* ¼ 1*;* 2*;* …*; k* by *pk* in the numerator of *A*. We

� � <sup>þ</sup> *pk*þ<sup>1</sup> *pk*þ<sup>1</sup>

<sup>3</sup> þ 1 which is less than or equal to the lower bound for *k*≥ 2

<sup>3</sup> <sup>þ</sup> 1 and min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � � for *<sup>k</sup>*≥2. Note that since the

*:* (6)

>2*k* � 1*:* (7)

*:* (8)

*p*1 <sup>¼</sup> <sup>11</sup> <sup>3</sup> with

� � in the numerator of *A*. We also

� � in the numerator of *A*. In this case, *A*

*<sup>A</sup>* <sup>¼</sup> <sup>2</sup>*k pk*

<sup>3</sup> *pk*þ<sup>1</sup>

• **Scenario 3**. *π<sup>k</sup>*þ<sup>1</sup> does not contain the uncertain edges which are visited in the ð Þ *k* � 1 th and the *k*th iterations. In this case, we show that *A* meets the proposed lower bound of the problem. Note that when *k*≤2, *π<sup>k</sup>*þ<sup>1</sup> ¼ *D*<sup>1</sup> in this scenario. Thus, the strategy ends in step 1 when *k*≤ 2. For *k*≥3, consider

� � by <sup>2</sup>*<sup>k</sup>*

condition 2 in step 2 at the ð Þ *k* � 2 th iteration. We have

2 P*<sup>k</sup>*�<sup>2</sup> *<sup>i</sup>*¼<sup>1</sup> *pi* � � <sup>þ</sup> *dk*�<sup>1</sup>

in the numerator since *pi* is nondecreasing in *i*. We obtain

ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>1</sup> by *pk*�<sup>1</sup> in the numerator of *<sup>A</sup>*. We obtain

<sup>3</sup> þ 3, which is less than or equal to the lower bound for *k*≥ 3.

Now, we can replace 2ð Þ *<sup>k</sup>* � <sup>2</sup> *pk*�<sup>1</sup> by <sup>2</sup>*k*�<sup>2</sup>

or equal to the lower bound, the proof is complete.

*pk*�<sup>1</sup>

Since *π<sup>k</sup>*þ<sup>1</sup> does not contain the uncertain edges which are visited in the ð Þ *k* � 1 th and the *<sup>k</sup>*th iterations, *<sup>π</sup><sup>k</sup>*þ<sup>1</sup> is equivalent to *Dk*�1. Hence we can replace *dk*�<sup>1</sup> by *pk*þ<sup>1</sup> in the numerator above. We can also replace *pi* values for ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>2</sup> by *pk*�<sup>1</sup>

ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>4</sup> *pk*�<sup>1</sup> <sup>þ</sup> *pk*þ<sup>1</sup>>ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> *pk*�<sup>1</sup>. Thus, *pk*þ<sup>1</sup>>3*pk*�<sup>1</sup>. Now, we replace *pi* values for

*<sup>A</sup>* <sup>¼</sup> ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>2</sup> *pk*�<sup>1</sup> <sup>þ</sup> <sup>2</sup>*pk* <sup>þ</sup> *pk*þ<sup>1</sup> *pk*þ<sup>1</sup>

replace *pk* by *pk*þ<sup>1</sup>, since *pi* is nondecreasing in *<sup>i</sup>*. In this case, *<sup>A</sup>* would be at most

As an illustrative example for the pessimistic strategy, consider the instance given in **Figure 2** which represents a part of the Gulf Coast area of the United States. In **Figure 2**, the nodes represent the cities, and the numbers on the edges denote the edge travel times (per hour) in a post-disaster scenario. The edges (2,6) and (5,6) are the uncertain edges whose costs are not known at the

beginning. The traveling agent is initially at node 1 and node 6 is the destination node. Path 1-3-6 is the shortest deterministic path (*D*1), and path 1-2-6 is the shortest optimistic path (*π*1) at the initial stage, i.e., *d*<sup>1</sup> ¼ 11 and *p*<sup>1</sup> ¼ 3. When step 1 of the pessimistic strategy is implemented, the agent compares *<sup>d</sup>*<sup>1</sup>

Since we showed that the competitive ratio of the pessimistic strategy is less than

<sup>3</sup> *pk*þ<sup>1</sup>

strategy ends in step 3, min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � � equals 2*<sup>k</sup>* � 1.

ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>4</sup> *pk* <sup>þ</sup> *pk*þ<sup>1</sup>>ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> *pk*; hence, *pk*þ<sup>1</sup>>3*pk*.

*DOI: http://dx.doi.org/10.5772/intechopen.88741*

Now, we can replace 2*k pk*

would be at most <sup>2</sup>*<sup>k</sup>*

since we are comparing <sup>2</sup>*<sup>k</sup>*

obtain

2*k*�2

**253**

and then go back to O and take *Di*þ1. Otherwise, return to O and go to the beginning of step 2.

• **Step 3**. Take *π<sup>k</sup>* and observe *ck*. Then compare

$$A = \frac{2\left(\sum\_{i=1}^{k} p\_i\right) + p\_{k+1}}{p\_{k+1}}\tag{3}$$

and

$$B = \frac{2\left(\sum\_{i=1}^{k-1} p\_i\right) + p\_k + c\_k}{p\_{k+1}}.\tag{4}$$

If *A* <*B* return to O and take the shortest path ð Þ *π<sup>k</sup>*þ<sup>1</sup> ; otherwise, travel through the uncertain edge in the *k*th uncertain path and reach D.

Below we show that our strategy is optimal by using the inequalities which are presented in different steps of the pessimistic strategy.

Theorem 1.2 The pessimistic strategy is optimal for the S-*k*-CTP-U.

**Proof**. Note that if the strategy ends in either step 1 or 2, the competitive ratio would be less than or equal to the lower bound. Hence, we just need to analyze the cases where the strategy ends in step 3. Note that the competitive ratio of the strategy would not exceed minf g *A; B* in step 3. Thus, it is enough to show that either *A* or *B* does not exceed the proposed lower bound of the problem, if the strategy ends in step 3. We consider three different scenarios for *π<sup>k</sup>*þ<sup>1</sup> to show the optimality of the pessimistic strategy, if the strategy ends in step 3.

• **Scenario 1**. *π<sup>k</sup>*þ<sup>1</sup> contains the uncertain edge which is visited in the *k*th iteration. In this case, we show that *B* meets the proposed lower bound of the problem. Since both *π<sup>k</sup>*þ<sup>1</sup> and *π<sup>k</sup>* (i.e., *π<sup>k</sup>*þ<sup>1</sup> ≥*πk*) contain the *k*th visited uncertain edge, *pk* <sup>þ</sup> *ck* equals *pk*þ<sup>1</sup>. Hence we can replace *pk* <sup>þ</sup> *ck* by *pk*þ<sup>1</sup> in the numerator of *<sup>B</sup>*. We can also replace *pi* values for ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>1</sup> by *pk*þ<sup>1</sup> in the numerator of *B*, since *pi* is nondecreasing in *i*. In this case, *B* would be at most 2*k* � 1 which equals the lower bound of the problem.

Here, we note that *π<sup>k</sup>*þ<sup>1</sup> does not contain the *k*th visited uncertain edge in the next two scenarios.

• **Scenario 2**. *π<sup>k</sup>*þ<sup>1</sup> contains the uncertain edge which is visited in the ð Þ *k* � 1 th iteration. Note that *k*≥2 in this scenario, since *π<sup>k</sup>*þ<sup>1</sup> does not contain the *k*th visited uncertain edge and contains the ð Þ *k* � 1 th visited uncertain edge. In this case, we show that *A* meets the proposed lower bound of the problem. Consider condition 1 in step 2 at the ð Þ *k* � 1 th iteration. Since we have assumed that the strategy ends in step 3, we have

$$\frac{2\left(\sum\_{i=1}^{k-2} p\_i\right) + p\_{k-1} + c\_{k-1}}{p\_k} > 2k - 1. \tag{5}$$

*New Variations of the Online* k*-Canadian Traveler Problem: Uncertain Costs at Known Locations DOI: http://dx.doi.org/10.5772/intechopen.88741*

Since, both *πk*þ<sup>1</sup> and *πk*�<sup>1</sup> (i.e., *πk*þ<sup>1</sup> ≥ *πk*�1) contain the ð Þ *k* � 1 th visited uncertain edge, *pk*�<sup>1</sup> <sup>þ</sup> *ck*�<sup>1</sup> is less than or equal to *pk*þ1. Hence, we can replace *pk*�<sup>1</sup> <sup>þ</sup> *ck*�<sup>1</sup> by *pk*þ<sup>1</sup> in the numerator above. We can also replace *pi* values for ð Þ *i* ¼ 1*;* 2*;* …*; k* � 2 by *pk* in the numerator since *pi* is nondecreasing in *i*. We obtain ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>4</sup> *pk* <sup>þ</sup> *pk*þ<sup>1</sup>>ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> *pk*; hence, *pk*þ<sup>1</sup>>3*pk*.

Now, we replace *pi* values for ð Þ *i* ¼ 1*;* 2*;* …*; k* by *pk* in the numerator of *A*. We obtain

$$A = \frac{2k\left(p\_k\right) + p\_{k+1}}{p\_{k+1}}.\tag{6}$$

Now, we can replace 2*k pk* � � by <sup>2</sup>*<sup>k</sup>* <sup>3</sup> *pk*þ<sup>1</sup> � � in the numerator of *A*. In this case, *A* would be at most <sup>2</sup>*<sup>k</sup>* <sup>3</sup> þ 1 which is less than or equal to the lower bound for *k*≥ 2 since we are comparing <sup>2</sup>*<sup>k</sup>* <sup>3</sup> <sup>þ</sup> 1 and min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � � for *<sup>k</sup>*≥2. Note that since the strategy ends in step 3, min *<sup>d</sup>*1*=p*1*;* <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> � � equals 2*<sup>k</sup>* � 1.

• **Scenario 3**. *π<sup>k</sup>*þ<sup>1</sup> does not contain the uncertain edges which are visited in the ð Þ *k* � 1 th and the *k*th iterations. In this case, we show that *A* meets the proposed lower bound of the problem. Note that when *k*≤2, *π<sup>k</sup>*þ<sup>1</sup> ¼ *D*<sup>1</sup> in this scenario. Thus, the strategy ends in step 1 when *k*≤ 2. For *k*≥3, consider condition 2 in step 2 at the ð Þ *k* � 2 th iteration. We have

$$\frac{2\left(\sum\_{i=1}^{k-2} p\_i\right) + d\_{k-1}}{p\_{k-1}} > 2k - 1. \tag{7}$$

Since *π<sup>k</sup>*þ<sup>1</sup> does not contain the uncertain edges which are visited in the ð Þ *k* � 1 th and the *<sup>k</sup>*th iterations, *<sup>π</sup><sup>k</sup>*þ<sup>1</sup> is equivalent to *Dk*�1. Hence we can replace *dk*�<sup>1</sup> by *pk*þ<sup>1</sup> in the numerator above. We can also replace *pi* values for ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>2</sup> by *pk*�<sup>1</sup> in the numerator since *pi* is nondecreasing in *i*. We obtain ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>4</sup> *pk*�<sup>1</sup> <sup>þ</sup> *pk*þ<sup>1</sup>>ð Þ <sup>2</sup>*<sup>k</sup>* � <sup>1</sup> *pk*�<sup>1</sup>. Thus, *pk*þ<sup>1</sup>>3*pk*�<sup>1</sup>. Now, we replace *pi* values for

ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>1</sup> by *pk*�<sup>1</sup> in the numerator of *<sup>A</sup>*. We obtain

$$A = \frac{(2k-2)p\_{k-1} + 2p\_k + p\_{k+1}}{p\_{k+1}}.\tag{8}$$

Now, we can replace 2ð Þ *<sup>k</sup>* � <sup>2</sup> *pk*�<sup>1</sup> by <sup>2</sup>*k*�<sup>2</sup> <sup>3</sup> *pk*þ<sup>1</sup> � � in the numerator of *A*. We also replace *pk* by *pk*þ<sup>1</sup>, since *pi* is nondecreasing in *<sup>i</sup>*. In this case, *<sup>A</sup>* would be at most 2*k*�2 <sup>3</sup> þ 3, which is less than or equal to the lower bound for *k*≥ 3.

Since we showed that the competitive ratio of the pessimistic strategy is less than or equal to the lower bound, the proof is complete.

As an illustrative example for the pessimistic strategy, consider the instance given in **Figure 2** which represents a part of the Gulf Coast area of the United States. In **Figure 2**, the nodes represent the cities, and the numbers on the edges denote the edge travel times (per hour) in a post-disaster scenario. The edges (2,6) and (5,6) are the uncertain edges whose costs are not known at the beginning. The traveling agent is initially at node 1 and node 6 is the destination node. Path 1-3-6 is the shortest deterministic path (*D*1), and path 1-2-6 is the shortest optimistic path (*π*1) at the initial stage, i.e., *d*<sup>1</sup> ¼ 11 and *p*<sup>1</sup> ¼ 3. When step 1 of the pessimistic strategy is implemented, the agent compares *<sup>d</sup>*<sup>1</sup> *p*1 <sup>¼</sup> <sup>11</sup> <sup>3</sup> with

• Condition 2. Note that immediately after the agent observes *ci*, *Di*þ<sup>1</sup> and *di*þ<sup>1</sup>

þ *di*þ<sup>1</sup>

<sup>þ</sup> *pk*þ<sup>1</sup>

þ *pk* þ *ck*

*pk*þ<sup>1</sup>

*pk*þ<sup>1</sup>

If *A* <*B* return to O and take the shortest path ð Þ *π<sup>k</sup>*þ<sup>1</sup> ; otherwise, travel through

Below we show that our strategy is optimal by using the inequalities which are

**Proof**. Note that if the strategy ends in either step 1 or 2, the competitive ratio would be less than or equal to the lower bound. Hence, we just need to analyze the cases where the strategy ends in step 3. Note that the competitive ratio of the strategy would not exceed minf g *A; B* in step 3. Thus, it is enough to show that either *A* or *B* does not exceed the proposed lower bound of the problem, if the strategy ends in step 3. We consider three different scenarios for *π<sup>k</sup>*þ<sup>1</sup> to show the

and then go back to O and take *Di*þ1. Otherwise, return to O and go to the

<2*k* � 1*,* (2)

*:* (4)

>2*k* � 1*:* (5)

(3)

*pi*þ<sup>1</sup>

2 P*<sup>k</sup> <sup>i</sup>*¼<sup>1</sup> *pi* � �

2 P*<sup>k</sup>*�<sup>1</sup> *<sup>i</sup>*¼<sup>1</sup> *pi* � �

Theorem 1.2 The pessimistic strategy is optimal for the S-*k*-CTP-U.

optimality of the pessimistic strategy, if the strategy ends in step 3.

most 2*k* � 1 which equals the lower bound of the problem.

that the strategy ends in step 3, we have

2 P*<sup>k</sup>*�<sup>2</sup> *<sup>i</sup>*¼<sup>1</sup> *pi* � �

• **Scenario 1**. *π<sup>k</sup>*þ<sup>1</sup> contains the uncertain edge which is visited in the *k*th

iteration. In this case, we show that *B* meets the proposed lower bound of the problem. Since both *π<sup>k</sup>*þ<sup>1</sup> and *π<sup>k</sup>* (i.e., *π<sup>k</sup>*þ<sup>1</sup> ≥*πk*) contain the *k*th visited

uncertain edge, *pk* <sup>þ</sup> *ck* equals *pk*þ<sup>1</sup>. Hence we can replace *pk* <sup>þ</sup> *ck* by *pk*þ<sup>1</sup> in the numerator of *<sup>B</sup>*. We can also replace *pi* values for ð Þ *<sup>i</sup>* <sup>¼</sup> <sup>1</sup>*;* <sup>2</sup>*;* …*; <sup>k</sup>* � <sup>1</sup> by *pk*þ<sup>1</sup> in the numerator of *B*, since *pi* is nondecreasing in *i*. In this case, *B* would be at

Here, we note that *π<sup>k</sup>*þ<sup>1</sup> does not contain the *k*th visited uncertain edge in the

• **Scenario 2**. *π<sup>k</sup>*þ<sup>1</sup> contains the uncertain edge which is visited in the ð Þ *k* � 1 th iteration. Note that *k*≥2 in this scenario, since *π<sup>k</sup>*þ<sup>1</sup> does not contain the *k*th visited uncertain edge and contains the ð Þ *k* � 1 th visited uncertain edge. In this

Consider condition 1 in step 2 at the ð Þ *k* � 1 th iteration. Since we have assumed

<sup>þ</sup> *pk*�<sup>1</sup> <sup>þ</sup> *ck*�<sup>1</sup>

case, we show that *A* meets the proposed lower bound of the problem.

*pk*

2 P*<sup>i</sup> <sup>i</sup>*¼<sup>1</sup> *pi* � �

*A* ¼

*B* ¼

the uncertain edge in the *k*th uncertain path and reach D.

presented in different steps of the pessimistic strategy.

• **Step 3**. Take *π<sup>k</sup>* and observe *ck*. Then compare

are computable. Check if

*Probability, Combinatorics and Control*

beginning of step 2.

next two scenarios.

**252**

and

• M-*k*-CTP-U-f. In this problem, the agents have to incur a cost of at least

*L* <sup>þ</sup> 1.

*L* <sup>þ</sup> 1.

Now, we present the rest of our proof for each problem separately:

which the first arrival of the agents at D is via *D*<sup>1</sup> is at least *<sup>d</sup>*<sup>1</sup>

• M-*k*-CTP-U-f. We present our proof by considering two cases:

*L*

which the first arrival of the agents at D is via *D*<sup>1</sup> is at least *<sup>d</sup>*<sup>1</sup>

*L*

matches the proposed lower bound of the problem.

• M-*k*-CTP-U-l. We present our proof by considering two cases:

which the last arrival of the agents at D is via *D*<sup>1</sup> is at least *<sup>d</sup>*<sup>1</sup>

*L*

• M-*k*-CTP-U-l. In this problem, it takes a cost of at least 2 *<sup>k</sup>*

*L*

*New Variations of the Online* k*-Canadian Traveler Problem: Uncertain Costs at Known Locations*

backtrack to O. The agents have to incur *p*<sup>1</sup> to learn the costs of the remaining uncertain edges and deliver at least one of the agents to D. Since the cost of the shortest path is at least *p*<sup>1</sup> in the adverse instance, the competitive ratio of deterministic strategies when none of the agents take *D*<sup>1</sup> would be at least

costs of all of the *k* uncertain edges and backtrack the agents to O. It takes at least *p*<sup>1</sup> for all of the agents to take the shortest path and arrive at D. Since the cost of the shortest path is *p*<sup>1</sup> in the adverse instance, the competitive ratio of deterministic strategies when none of the agents take *D*<sup>1</sup> would be at least

Note that since we are considering the arrivals of the agents at D via the uncertain paths, the performance of the strategies will not be improved if one or more

<sup>þ</sup> 1. In this case, the competitive ratio of the strategies in

<sup>þ</sup> <sup>1</sup> . The competitive ratio of

<sup>þ</sup> 1, which is greater than the proposed lower

<sup>þ</sup> 1, which matches the proposed

*L*

<sup>þ</sup> 1. In this case, the competitive ratio of deterministic

<sup>þ</sup> <sup>1</sup> . The competitive ratio of the strategies in

<sup>þ</sup> 1. In this case, the competitive ratio of the strategies in

<sup>þ</sup> <sup>1</sup> . The competitive ratio of

<sup>þ</sup> 1. In this case, the competitive ratio of deterministic

deterministic strategies in which the last arrival of the agents at D is via the

*L*

strategies in which the last arrival of the agents at D is via the uncertain

deterministic strategies in which the first arrival of the agents at D is via the

*L*

strategies in which the first arrival of the agents at D is via the uncertain

number of uncertain edges and

*L*

*p*1

*p*1

*p*1

<sup>þ</sup> 1, which matches the proposed

, which

which is greater

, which is

*<sup>p</sup>*<sup>1</sup> to explore the

*<sup>p</sup>*<sup>1</sup> to discover the costs of *<sup>L</sup> <sup>k</sup>*

*DOI: http://dx.doi.org/10.5772/intechopen.88741*

, which is equal to 2 *<sup>k</sup>*

, which is equal to 2 *<sup>k</sup>*

greater than or equal to min *d*1*=p*1*;* 2 *<sup>k</sup>*

uncertain paths would be at least 2 *<sup>k</sup>*

lower bound of the problem.

paths would be at least 2 *<sup>k</sup>*

bound of min *d*1*=p*1*;* 2 *<sup>k</sup>*

than or equal to min *d*1*=p*1*;* 2 *<sup>k</sup>*

lower bound of the problem.

uncertain edges would be at least 2 *<sup>k</sup>*

agents take *D*1. The proof is complete.

2 *<sup>k</sup> L*

<sup>2</sup> *<sup>k</sup>* ð Þ b c*<sup>L</sup> <sup>p</sup>*1þ*p*<sup>1</sup> *p*1

<sup>2</sup> *<sup>k</sup>* ð Þ d e*<sup>L</sup> <sup>p</sup>*1þ*p*<sup>1</sup> *p*1

◦ Case 1. *<sup>d</sup>*<sup>1</sup>

◦ Case 2. *<sup>d</sup>*<sup>1</sup>

◦ Case 1. *<sup>d</sup>*<sup>1</sup>

◦ Case 2. *<sup>d</sup>*<sup>1</sup>

**255**

*p*1 ≥2 *<sup>k</sup> L*

*p*1 < 2 *<sup>k</sup> L*

*p*1 ≥2 *<sup>k</sup> L*

*p*1 < 2 *<sup>k</sup> L*

**Figure 2.**

*A scenario from the Gulf Coast area of the United States network with Atlanta as the source node and Wilmington as the destination node.*

<sup>2</sup>*<sup>k</sup>* � <sup>1</sup> <sup>¼</sup> 3. Since <sup>11</sup> <sup>3</sup> >3, the strategy enters step 2. Next, the agent takes the shortest optimistic path *π*<sup>1</sup> and arrives at node 2 after traversing edge (1,2). We assume that the costs of the uncertain edges (2,6) and (5,6) are 3 and 2, respectively. When the agent arrives at node 2, she learns the traveling time of edge (2,6), i.e., *<sup>c</sup>*<sup>1</sup> <sup>¼</sup> 3. Then she checks if *<sup>p</sup>*1þ*c*<sup>1</sup> *p*2 <sup>&</sup>lt; <sup>2</sup>*<sup>k</sup>* � 1. Since <sup>6</sup> <sup>6</sup> < 3, the agent takes edge (2,6) to arrive at node 6 and the strategy ends. Note that the cost of the offline optimum is 6. Therefore, the competitive ratio of the pessimistic strategy is one in the described scenario.
