Online traveling salesman problem with time cost and non-zealous server

Abstract

Considering that the time of meeting the demands is very important for emergency vehicle and emergency vehicle can’t reject any request, we introduce a more realistic cost form into online traveling salesman problem(OL-TSP). When a new request arrives, if the salesman can’t serve the request immediately, per-unit-time cost will be generated. The goal is to minimize server’s total costs(travel makespan plus the per-unit-time costs). We consider the server is a non-zealous server and show that neither deterministic nor randomized online algorithms can achieve constant competitive ratio for OL-TSP on general metric space. While on truncated line segment and uniform metric space, we prove lower bounds, and present competitive online algorithms. Especially for the case with uniform metric space, we prove an optimal Greedy algorithm.

Appendices

Appendix A. Case 1.2.1 in Theorem 3

The added cost of the online server includes four parts:

• 1. The added penalties of the requests between node $-y-1$ and node $-L$. For each of $L-y$ requests, the increased cost is equal to $2y_{1}B$. The increased costs of the $L-y$ requests are equal to $2(L-y)y_{1}B$;

• 2. The costs of the requests between node 1 and node y. y requests are released during the time interval $[2L+1,2L+y]$ one by one, the requests are served during the time interval $[4L+2y_{1}+1,4L+2y_{1}+y]$ by online server. The cost is equal to $(2L+2y_{1})B$ for each request. The costs of the y requests are equal to $2(L+y_{1})yB$;

• 3. The costs of the requests between node $y+1$ and node L. $L-y$ requests are released during the time interval $[2L+2y_{1}+y+1,3L+2y_{1}]$ one by one, the $L-y$ requests are served during the time interval $[4L+2y_{1}+y+1,5L+2y_{1}]$ by online server. The cost is equal to 2LB for each request. The costs of the $L-y$ requests is equal to $2(L-y)LB$;

• 4. The added traveling time of the online server. The increased traveling time of the online server is equal to $2L+2y_{1}$.

The total added cost of the online server is equal to $2L + 2{y_1} + 2(L - y){y_1}B + 2(L + {y_1})yB + 2(L - y)LB = 2L + 2{y_1} + (2{L^2} + 2L{y_1})B$.

Appendix B. Case 1.2.2 in Theorem 3

When $y_{2}\le y$, the added cost of the online server includes seven parts:

• 1. The added costs of the requests between node $-y-1$ and node $-L$. For each of $L-y$ requests, the increased cost is equal to $2(y+y_{2})B$. The increased costs of the $L-y$ requests are equal to $2(L-y)(y+y_{2})B$;

• 2. The costs of the requests between node 1 and node $y_{2}$. $y_{2}$ requests are released during the time interval $[2L+1,2L+y_{2}]$ one by one, the $y_{2}$ requests are served during the time interval $[2L+2y+1,2L+2y+y_{2}]$ by online server. The cost is equal to 2yB for each request. The costs of the $y_{2}$ requests are equal to $2yy_{2}B$;

• 3. The costs of the requests between node $y_{2}+1$ and node y. $y-y_{2}$ requests are released during the time interval $[2L+2y_{2}+1,2L+y]$ one by one, the $y-y_{2}$ requests are served during the time interval $[4L+3y_{2}+2y+1,4L+2y_{2}+3y]$ by online server. The cost is equal to $2(L+y+y_{2})B$ for each request. The costs of the $y-y_{2}$ requests are equal to $2(L+y+y_{2})(y-y_{2})B$;

• 4. The costs of the requests between node $-1$ and node $-y_{2}$. $y_{2}$ requests are released during the time interval $[2L+2y+1,2L+2y+y_{2}]$ one by one, the $y_{2}$ requests are served during the time interval $[2L+2y+2y_{2}+1,2L+2y+3y_{2}]$ by online server. The cost is equal to $2y_{2}B$ for each request. The costs of the $y_{2}$ requests are equal to $2y_{2}^{2}B$;

• 5. The costs of the requests between node 1 and node $y_{2}$. $y_{2}$ requests are released during the time interval $[2L+2y+2y_{2}+1,2L+2y+3y_{2}]$ one by one, the requests are served during the time interval $[4L+2y+2y_{2}+1,4L+2y+3y_{2}]$ by online server. The cost is equal to 2LB for each request. The costs of the $y_{2}$ requests are equal to $2y_{2}LB$;

• 6. The costs of the requests between node $y+1$ and node L. $L-y$ requests are released during the time interval $[2L+3y+2y_{2}+1,3L+2y+2y_{2}]$ one by one, the $L-y$ requests are served during the time interval $[4L+3y+2y_{2}+1,5L+2y+2y_{2}]$ by online server. The cost is equal to 2LB for each request. The costs of the $L-y$ requests are equal to $2(L-y)LB$;

• 7. The added traveling time of the online server. The increased traveling time of the online server is equal to $2L(L+y+y_{2})$.

The total added cost of the online server is equal to $2L + 2y + 2{y_2} + (2(L - y)(y + {y_2})B + 2y{y_2}B + 2(L + y + {y_2})(y - {y_2})B + 2{y_2}^2B + 2{y_2}LB + 2(L - y)LB) = 2(L + y + {y_2}) + 2(L + y + {y_2})LB$.

When $y_{2} > y$, the added cost of the online server includes five parts:

• 1. The added costs of the requests between node $-y-1$ and node $-L$. For each of $L-y$ requests, the increased cost is equal to $2(y+y_{2})B$. The increased costs of the $L-y$ requests are equal to $2(L-y)(y+y_{2})B$;

• 2. The costs of the requests between node 1 and node y. y requests are released during the time interval $[2L+1,2L+y]$ one by one, the $y_{2}$ requests are served during the time interval $[2L+2y+1,2L+3y]$ by online server. The cost is equal to 2yB for each request. The costs of the y requests are equal to $2y^{2}B$;

• 3. The costs of the requests between node $-1$ and node $-y$. y requests are released during the time interval $[2L+2y+1,2L+3y]$ one by one, the y requests are served during the time interval $[2L+2y+2y_{2}+1,2L+3y+2y_{2}]$ by online server. The cost is equal to $2y_{2}B$ for each request. The costs of the y requests are equal to $2yy_{2}B$;

• 4. The costs of the requests between node 1 and node L. L requests are released during the time interval $[2L+2y+2y_{2}+1,3L+2y+2y_{2}]$ one by one, the L requests are served during the time interval $[4L+2y+2y_{2}+1,5L+2y+2y_{2}]$ by online server. The cost is equal to 2LB for each request. The costs of the L requests are equal to $2L^{2}B$;

• 5. The added traveling time of the online server. The increased traveling time of the online server is equal to $2L(L+y+y_{2})$.

The total added cost of the online server is equal to $2L + 2y + 2{y_2} + (2(L - y)(y + {y_2})B + 2{y^2}B + 2y{y_2}B + 2{L^2}B) = 2(L + y + {y_2}) + 2(L + y + {y_2})LB$.

Appendix C. Case 2 in Lemma 2

The cost of the online server after time $t+2L$ includes six parts:

• 1. The costs of the requests between node $-x-1$ and node $-L$. $L-x$ requests are released during the time interval $[t+2x+1,t+x+L]$ one by one, the $L-x$ requests are served during the time interval $[t+2L+1,t-x+3L]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the $L-x$ requests are equal to $2(L-x)^{2}B$;

• 2. The costs of the requests between node $-x+1$ and node $-x+y$. y requests are released during the time interval $[t+2L+1,t+2L+y]$ one by one, the y requests are served during the time interval $[t+4L-2x+1,t+4L-2x+y]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the y requests are equal to $2(L-x)yB$;

• 3. The costs of the requests between node $-x$ and node $-x-y_{1}$. $y_{1}+1$ requests are released during the time interval $[t+2L+2y,t+2L+2y+y_{1}]$ one by one, the $y_{1}+1$ requests are served during the time interval $[t+4L-2x-y_{1},t+4L-2x]$ by online server. The costs of these requests are equal to $2(L - x - y)B,2(L - x - y - 1)B, \cdots \cdots 2(L - x - y - {y_1})B$ respectively which forms an arithmetic sequence. The costs of the $y_{1}+1$ requests are equal to $(2(L - x - y)({y_1} + 1) - ({y_1}^2 + {y_1}))B$;

• 4. The costs of the requests between node $-x+y+1$ and node L. $L+x-y$ requests are released during the time interval $[t + 2L + 3y + 2{y_1} + 1,t + 3L + 2y + 2{y_1} + x]$ one by one, the $L+x-y$ requests are served during the time interval $[t + 4L + y - 2x + 1,t + 5L - x ]$ by online server. The cost is equal to $2(L - y - {y_1} - x)B$ for each request. The costs of the $L+x-y$ requests are equal to $2(L - y - {y_1} - x)(L + x - y)B$;

• 5. The costs of the requests between node $x+y+y_{1}-1$ and the origin. $x+y+y_{1}$ requests are released during the time interval $[t + 4L + y + {y_1} + 1,t + 4L + 2y + 2{y_1} + x]$ one by one, the $x+y+y_{1}$ requests are served during the time interval $[t + 6L - y - {y_1} - 2x + 1, t+6L-x]$ by online server. The cost is equal to $2(L - y - {y_1} - x)B$ for each request. The costs of the $x+y+y_{1}$ requests are equal to $2(L - y - {y_1} - x)(x + y + {y_1})B$;

• 6. The traveling time of the online server. The traveling time of the online server is equal to $4L-x$.

The total cost of the online server is equal to ${c_{online}} = 4L - x + (2(L - x)(L - x + y) + 2(L - x - y)({y_1} + 1) - ({y_1}^2 + {y_1}) + 2(L - y - {y_1} - x)(L + 2x + {y_1}))B$.

As shown in Fig. 6, the cost of the offline server is equal to ${c_{offline}} = 2y + 2{y_1} + x + 2L$.

The cost ratio is equal to $f({y_1}) = \frac{{{c_{online}}}}{{{c_{offline}}}}$,

$$\begin{aligned} f'({y_1}) = \frac{{ - 2{c_{online}} + (2L - 8x - 4y - 6{y_1} - 1)B{c_{offline}}}}{{{c_{offline}}^2}}\end{aligned}$$

.

Let $g({y_1}) = - 2{c_{online}} + (2L - 8x - 4y - 6{y_1} - 1)B{c_{offline}}$, $g'({y_1}) = - 6B{c_{offline}} < 0$.

Because $0 \le {y_1} \le L - x - y - 1$, when $y_{1}=0$, $g(y_{1})$ can get the maximum value.

Let $h(x) = g(0) = - 2(4L - x + 2((L - x)(L - x + y) + (L - x - y)(L + 2x + 1))B) + (2y + x + 2L)(2L - 8x - 4y - 1)B$

$h'(x) = ( - 16L - 20x - 26y - 3)B + 2 < 0$, when $x=0$, h(x) can get the maximum value.

$h(0) = - 8L + ( - 4{L^2} - 4yL - 8{y^2} - 6L + 2y)B < 0$, $g(0) \le h(0) < 0$, $f'({y_1}) < 0$, $f(y_{1})$ is a decreasing function of $y_{1}$. When $y_{1}=0$, $\frac{{{c_{online}}}}{{{c_{offline}}}}$ can get the maximum value. Similar to the above analysis, when $y=1$, $\frac{{{c_{online}}}}{{{c_{offline}}}}$ can get the maximum value.

When $y_{1}=0, y=1$, $\frac{{{c_{online}}}}{{{c_{offline}}}} = \frac{{4L - x + (2(L - x)(L + 2x) + (L - 7x - 2))B}}{{2L + x + 2}}$

$$\begin{aligned} {\rho _0} - \frac{{{c_{online}}}}{{{c_{offline}}}} = \frac{{2(4L - x) + 8(3x + 1)(2L + x)B}}{{(2L + x + 2)(2L + x)}} > 0\end{aligned}$$

Appendix D. Case 3.1.1 in Lemma 2

The cost of the online server after time $t+2L$ includes seven parts:

• 1. The costs of the requests between node $-x-1$ and node $-L$. $L-x$ requests are released during the time interval $[t+2x+1,t+x+L]$ one by one, the $L-x$ requests are served during the time interval $[t+2L+1,t-x+3L]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the $L-x$ requests are equal to $2(L-x)^{2}B$;

• 2. The costs of the requests between node $-x+1$ and node $-x+y$. y requests are released during the time interval $[t+2L+1,t+2L+y]$ one by one, the y requests are served during the time interval $[t+4L-2x+1,t+4L-2x+y]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the y requests are equal to $2(L-x)yB$;

• 3. The costs of the requests between node $-x$ and node $-L+y+1$. $L-x-y$ requests are released during the time interval $[t+2L+2y,t+3L+y-x-1]$ one by one, the $L-x-y$ requests are served during the time interval $[t+3L+y-x+1,t+4L-2x]$ by online server. The costs of these requests are equal to $2(L - x - y)B,2(L - x - y - 1)B, \cdots \cdots 2B$ respectively which forms an arithmetic sequence. The costs of the $L-x-y$ requests are equal to $({(L - x - y)^2} + (L - x - y))B$;

• 4. The costs of the requests between node $-L+y-1$ and node $-L+y-z$. z requests are released during the time interval $[t + 3L + y - x + 1,t + 3L + y - x + z]$ one by one, the z requests are served during the time interval $[t + 5L + y - 3x + 1,t + 5L + y - 3x + z]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the z requests are equal to $2(L-x)zB$;

• 5. The costs of the requests between node $-L+y$ and node $y-x-z-1$. $L-x-z$ requests are released during the time interval $[t+3L+y-x+2z,t+4L+y-2x+z-1]$ one by one, the $L-x-z$ requests are served during the time interval $[t+4L+y-2x+z+1,t+5L-3x+y]$ by online server. The costs of these requests are equal to $2(L - x - z)B,2(L - x - z - 1)B, \cdots \cdots 2B$ respectively which forms an arithmetic sequence. The costs of the $L-x-z$ requests are equal to $({(L - x - z)^2} + (L - x - z))B$;

• 6. The costs of the requests between node $y-x-z+1$ and the origin. $x+z-y$ requests are released during the time interval $[t + 4L - 2x + y + z + 1,t + 4L - x + 2z]$ one by one, the $x+z-y$ requests are served during the time interval $[t + 6L - 4x + z + y + 1,t + 6L - 3x + 2z]$ by online server. The cost is equal to $2(L-x)B$ for each request. The costs of the $x+z-y$ requests are equal to $2(L - x)(x + z - y)B$;

• 7. The traveling time of the online server. The traveling time of the online server is equal to $4L-3x+2z$.

The total cost of the online server is equal to ${c_{online}} = 4L + 3x - 2z + (2(L - x)(2z + L) + {(L - x - y)^2} + (L - x - y) + {(L - x - z)^2} + (L - x - z))B$.

As shown in Fig. 7, the cost of the offline server is equal to ${c_{offline}} = 2L + 2z - x$.

Let $f(y,z) = \rho - \frac{{{c_{online}}}}{{{c_{offline}}}} = \frac{A}{{x{c_{offline}}}}$, $f{'_z}(y,z) = \frac{{ - 2xA + A'x{c_{offline}}}}{{x{c_{offline}}}}$.

Let $g(z) = - 2xA + A'x{c_{offline}}$, $g'(z) = - 2xB{c_{offline}} < 0$.

Because $z \le y$, when $z=y$, g(z) can get the minimum value.

$g'(y) = 2x(2L + 2y - x)B > 0$, when $y=1$, g(y) can get the minimum value.

$ming(y)=g(1)> g(0) = {x^2}(4{L^2} + 2{x^2} - 6xL + 6L - 5x)B > 0$, we can get $g(z) > 0$.

f(y, z) is an increasing function of z, when $z=1$, f(y, z) can get the minimum value. For f(y, 1), with the similar analysis, we can get f(y, 1) is an increasing function of y, and

$minf(y,z)=f(1,1)> f(0,0) = \frac{{(4{{(L - x)}^2}L - 2x({{(L - x)}^2} + (L - x))B + 4{{(L - x)}^2}}}{{x(2L - x)}} > 0$.