Bicriteria two-machine flowshop scheduling: approximation algorithms and their limits

Abstract

We consider bicriteria flowshop scheduling problems with two machines to simultaneously minimize the makespan and the total completion time without any prioritization between the two objectives. An approximation relative to the optimal value of the problem with regard to each objective is adopted, even though a schedule with both objectives reaching their minimum at the same time may not exist. We consider several problems with ordered aspects on processing times such as job ordered and machine ordered. For each problem, we propose a fast $(\rho _1,\rho _2)$-approximation algorithm, where $\rho _1$ and $\rho _2$ indicate the approximation ratios with regard to the makespan and the total completion time, respectively, and we explore the problem’s inapproximability region that cannot be reached by any algorithm.

Appendices

Exact expressions for specific ratios

$$\begin{aligned}{} & {} {\mathcal {L}}(\rho _1, b): = \nonumber \\{} & {} \quad \frac{\rho _1 b^2 + (3\rho _1 -1)b + 2(\rho _1-1)}{(2\rho _1^2 + \rho _1 -2)b^2 + (4\rho _1^2 - \rho _1-3)b +2(\rho _1-1)\rho _1} \end{aligned}$$

(49)

where b is the function of $\rho _1$

$$\begin{aligned}{} & {} b(\rho _1):=\frac{\rho _1^3-3\rho _1+2 - \sqrt{\rho _1^3+\rho _1^2- 3\rho _1 +1}}{-(\rho _1^3+ \rho _1^2 - 2\rho _1+1)}.\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(50)

$$\begin{aligned}{} & {} R(\mu ,\kappa ):= (1-2\mu ) \frac{(2 \mu + 1)\kappa ^2 + (6\mu -5)\kappa }{(1-\mu )^2\kappa ^2 + (4\mu ^2-9\mu + 3)\kappa - 5\mu ^2+11\mu -4}.\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(51)

$$\begin{aligned}{} & {} \kappa ^*(\mu ) = \frac{-(2 \mu + 1)(-5\mu ^2 + 11\mu - 4) - 2 \sqrt{(-5\mu ^2 + 11\mu - 4)(-8\mu ^4 + 4 \mu ^3 +38\mu ^2 - 37 \mu +9)}}{2\mu ^3+3\mu ^2-19\mu +8}\qquad \qquad \qquad \qquad \qquad \end{aligned}$$

(52)

Proofs of lemmas

Lemma 4

Suppose there are in schedule $\sigma ^T$ two jobs $[h_1]$ and $[h_2]$ such that $h_1<h_2$ and $A_{[h_1]} \ge A_{[h_2]}$. Let $z_{[h_1]}:= C_{[h_1-1]}(\sigma ^T)-C_{1,[h_1-1]}(\sigma ^T)$ for $h_1 \ge 2$ and $z_{[1]}:=0$.

If $C_{[h_1]}(\sigma ^T)= C_{1,[h_1]}(\sigma ^T) + B_{[h_1]}$, then

• $B_{[h_2]}\ge B_{[h_1]}$ and

• $\max \left\{ z_{[h_1]},A_{[h_2]}\right\} + B_{[h_2]} \ge A_{[h_1]} + B_{[h_1]}$.

Proof

Suppose that it is not true, i.e., that interchanging jobs $[h_1]$ and $[h_2]$ in $\sigma _0$ decreases the total completion time, contradicting the optimality of $\sigma _0$. $\square$

Lemma 5

For any $\eta \le |J_1(\kappa )|$, we have

$$\begin{aligned} \sum C_j^*\ge & {} A_n + B_n + \left( \kappa + \eta + 1\right) L_{11}(\kappa ) \\{} & {} + \ \left( \frac{(\kappa +3)\kappa }{2} -\frac{(\eta -1)\eta }{2}\right) A_{n-\kappa } \\{} & {} + \sum _{j=n-\kappa }^{n-1} B_j \end{aligned}$$

Proof

Consider the completion time of each job in $\sigma _0$.

$$\begin{aligned} C^*_{n-\kappa +h}&= L_{11}(\kappa ) + \sum ^{n-\kappa +h}_{l=n-\kappa } A_{l} + B_{n-\kappa +h}{} & {} \\&\ge L_{11}(\kappa ) + (h+1) A_{n-\kappa } + B_{n-\kappa +h} \\&\quad \text {for } h=0,\ldots ,\kappa -1, \\ C^*_{n-\kappa -h}&= L_{11}(\kappa ) - \sum ^{n-\kappa -1}_{l=n-\kappa -h+1} A_{l} + B_{n-\kappa -h}{} & {} \\&\ge L_{11}(\kappa ) - (h-1) A_{n-\kappa } + B_{n-\kappa -h} \\&\quad \text {for } h=1,\ldots ,|J_{1}(\kappa )| \end{aligned}$$

For any $\eta \le |J_1(\kappa )|$, by adding the inequalities above, we obtain

$$\begin{aligned} \sum _{j=1}^{n-1} C_j^*&\ge \left( \kappa +\eta \right) L_{11}(\kappa ) + \left( \frac{(\kappa +1)\kappa }{2} -\frac{(\eta -1)\eta }{2}\right) A_{n-\kappa } \nonumber \\&\quad + \sum _{j=\eta }^{n-1} B_j{} & {} \end{aligned}$$

(53)

We also have

$$\begin{aligned} C_n^*{} & {} = L_{11}(\kappa ) + L_{12}(\kappa ) + A_n + B_n \nonumber \\{} & {} \ge L_{11}(\kappa )+ \kappa A_{n-\kappa } + A_n + B_n \end{aligned}$$

(54)

Thus, we add (53) and (54) to obtain

$$\begin{aligned} \sum C_j^*\ge & {} A_n + B_n + \left( \kappa + \eta + 1 \right) L_{11}(\kappa ) \\{} & {} + \left( \frac{(\kappa +3)\kappa }{2} -\frac{(\eta -1)\eta }{2}\right) A_{n-\kappa } + \sum _{j=n-\kappa }^{n-1} B_j. \end{aligned}$$

and the proof is complete. $\square$

Lemma 6

For $\mu \in \left[ \frac{1}{3},\frac{1}{2} \right)$, if $C_{\max }(\sigma _0)> (1+\mu ) C_{\max }^*$, then $B_n \ge B_j$, for $j\in J$.

Proof

Suppose it is not true, then, from (18), job $j_{\max }$ satisfies $B_{j_{\max }} > B_n$, and thus

$$\begin{aligned} C_{\max }^* \ge A_{j_{\max }} + \max \left\{ B_{j_{\max }}, A_n\right\} + B_n > 3 B_n \end{aligned}$$

since $A_{j_{\max }} \ge B_{j_{\max }} > B_n$ and $A_n \ge B_n$. From (19) it follows that

$$\begin{aligned} C_{\max }(\sigma _0) = L_1 + A_n + B_n < \left( 1+\frac{1}{3}\right) C_{\max }^* \le (1+\mu ) C_{\max }^* \end{aligned}$$

contradicting the assumption $C_{\max }(\sigma _0)> (1+\mu ) C_{\max }^*$. $\square$

Lemma 7

If $C_{\max }(\sigma _0) > (1+\mu )C_{\max }^*$, then the total completion time of $\sigma (\kappa )$ is

$$\begin{aligned}{} & {} \sum C_j(\sigma (\kappa ))\\{} & {} \le \sum C_j^* \\{} & {} \quad + \kappa \left( (A_n + B_n) - \frac{\kappa +3}{2}A_{n-\kappa } + \frac{\kappa -1}{\kappa } \sum _{j=n-\kappa }^{n-2} B_j\right) \end{aligned}$$

where $\mu \in \left[ \frac{1}{3},\frac{1}{2}\right)$ and $\kappa \ge 2$.

Proof

We first prove the following statement. If $C_{\max }(\sigma _0) > (1+\mu )C_{\max }^*$ for $\mu \in \left[ \frac{1}{3},\frac{1}{2}\right)$ and $\kappa \ge 2$, then

$$\begin{aligned} L_{12}(\kappa ) \le B_n. \end{aligned}$$

(55)

We prove it by contradiction. If $L_{12}(\kappa ) > B_n$, then, by (19) and (26), with $\mu \ge \frac{1}{3}$,

$$\begin{aligned} C_{\max }(\sigma (\kappa ))\le & {} L_{1} + A_n + L_{22}(\kappa ) < C_{\max }^* + (1-2\mu ) C_{\max }^* \\= & {} (2-2\mu ) C_{\max }^* \le (1+\mu ) C_{\max }^* \end{aligned}$$

which contradicts (17), i.e., $C_{\max }(\sigma (\kappa -1)) > (1+\mu )C_{\max }^*$. Thus, $L_{12}(\kappa ) \le B_n$.

Table 7 Approximation ratios of simple priority rules for $F2 \mid \beta \mid (C_{\max }, \sum C_j)$

Then, the total completion time of $\sigma (\kappa )$ can be expressed as follows.

Based on (55) and (15), we obtain

$$\begin{aligned} \sum C_j(\sigma (\kappa ))= & {} \sum _{j=1}^{n-\kappa -1} ((n+1-j) A_j + B_j) \\{} & {} + (\kappa +1)(A_n + B_n) + \sum _{j=n-\kappa }^{n-1} (n-j) B_j \\= & {} \sum C_j^* + \kappa (A_n + B_n) - \sum _{j=n-\kappa }^{n-1} (n+1-j) A_j \\{} & {} + \sum _{j=n-\kappa }^{n-2} (n-1-j) B_j\\\le & {} \sum C_j^* + \kappa (A_n + B_n) - \frac{(\kappa +3)\kappa }{2}A_{n-\kappa } \\{} & {} + (\kappa -1) \sum _{j=n-\kappa }^{n-2} B_j \end{aligned}$$

since $\sum _{j=n-\kappa }^{n-2} (n-1-j) B_j \le \sum _{j=n-\kappa }^{n-2} (n-1-(n-\kappa )) B_j = (\kappa -1)\sum _{j=n-\kappa }^{n-2} B_j$. $\square$

Lemma 8

Given $\mu \in \left[ \frac{1}{3}, 1\right]$, if $C_{\max }(\sigma _0) > (1+\mu ) C_{\max }^*$, we have

$$\begin{aligned} L_{23} + L_{22}(\kappa ) - B_{n_1} > \kappa + \pi . \end{aligned}$$

Proof

If $\kappa = 1$, then $L_{23} + L_{22}(\kappa ) - B_{n_1} = L_{23} \ge 1 + \pi = \kappa + \pi$. Thus, we only need to consider $\kappa \ge 2$. Suppose that $L_{23} + L_{22}(\kappa ) - B_{n_1} \le \kappa + \pi$. Then,

$$\begin{aligned} C_{\max }(\sigma (\kappa )) = n + B_{n_1} \le C_{\max }^* \end{aligned}$$

Note that, for $\kappa \ge 2$, $L_{23} + L_{22}(\kappa -1) - B_{n_1} > \kappa + \pi -1$; otherwise $C_{\max }(\sigma (\kappa -1)) = n + B_{n_1} \le C_{\max }^*$ contradicts (41) from the definition of $\kappa$, i.e., $C_{\max }(\sigma (\kappa -1)) >(1+\mu ) C_{\max }^*$. Then,

$$\begin{aligned} C_{\max }(\sigma (\kappa -1))\le & {} n-\pi -\kappa + 1 + L_{23} + L_{22}(\kappa -1)\\\le & {} n + 1 + B_{n_1} - B_{n-\pi -\kappa } \\\le & {} n + 1 \le C_{\max }^* + \frac{1}{n} C_{\max }^*\\\le & {} \frac{4}{3} C_{\max }^* \le (1+\mu ) C_{\max }^* \end{aligned}$$

which leads to a contradiction of $C_{\max }(\sigma (\kappa -1)) >(1+\mu ) C_{\max }^*$. $\square$

Approximation ratios for simple priority rules

We consider the performance analysis of some priority rules for bicriteria fully ordered or semi-ordered flowshop problems on two machines.

One can easily design instances to claim that LPT according to no matter $A_j$, $B_j$, or $A_j+B_j$ may be arbitrarily bad for the problem minimizing the total completion time. Take $F2 \mid A_j=A \mid (C_{\max },\sum C_j)$ for example. Consider the instance with q jobs having $B_j=0$, for $j=1,\ldots ,q-1$, and $B_{q}=q$ as well as $0 < A\le 1$. In schedule $\sigma ^T$, job q is processed in the end, while in schedule $\sigma ^M$ job q is processed in the beginning, following the LPT rule. When q goes to $\infty$ and A goes to 0, the ratio of ${\sum C_j(\sigma ^M)}/{\sum C_j^*}$ goes to infinity.

Hence, we only discuss the performance of SPT rule on $A_j$, $B_j$ and $A_j + B_j$ and summarize the results in Table 7. Furthermore, the last two columns of Table 7 indicate the approximation ratio of $C_{\max }$ (or $\sum C_j$) for an optimal schedule only minimizing $\sum C_j$ (or $C_{\max }$). Specifically, given an arbitrary optimal schedule for $F2 \mid \beta \mid C_{\max }$, we show that it can be arbitrarily bad for $F2 \mid \beta \mid \sum C_j$ accordingly regardless of $\beta$. On the other hand, an optimal schedule for $F2 \mid \beta \mid \sum C_j$ may have a different performance ratio regarding the makespan when different $\beta$ is chosen. If the problem to minimize the total completion time is polynomial time solvable, we can investigate properties of optimal schedules and find the approximation ratio regarding the makespan. Even if a problem to minimize the total completion time is NP-hard, we may be able to identify the approximation ratio regarding the makespan from its lower bound. From the inapproximability instance for $F2 \mid A_j=A \mid (C_{\max }, \sum C_j)$ in Sect. 2, we consider the problem instance having $q+1$ jobs with $(A_j,B_{j})=(a,0)$ for $j=1,\ldots ,q$, and $(A_{q+1},B_{q+1}) = (a,N)$, where $0\le a \le N/q$. In this instance, the optimal total completion time schedule has the makespan twice of the optimal makespan. Note that an arbitrary schedule for $F2 \mid \mid C_{\max }$ without unnecessary idle times is a 2-approximation schedule Lee et al. (2019). Thus, the tight approximation ratio of the optimal total completion time schedule is (2, 1).

Generally, for the machine ordered flowshop, we consider an arbitrary schedule $\sigma$ with a critical job l. Then, we obtain the approximation ratio of the makespan.

$$\begin{aligned} C_{\max }(\sigma )= & {} \sum _{j=1}^l A_j + \sum _{j=l}^n B_j \le \max \left\{ \sum _{j=1}^n A_j, \sum _{j=1}^n B_j \right\} \nonumber \\{} & {} + \min \{A_l, B_l\} \le C_{\max }^* + \frac{1}{2} C_{\max }^* = \frac{3}{2} C_{\max }^* \end{aligned}$$

(56)

Some ratios (with respect to either $C_{\max }$ or $\sum C_j$) can be straightforwardly derived from the literature. We analyze three nontrival cases as follows.

• SPT-B for $F2 \mid A_j \le B_j \mid \sum C_j$

• SPT-AB for $F2 \mid A_j \le B_j \mid C_{\max }$

• SPT-AB for $F2 \mid A_j \ge B_j \mid \sum C_j$

These results are combined with previously known results to establish the following tight approximation ratios of $F2 \mid \beta \mid (C_{\max }, \sum C_j)$ with various $\beta$.

Theorem 5

Some tight approximation ratios are obtained as follows:

• (i) SPT-B provides a $\left( \frac{3}{2},\frac{5}{3}\right)$-approximation schedule for $F2 \mid A_j \le B_j \mid (C_{\max }, \sum C_j)$.

• (ii) SPT-AB provides a $\frac{4}{3}$-approximation schedule for $F2 \mid A_j \le B_j \mid C_{\max }$.

• (iii) SPT-AB provides a $\sqrt{2}-$approximation schedule for $F2 \mid A_j \ge B_j \mid \sum C_j$.

Proof

See Online supplement.

We do not have exact approximation ratios regarding SPT-AB for $F2 \mid j-ordrd \mid \sum C_j$ and $F2 \mid A_j \le B_j \mid \sum C_j$. It is known that the approximation ratio regarding SPT-AB for the most general case ($F2 \mid \mid \sum C_j$) is 2. As for $F2 \mid j-ordrd \mid \sum C_j$, its special case $F2 \mid A_j = A \mid \sum C_j$ has the approximation ratio of 4/3. Thus, the approximation ratio $\rho _2 \in [4/3, \ 2]$. As for $F2 \mid A_j \le B_j \mid \sum C_j$, we consider an instance with $(A_j,B_j)= (1,1)$ for $j=1,\dots , (\sqrt{2} -1) n -1$ and $(A_j,B_j)= (0,2)$ for $j= (\sqrt{2} -1) n,\dots , n$. Then, we can show that the approximation ratio of SPT-AB is no less than $\sqrt{2}$ for $F2 \mid A_j \le B_j \mid \sum C_j$. Thus, the approximation ratio $\rho _2 \in [\sqrt{2}, \ 2]$.