Single-Machine Scheduling with Multi-Criteria and Due- Windows

Theorem 1.

³⁵ The problem of minimizing maximum earliness subject to no machine idle time, denoted as $1\Vert E_{\mathit{max}},$ is solved by sequencing the jobs according to the Minimum Slack Time $\left(\mathit{MST}\right)-$ rule, that is, in order of non-decreasing $s_i=d_i^{\left(1\right)}-p_i.$

Theorem 2.

³⁵ The problem of minimizing maximum tardiness subject to no machine idle time, denoted as $1\Vert T_{\mathit{max}}$ , is solved by sequencing the jobs according to the Earliest Due Date $\left(\mathit{EDD}\right)-$ rule, that is, in order of non-decreasing $d_i^{\left(2\right)}.$

Definition 1.

³⁶ In the Latest Processing Time $\left(\mathit{LPT}\right)-$ rule: Sequencing all jobs in non-increasing order of the processing times $p_i$ i.e. $(p_1\ge p_2\ge \cdots \ge p_n).$

Definition 2.

³⁷ The term “optimize” in a multi-objective decision making problem refers to a solution around which there is no way of improving any objective without worsening at least one other objective.

Definition 3.

³⁵ A feasible schedule $\sigma$ is Pareto optimal with respect to the objective functions $f_1,f_2,\dots ,f_n$ if there is no feasible schedule $\pi$ with $f_i\left(\pi \right)\le f_i\left(\sigma \right)$ for $i=1,\dots ,n$ , where at least one of the inequalities is strict.

Definition 4.

³⁸ A solution s dominates $s^{\prime }$ if the $z=f\left(s\right)$ dominates $z^{\prime }=f\left(s^{\prime }\right)$ , this is, $f_i\left(s\right)\le f_i\left(s^{\prime }\right)$ for all $i$ and $f_i\left(s\right)<f_i\left(s^{\prime }\right)$ for at least one $i$ .

Theorem 3.

³⁵ The $1\left|\right|f_{\mathit{max}}$ problem is solved as follows: while there are unassigned jobs, assign the job that has minimum cost when scheduled in the last unassigned position to that position.

Case (1).

If $d_i^{\left(1\right)}>C_i,\forall i\in N$ then the $\mathit{MST}-$ sequence is an $\mathit{ES}$ for the problem $\left(P\right)$ , as shown in Table 1.

Proof.

Since $d_i^{\left(1\right)}>C_i,$ for all $i\in N,$ we obtain that all the jobs are early i.e. $(T_i=0,\forall i\in N.$ So $T_{\mathit{max}}=0)$ , and $(V_i=0,\forall i\in N$ . So $V_{\mathit{max}}=0)$ , and $E_i=\{d_i^{\left(1\right)}-C_i,\forall i\in N\}.$

So $E_{\mathit{max}}=max\{d_i^{\left(1\right)}-C_i\},\forall i\in N,$ and the problem (P) can be state as: $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=F(E_{\mathit{max}},0)=F(\mathit{max}\{d_i^{\left(1\right)}-C_i\},0)$ .

Since the $\mathit{MST}-$ sequence yields optimal solution for $E_{\mathit{max}},$ it follows that the $\mathit{MST}-$ sequence is an $\mathit{ES}$ for the problem $\left(P\right)$ .

Case (2).

If $d_i^{\left(2\right)}<C_i,\forall i\in N\backslash \left\{1\right\}(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}])$ for the problem $\left(P\right)$ , then we obtain three cases:

Proof (a).

Since $d_i^{\left(2\right)}<C_i$ except $(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ this means that all the jobs are not early i.e. $(E_i=0,\forall i\in N$ . So $E_{\mathit{max}}=0)$ ,

For $f_1$ problem: ${\mathit{ET}}_{\mathit{max}}=T_{\mathit{max}}=\mathit{max}\left\{T_i\right\}=\mathit{max}\{\mathit{max}\{C_i-d_i^{\left(2\right)},0\}\}=\mathit{max}\{C_i-d_i^{\left(2\right)}\},i\in N.$

Therefore, $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=F(T_{\mathit{max}}^{\ast },V_{\mathit{max}})=(\mathit{max}\{C_i-d_i^{\left(2\right)}\},V_{\mathit{max}}),i\in N$ .

For $f_2$ problem: If $d_i^{\left(2\right)}<C_i<d_i^{\left(2\right)}+p_i$ except $(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ , then $V_i=C_i-d_i^{\left(2\right)},(V_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\{C_i-d_i^{\left(2\right)}\}=T_{\mathit{max}}),i\in N\backslash \left\{1\right\}.$

So, the problem $\left(P\right)$ can be state as the following:

Now, since the $\mathit{EDD}‐$ sequence gives optimal solution for $T_{\mathit{max}},$ then the $\mathit{EDD}‐$ sequence is an $\mathit{ES}$ .

Proof (b).

Suppose that $\sigma$ be the $\mathit{EDD}‐$ sequence.

Now, since $d_i^{\left(2\right)}+p_i\le C_i⟹d_i^{\left(2\right)}<C_i,(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ ,

which implies all the jobs are not early i.e. $(E_i=0,\forall i\in N$ . So $E_{\mathit{max}}=0)$ ,

For $f_1$ problem: ${\mathit{ET}}_{\mathit{max}}=T_{\mathit{max}}=\mathit{max}\left\{T_i\right\}=\mathit{max}\{\mathit{max}\{C_i-d_i^{\left(2\right)},0\}\}=\mathit{max}\{C_i-d_i^{\left(2\right)}\},i\in N.$

For $f_2$ problem: If $d_i^{\left(2\right)}+p_i\le C_i(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\},$ then $V_i=p_i,$ and $p_i\le T_i,\forall i\in N\backslash \left\{1\right\},$ i.e. $(V_{\mathit{max}}=\mathit{max}\{\mathit{min}\{T_i,p_i\}\}=\mathit{max}\left\{p_i\right\}=p_{\mathit{max}},\forall i\in N\backslash \left\{1\right\})$ .

So, the problem (P) can be presented as the following:

Therefore, the sequence $\sigma$ is an $\mathit{ES}$ by Theorem (2).

Proof (c).

Suppose that $\sigma$ be the $\mathit{EDD}‐$ sequence.

Since $d_i^{\left(2\right)}<C_i,(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ , this leads to all the jobs are not early i.e. $(E_i=0,\forall i\in N$ . So $E_{\mathit{max}}=0)$ ,

For $f_1$ problem: ${\mathit{ET}}_{\mathit{max}}=T_{\mathit{max}}=\mathit{max}\left\{T_i\right\}=\mathit{max}\{\mathit{max}\{C_i-d_i^{\left(2\right)},0\}\}=\mathit{max}\{C_i-d_i^{\left(2\right)}\},i\in N$ .

For $f_2$ problem: If $d_i^{\left(2\right)}<C_i<d_i^{\left(2\right)}+p_i$ for $(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ , then $V_i=C_i-d_i^{\left(2\right)},(C_1\in [d_1^{\left(1\right)},d_1^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ and if $d_i^{\left(2\right)}+p_i\le C_i,\forall i\in N\backslash \left\{1\right\}$ , then $V_i=p_i$ , i.e. $(V_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\{\mathit{min}\left\{\right\{C_i-d_i^{\left(2\right)}\},p_i\}\}),$ $i\in N\backslash \left\{1\right\}$ .

Therefore, the problem $\left(P\right)$ can be introduced as the following:

Thus, the sequence $\sigma$ is an $\mathit{ES}$ by Theorem (2).

Case (3).

If we have a ${\sigma }^{\prime }$ be any sequence of jobs such that $d_i^{\left(1\right)}\le C_i\le d_i^{\left(2\right)},\forall i\in N$ , then $\sigma$ is an $\mathit{ES}$ for the problem $\left(P\right)$ , as shown in example (1).

Proof.

Suppose that ${\sigma }^{\prime }$ be any sequence of jobs such that $d_i^{\left(1\right)}\le C_i\le d_i^{\left(2\right)},\forall i\in N,$ this means all the jobs $\mathit{JIT}$ of due-windows.

For $f_1$ problem: $(E_i=0,\forall i\in N$ , so $E_{\mathit{max}}=0)$ and $(T_i=0,\forall i\in N$ , so $T_{\mathit{max}}=0)$ which yields ${\mathit{ET}}_{\mathit{max}}=0$ .

For $f_2$ problem: $(V_i=0,\forall i\in N$ , so $V_{\mathit{max}}=0)$ .

Thus, the problem $\left(P\right)$ can be written as the following:

$F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(0,0)$ . So, any sequence $\sigma$ with the condition $d_i^{\left(1\right)}\le C_i\le d_i^{\left(2\right)},\forall i\in N,$ is an $\mathit{ES}$ for the problem $\left(P\right)$ .

Proposition 1.

If $\mathit{MST}‐$ rule, $\mathit{EDD}‐$ rule, $\mathit{LA}$ and $\mathit{LPT}‐$ sequence are identical in one sequence, then this sequence gives only one $\mathit{ES}$ for problem $\left(P\right)$ , as shown in Table 1.

Proof.

Suppose that $\sigma =\mathit{MST}=\mathit{EDD}=\mathit{LA}=\mathit{LPT}$ have the same sequence,

Since the $\mathit{MST}‐$ rule ensures the minimum value of $E_{\mathit{max}}$ , schedule $\sigma$ is an optimal for $E_{\mathit{max}}$ according to Theorem (1). In addition, by applying the $\mathit{EDD}‐$ rule, $T_{\mathit{max}}$ is minimum, which confirms that $\sigma$ is optimal for $T_{\mathit{max}}$ based on Theorem (2). Furthermore, the $\mathit{LA}$ guarantees the minimization of $V_{\mathit{max}}$ , which making $\sigma$ optimal for $V_{\mathit{max}}$ as established in Theorem (3).

To establish the uniqueness of schedule $\sigma$ , let $\mu$ represent any arbitrary schedule. Then it holds that:

$E_{\mathit{max}}(\sigma =\mathit{MST})\le E_{\mathit{max}}\left(\mu \right)$ , $T_{\mathit{max}}(\sigma =\mathit{EDD})\le T_{\mathit{max}}\left(\mu \right)$ and $V_{\mathit{max}}(\mathrm{\sigma }=\mathrm{LA})\le V_{\mathit{max}}\left(\mu \right)$ . Consequently, for the problem (P) the combines solution $({\mathit{ET}}_{\mathit{max}}\left(\sigma \right),V_{\mathit{max}}\left(\sigma \right))$ strictly dominates the solution $({\mathit{ET}}_{\mathit{max}}\left(\mu \right),V_{\mathit{max}}\left(\mu \right))$ .

Case (4).

In the common due-windows for the problem $\left(P\right)$ , if $d^{\left(1\right)}\ge C_n$ and we have the $\mathit{LPT}‐$ sequence then $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(d^{\left(1\right)}-C_1,0)$ and the $\mathit{MST}‐$ sequence is $\mathit{ES}$ , as shown in Table 1.

Proof.

Suppose that $\sigma$ be $\mathit{LPT}‐$ sequence and $\mathit{MST}‐$ sequence.

Now, since $d^{\left(1\right)}>C_i,\forall i\in N,$ this means that all the jobs are early. $(T_i=0$ and $V_i=0,\forall i\in N$ . So $T_{\mathit{max}}=0$ and $V_{\mathit{max}}=0)$ , respectively.

Since we have common due-windows, this means $[d_i^{\left(1\right)},d_i^{\left(2\right)}]=[d^{\left(1\right)},d^{\left(2\right)}],\forall i\in N.$

$E_i=d^{\left(1\right)}-C_i,\forall i\in N,E_{\mathit{max}}=\mathit{max}\{\mathit{max}\{d^{\left(1\right)}-C_i,0\}\}=\mathit{max}\{d^{\left(1\right)}-C_i\}=d^{\left(1\right)}-C_1.$ So, the $\mathit{MST}‐$ sequence gives optimal solution for $E_{\mathit{max}}$ by Theorem (1).

Thus, the problem $\left(P\right)$ can be written as the following:

Hence, the sequence $\sigma$ is $\mathit{ES}$ for the problem (P).

Lemma (1).

The $\mathit{ES}$ for $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(d^{\left(1\right)}-C_1,0)$ , where $d^{\left(1\right)}\ge C_n$ be $(d^{\left(1\right)}-p_{\mathit{max}},0)$ .

Proof.

It is clear by Definition (1).

Case (5).

For the problem $\left(P\right)$ with the common due-window $[d_i^{\left(1\right)},d_i^{\left(2\right)}]=[d^{\left(1\right)},d^{\left(2\right)}],\forall i\in N$ , if $d^{\left(2\right)}<C_i,(C_1\in [d^{\left(1\right)},d^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ , then conclude three cases:

Proof (a).

Let $d^{\left(2\right)}<C_i,(C_1\in [d^{\left(1\right)},d^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\},$ it follows that all the jobs are not early.

For $f_1$ problem: $E_i=0,\forall i\in N$ , therefore $E_{\mathit{max}}=0$ ,

For $f_2$ problem: $V_i=\mathit{min}\{T_i,p_i\}=T_i,\forall i\in N\backslash \left\{1\right\}$ , i.e. $V_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\left\{T_i\right\}=T_{\mathit{max}}=C_n-d^{\left(2\right)}$ .

Therefore, the problem $\left(P\right)$ can introduced by:

Proof (b).

If $d^{\left(2\right)}+p_i\le C_i,(C_1\in [d^{\left(1\right)},d^{\left(2\right)}]),\forall i\in N\backslash \left\{1\right\}$ , then all the jobs are not early.

For $f_1$ problem: $E_{\mathit{max}}=0$ and $T_{\mathit{max}}=C_n-d^{\left(2\right)}$ .

For $f_2$ problem: $V_i=\mathit{min}\{T_i,p_i\}=p_i,\forall i\in N\backslash \left\{1\right\}$ , i.e. $V_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\left\{p_i\right\}=p_{\mathit{max}}$ .

Therefore, the problem $\left(P\right)$ can be formulated as:

Proof (c).

If $d^{\left(2\right)}<C_i<d^{\left(2\right)}+p_i$ or $d^{\left(2\right)}+p_i\le C_i,i\in N\backslash \left\{1\right\}$ .

For $f_1$ problem: $E_{\mathit{max}}=0$ and $T_{\mathit{max}}=C_n-d^{\left(2\right)}$ .

For $f_2$ problem: $V_{\mathit{max}}=\underset{i\in N\backslash \left\{1\right\}}{\mathit{max}}\left\{V_i\right\}=\underset{i\in N\backslash \left\{1\right\}}{\mathit{max}}\left\{\underset{i\in N\backslash \left\{1\right\}}{\mathit{min}}\left\{\right\{C_i-d_i^{\left(2\right)}\},p_i\}\right\}$ , $\forall i\in N$ .

Therefore, the problem $\left(P\right)$ can be written as:

Case (6).

For the problem $\left(P\right)$ , if we have common processing time $p_i=p,\forall i\in N$ with the condition $d_n^{\left(1\right)}\ge C_n=\mathit{np},\forall i\in N$ and we have the $\mathit{MST}‐$ sequence then this sequence is an $\mathit{ES}$ and $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(d_n^{\left(1\right)}-\mathit{np},0)$

Proof.

Suppose that $\sigma$ be $\mathit{MST}‐$ sequence which yields $(s_1\le s_2\le \cdots \le s_n)=(d_1^{\left(1\right)}-p\le d_2^{\left(1\right)}-p\le \cdots \le d_n^{\left(1\right)}-p)$ .

Assume $d_i^{\left(1\right)}\ge C_n=\mathit{np},$ for all $i\in N,$ this implies that all the jobs are early such that $T_i=0$ and $V_i=0,\forall i\in N$ , which gives $T_{\mathit{max}}=0$ and $V_{\mathit{max}}=0$ .

For each job $i\in N$ the earliness is $E_i=d_i^{\left(1\right)}-C_i=d_i^{\left(1\right)}-\mathit{ip}$ . Thus, $E_{\mathit{max}}=\mathit{max}\{d_i^{\left(1\right)}-C_i,0\}=\mathit{max}\{d_i^{\left(1\right)}-\mathit{ip}\}=d_n^{\left(1\right)}-\mathit{np}$ .

Therefore, the problem $\left(P\right)$ is presented by:

By Theorem (1) this shows that the sequence $\sigma$ is an $\mathit{ES}$ for the problem $\left(P\right)$ .

Case (7).

For the problem $\left(P\right)$ , if we have common processing time $p_i=p$ and $d_i^{\left(2\right)}<C_i=\mathit{ip},\forall i\in N\backslash \left\{1\right\},(p\in [d_1^{\left(1\right)},d_1^{\left(2\right)}])$ , then we have three cases:

Proof (a).

Since $p_i=p,\forall i\in N(i.e.C_n=\sum _{i=1}^n{p}_i=\sum _{i=1}^{n}p=\mathit{np})$ , suppose that $\sigma$ be an $\mathit{EDD}‐$ sequence which yields $(d_1^{\left(2\right)}\le d_2^{\left(2\right)}\le \dots \le d_n^{\left(2\right)})$ , and since $d_i^{\left(2\right)}<C_i=\mathit{ip},\forall i\in N\backslash \left\{1\right\}$ , that gives all the jobs are not early.

For $f_1$ problem: $E_i=0,\forall i\in N,\text{so}{E}_{\mathit{max}}=0$ and $T_i=C_i-d_i^{\left(2\right)}=\mathit{ip}-d_i^{\left(2\right)},\forall i\in N,\text{so}{T}_{\mathit{max}}=\mathit{max}\{\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)},0\}\}=\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)}\}=\mathit{np}-d_{\mathit{max}}^{\left(2\right)}$ ,

For $f_2$ problem: Since $d_i^{\left(2\right)}<C_i<d_i^{\left(2\right)}+p,\forall i\in N\backslash \left\{1\right\},$ this means $V_i=\mathit{min}\{T_i,p_i\}=T_i,\forall i\in N$ , therefore $V_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\left\{T_i\right\}=\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)}\}=\mathit{np}-d_{\mathit{max}}^{\left(2\right)}$ .

Therefore, the problem $\left(P\right)$ can be state as:

Hence, the sequence $\sigma$ is $\mathit{ES}$ for the problem (P).

Proof (b).

Since $p_i=p,\forall i\in N,(i.e.C_n=\sum _{i=1}^n{p}_i=\sum _{i=1}^{n}p=\mathit{np})$ ,

Suppose that $\sigma$ is an $\mathit{EDD}‐$ sequence which yields $(d_1^{\left(2\right)}\le d_2^{\left(2\right)}\le \dots \le d_n^{\left(2\right)})$ and since $d_i^{\left(2\right)}<C_i=\mathit{ip},\forall i\in N\backslash \left\{1\right\}$ , this implies all the jobs are not early.

For $f_1$ problem: $E_i=0,\forall i\in N,\text{so}{E}_{\mathit{max}}=0$ and $T_i=C_i-d_i^{\left(2\right)}=\mathit{ip}-d_i^{\left(2\right)},\forall i\in N,\text{so}{T}_{\mathit{max}}=\mathit{max}\{\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)},0\}\}=\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)}\}=\mathit{np}-d_{\mathit{max}}^{\left(2\right)},$

For $f_2$ problem: since $d_i^{\left(2\right)}+p\le C_i,\forall i\in N\backslash \left\{1\right\},$ which means $V_i=\mathit{min}\{T_i,p\}=p,\forall i\in N,\text{so}{V}_{\mathit{max}}=\mathit{max}\left\{V_i\right\}=\mathit{max}\left\{p\right\}=p.$

So, we can be state the problem $\left(P\right)$ is stated as:

Hence, the sequence $\sigma$ is an $\mathit{ES}$ for the problem $\left(P\right)$ .

Proof (c).

Since $p_i=p,\forall i\in N,(i.e.C_n=\sum _{i=1}^n{p}_i=\sum _{i=1}^{n}p=\mathit{np})$ , suppose that $\sigma$ is an $\mathit{EDD}‐$ sequence which yields $(d_1^{\left(2\right)}\le d_2^{\left(2\right)}\le \cdots \le d_n^{\left(2\right)}),$ and since $d_i^{\left(2\right)}<C_i=\mathit{ip},\forall i\in N\backslash \left\{1\right\}$ , this means all the jobs are not early.

For $f_1$ problem: $E_i=0,\forall i\in N,$ therefore $E_{\mathit{max}}=0$ and $T_i=C_i-d_i^{\left(2\right)}=\mathit{ip}-d_i^{\left(2\right)},\forall i\in N,$ so $T_{\mathit{max}}=\mathit{max}\{\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)},0\}\}=\mathit{max}\{\mathit{ip}-d_i^{\left(2\right)}\}=\mathit{np}-d_{\mathit{max}}^{\left(2\right)}$ .

For $f_2$ problem: if $d_i^{\left(2\right)}<C_i<d_i^{\left(2\right)}+p$ or $d_i^{\left(2\right)}+p\le C_i,i\ne 1,$ then $V_i=\mathit{min}\{T_i,p\},\forall i\in N\backslash \left\{1\right\},$ so $V_{\mathit{max}}=\underset{i\in N\backslash \left\{1\right\}}{\mathit{max}}\left\{\underset{i\in N\backslash \left\{1\right\}}{\mathit{min}}\left\{\right\{C_i-d_i^{\left(2\right)}\},p\}\right\}$ .

Thus, the problem $\left(P\right)$ can be written as:

Hence, the sequence $\sigma$ is an $\mathit{ES}$ for the problem $\left(P\right)$ .

Case (8).

For the problem (P), If $p_i=p,\forall i\in N$ and $[d_i^{\left(1\right)},d_i^{\left(2\right)}]=[d^{\left(1\right)},d^{\left(2\right)}]$ with the condition $d^{\left(1\right)}>C_i$ , $\forall i\in N,$ then $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(d^{\left(1\right)}-C_1,0)$ and any sequence is an $\mathit{ES}$ .

Proof:

Suppose that $\sigma$ be any sequence, if $d^{\left(1\right)}>C_i,$ for every job $i\in N$ then this implies all the jobs are early. Consequently, $T_i=0\text{and}{V}_i=0,\forall i\in N.$ Thus, both the maximum tardiness and the maximum late of work are equal to zero. For each job $i$ , the earliness is given by: $E_i=d^{\left(1\right)}-C_i=d^{\left(1\right)}-\mathit{ip},\forall i\in N$ , so ${\mathit{ET}}_{\mathit{max}}=E_{\mathit{max}}=\mathit{max}\{d^{\left(1\right)}-C_i,0\}=d^{\left(1\right)}-\underset{i\in N}{\mathit{min}}{C}_i$ . Since all processing times are identical $(p_i=p)$ , the earliest completion time corresponds to the first scheduled job. Hence $\underset{i\in N}{\mathit{min}}{C}_i=C_1$ . It follows that $E_{\mathit{max}}=d^{\left(1\right)}-C_1$ , $V_{\mathit{max}}=0.$

Therefore, we can be state the problem $\left(P\right)$ as a following:

Hence, the sequence $\sigma$ is an $\mathit{ES}$ for the problem (P).

Lemma (2).

The $\mathit{ES}$ for $F({\mathit{ET}}_{\mathit{max}},V_{\mathit{max}})=(d^{\left(1\right)}-C_1,0)$ , where $d^{\left(1\right)}\ge C_n$ be $(d^{\left(1\right)}-p,0)$ .

Proof:

It is clear from Lemma (1).

Case (9).

For the problem $\left(P\right)$ , if $p_i=p$ and $[d_i^{\left(1\right)},d_i^{\left(2\right)}]=[d^{\left(1\right)},d^{\left(2\right)}]$ , $\forall i\in N$ and $d^{\left(1\right)}\ge p$ (i.e. if the jobs have identical of the processing times and due windows) with $C_1=p\in [d^{\left(1\right)},d^{\left(2\right)}]$ then we have three subcases: