Parameterized Complexity of Scheduling Chains of Jobs with Delays

Suppose $F$ is satisfiable by making variables $x_{i_{1}},\ldots,x_{i_{k}}$ true. For each $j$, with $1\leq j\leq k$, we let one true variable chain start at time $i_{j}$.

First we introduce a notion satisfying, intuitively, this will be the elements that make $F$ true. We define this top-down. First we call $F$ satisfying. For each element $F^{\prime}$ of $F$:

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  If $F^{\prime}$ is satisfying and $F^{\prime}$ is a conjunction $F^{\prime}=F_{1}\wedge\cdots\wedge F_{q}$, then all terms $F_{i}$ are satisfying.

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  If $F^{\prime}$ is satisfying and $F^{\prime}$ is a disjunction $F^{\prime}=F_{1}\vee\cdots\vee F_{q}$, then at least one $F_{i}$ is satisfied, say $F_{j}$. We say that $F_{j}$ is satisfying, and the other term $F_{i}$ with $i\neq j$ are not satisfying.

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  If $F^{\prime}$ is not satisfying, all its terms are not satisfying.

For each element $F^{\prime}$ of $F$ that is a disjunction $F^{\prime}=F_{1}\vee\cdots\vee F_{q}$, consider the disjunction chain $C$ associated with this element. If $F^{\prime}$ is not satisfying, then start this chain $C$ arbitrarily, say at its release time. If $F^{\prime}$ is satisfying, let $F_{j}$ be its term that is satisfying. Now, start this chain $C$ at time $\ell(F^{\prime})+(2j-2)s_{\max}(F^{\prime})$, where $s_{\max}(F^{\prime})$ is again the maximum interval size of the terms $F_{i}$.

Now we will verify that is a feasible schedule, that is, that we never use more than $m$ machines.

First consider a time step $\alpha$ from $i$ to $i+1$, with $0\leq i\leq n-1$. At this time step there are $m-1$ jobs of fill chains scheduled. Since the variables $x_{i_{1}},\ldots,x_{i_{k}}$ are different, the true variable chains start at different times. Hence, there is at most one job of a true variable chain scheduled at $\alpha$. Thus, there are at most $m$ jobs scheduled at time $\alpha$.

Consider a time step $\alpha$ from $i$ to $i+1$ with $i\geq n$, such that $i\neq\ell(F^{\prime})+n-1$ for all elements $F^{\prime}$ of $F$ that are literals. Notice that at those times no jobs of fill chains and variable check chains are scheduled. There are at most $k$ jobs of true variable chains scheduled at time step $\alpha$, since there are only $k$ true variable chains. And there are at most $t^{\prime}$ jobs of disjunction chains scheduled, since there are $t^{\prime}$ levels of disjunction. It follows that there are at most $k+t^{\prime}=m$ jobs scheduled at time step $\alpha$.

Consider a time step $\alpha$ from $i$ to $i+1$ with $i\geq n$, such that $i=\ell(F^{\prime})+n-1$ for an element $F^{\prime}$ of $F$ that is a literal. We distinguish three cases.

Suppose that $F^{\prime}$ is satisfying, and $F^{\prime}$ is a positive literal $F^{\prime}=x_{j}$. Then we know that there are $t^{\prime}$ machines used for the disjunction chains. Besides, there is one machine used for the variable check chains. Since we know that $F^{\prime}$ is satisfying, $x_{j}$ is true. Thus one of the true variable chains starts at time $j$. So this chain has no job scheduled from $i$ until $i+1$. So at most $k-1$ true variable chains have a job scheduled from $i$ until $i+1$. In total there are at most $t^{\prime}+1+k-1=m$ machines used.

Suppose that $F^{\prime}$ is satisfying, and $F^{\prime}$ is a negative literal, say $F^{\prime}=\neg x_{j}$. Then we know that there are $t^{\prime}$ machines used for the disjunction chains. There are $k$ machines used for the variable check chains. Since we know that $F^{\prime}$ is satisfying, $x_{j}$ is false. Thus none of the true variable chains starts at time $j$. Hence, no true variable chain has a job scheduled from $i$ until $i+1$. In total there are at most $m$ machines used from $i$ until $i+1$.

Suppose that $F^{\prime}$ is not satisfying. Let $F^{\prime\prime}$ be the last satisfying element on the path from $F$ to $F^{\prime}$. The disjunction chain that corresponds to $F^{\prime\prime}$ has no job that is scheduled form $i$ until $i+1$. Hence, there are at most $t^{\prime}-1$ machines used for the disjunction chains from time $i$ until $i+1$. It follows that there are $k+1$ machines for the true variable chains and the variable check chains. Since all true variable chains start at different times, those chains together have $k+1$ jobs that can be scheduled from $i$ until $i+1$. We conclude that we used at most $m$ machines from time $i$ until $i+1$.

For each true variable chain, if it starts at time $i$, then set $x_{i}$ to true. All other variables are set to false, i.e, $x_{i}$ is true, if and only if there is a true variable chain that starts at time $i$.

Note that all variable chains must start at different times, and they must start at times $0,1,\ldots n-1$. If two of these start at the same time $i$, then both have a job scheduled from $i$ until $i+1$, but there are $m-1$ fill chains that have a job scheduled from $i$ until $i+1$ as well, which is a contradiction with the total number of machines. Thus, we have set exactly $k$ variables to true.

We claim that this setting makes $F$ true. We define elements of $F$ to be demonstrated in the following way, recursively. We say that $F$ is demonstrated.

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  If a conjunction is demonstrated, then all its terms are demonstrated.

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  If a disjunction $F^{\prime}=F_{1}\vee\cdots\vee F_{q}$ is demonstrated, then we consider the corresponding disjunction chain $C$. If for every time step in $[l(F_{j}),r(F_{j})]$ a job of $C$ is scheduled, we say that $F_{j}$ is demonstrated.

After this top-down definition of demonstrated elements, we will now inductively show bottom-up that demonstrated elements are satisfied by the setting of variables described above.

Consider a demonstrated element $F^{\prime}$ that is a literal. By the definition of demonstrated, $t^{\prime}$ of the disjunction chains have a job scheduled at each time step of $[l(F^{\prime}),r(F^{\prime})]$. Consider the time step from $l(F^{\prime})+n-1$ until $l(F^{\prime})+n$. We know that $t^{\prime}$ machines process a job of a disjunction chain at this time step, so at most $k$ machines process jobs of true variable chains and variable check chains.

Suppose that $F^{\prime}$ is a positive literal, say $x_{i}$. Notice that one machine processes a check variable chain from $l(F^{\prime})+n-1$ until $l(F^{\prime})+n$. So at most $k-1$ machines process a job of a true variable chain. Hence there must be at least one true variable chain that does not have a job scheduled at this time. By construction of the true variable chains, such a chain must start at time $i$; and thus we set $x_{i}$ to true, i.e., our setting satisfies the formula consisting of the single positive literal $x_{i}$.

If $F^{\prime}$ is a negative literal $\neg x_{i}$, $k$ machines are processing check variable chains from $l(F^{\prime})+n-1$ until $l(F^{\prime})+n$; thus no true variable chain can have a job scheduled at this step. By construction of the true variable chains, this implies that no true variable chain starts at time $i$, and thus $x_{i}$ is set to false. Hence, $F^{\prime}$ is satisfied.

Now consider a demonstrated element $F^{\prime}$ and assume, by induction, that for all its terms $F_{i}$ holds: if $F_{i}$ is demonstrated, then $F_{i}$ is true.

Suppose that $F^{\prime}$ that is a conjunction. All its terms are demonstrated, and by the induction hypothesis, all its terms are true. Thus the conjunction is also true.

Suppose that $F^{\prime}$ is a disjunction. Since the corresponding disjunction chain has length $3s_{\max}(F^{\prime})$, there is at least one $F_{j}$ such that for every time step in $[l(F_{j}),r(F_{j})]$ a job of $C$ is scheduled. By definition, this $F_{j}$ is demonstrated. By induction, $F_{j}$ is true, and thus $F^{\prime}$ is true.

We conclude that $F$ is satisfied.

For each regular chain, let it start at time $\lfloor t/\tau\rfloor$, when its transformed instance starts at time $t$. This implies that for every job in the chain it will start at time $\lfloor t/\tau\rfloor$, when its transformed instance starts at time $t$. We know that for each time interval $[t\tau,(t+1)\tau]$, $\tau-m$ steps are used for extra jobs, so $m$ time steps are available for jobs of regular chains. Thus, at every time step $t$ in the original instance, at most $m$ jobs are scheduled.

We start with assigning a number $0,1,\ldots,\tau-1$ to every chain such that for every time step all the chains that overlap this time step have different number. We denote this assignment by $c$. We can do this as follows: go through time $0,1,2,\ldots$, and every time a chain is released, assign a number that is currently unused, if a deadline passes, the number of the corresponding chain becomes available again. We can do this with $\tau$ numbers, since by definition for every timestep there are at most $\tau$ chains that overlap this timestep.

For every regular chain $C$, let $C$ start at time $t\tau+c(C)$, where $t$ is the starting time of the corresponding original chain. Then chains of thickness number $i$ only have jobs starting at times $t$ with $t\equiv i\pmod{\tau}$. For every time $t$, all jobs that are scheduled to start at time $t$ in the original instance, will now be scheduled in the interval $[t\tau,(t+1)\tau]$ in the transformed instance. Those jobs are in different chains and those chains are assigned different numbers. Thus those jobs are scheduled at different times in the tranformed instance.

In the original instance, at each time $t$ at most $m$ chains have a job scheduled to start at $t$, so, in the transformed instance, for each interval $[t\tau,(t+1)\tau]$, at most $m$ regular chains have a job scheduled in this interval. Thus, we can schedule the $\tau-m$ extra chains in this time interval at the time steps where no job of a regular chain is scheduled.

A set of integers $S$ is said to be a Golomb ruler if all differences $a-b$ of two elements $a,b\in S$ are unique, that is, $s_{1}-s_{2}\neq s_{3}-s_{4}$ for $s_{1},s_{2},s_{3},s_{4}\in S$ with $s_{1}\neq s_{2}$ and $s_{3}\neq s_{4}$.

Erdös and Turán [9] gave the following explicit construction of a Golumb ruler. Let $p>2$ be a prime number. Then the set $\{2pk+(k^{2}\bmod p)\leavevmode\nobreak\ |\leavevmode\nobreak\ k\in\{0,1,\ldots,p-1\}\}$ is a Golomb ruler with $p$ elements. We can build a Golumb ruler of size $n$ in $O(n\sqrt{n})$ time: with help of the Sieve of Eratosthenes, we find a prime number $p$ between $n$ and $2n$ (such a number always exist, by the classic postulate of Bertrand (see [1]), and then follow the Erdös-Turán-construction with this value of $p$, and take the first $n$ elements of this set. Notice that the elements in this set are smaller than $4n^{2}$.

Suppose we have an input of independent set $G=(V,E)$ and $k$. Assume $V=\{v_{1},\ldots,v_{n}\}$ and let $m$ be the number of edges. First, build a Golumb ruler $S^{n}$ of size $n$. Denote the elements by $s_{0}\leq s_{1}\leq\ldots\leq s_{n-1}$. Notice that $s_{0}=0$. Write $c_{0}=s_{n-1}+1$.

We will construct an instance of Chain Scheduling with Exact Delays. We will make $k+1$ chains. We call one chain the start time forcing chain, the other $k$ chains are the vertex selection chains.

The start time forcing chain has release date $1$, deadline $c_{0}$ and total execution time (including delays) $c_{0}-1$. I.e., it must start at time 1. The chain will have a job starting at every time in $[1,c_{0}-1]$ except the times $s_{1},s_{2},\ldots,s_{n-1}$.

The vertex selection chains have release date $0$. They have deadline $c_{0}+T-1$ and total execution time $T$, where $T=(m\cdot k(k-1))(2c_{0}+1)+1$. So, they can start at times $0,1,\ldots,c_{0}-1$. The vertex selection chains start with a job and then a delay of $c_{0}-1$. Note that as a result of this, in order not to conflict with the start time forcing chain, they have to start at an element of $S^{n}$.

Now, for each edge $\{v_{i},v_{j}\}\in E$, and each ordered pair of vertex selection chains $C_{a},C_{b}$ with $a,b\in\{1,2,\ldots,k\}$ we dedicate an interval $I_{v_{i},v_{j},C_{a},C_{b}}$ of $2c_{0}+1$ time steps. More precisely, we have $m\cdot k(k-1)$ intervals $[c_{0}+i(2c_{0}+1),c_{0}+(i+1)(2c_{0}+1)]$ for $i=0,1,\ldots,m\cdot k(k-1)-1$. And to each interval we assign a unique label $I_{v_{i},v_{j},C_{a},C_{b}}$ where $v_{i}v_{j}\in E$ and $a,b\in\{1,2,\ldots,k\}$. In the interval $I_{v_{i},v_{j},C_{a},C_{b}}$ we will check whether the chains $C_{a}$ and $C_{b}$ did not select the edge $v_{i}v_{j}$, that is, whether $C_{a}$ does not start at $s_{i}$ or $C_{b}$ does not start at $s_{j}$.

We will now extend the vertex selection chains. Consider the interval $[c_{0},c_{0}+c_{0}+(m\cdot k(k-1))(2c_{0}+1)]$ from left to right. For each interval $I_{v_{i},v_{j},C_{a},C_{b}}$ that we encounter, we extend the vertex selection chains as follows.

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  For each chain $C$, with $C\neq C_{a}$, $C\neq C_{b}$, add a delay of $2c_{0}+1$.

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  Add the following gadget to the chain $C_{a}$: a delay of $c_{0}-s_{i}$, a job, and then a delay if $c_{0}+s_{i}$.

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  Add the following gadget to the chain $C_{b}$: a delay of $c_{0}-s_{j}$, a job, and then a delay if $c_{0}+s_{j}$.

Add one job at the end of all chains. See Figures 2, 3 and 4 for an example of the construction and a feasible schedule.

We claim that there is a feasible schedule, if and only if $G$ has an independent set of size at least $k$.

Suppose $v_{i_{1}},\ldots,v_{i_{k}}$ form an independent set in $G$. Let the vertex selection chain $C_{a}$ start at times $s_{i_{a}}$ for $a=1,2,\ldots,k$.

Notice that for the first $c_{0}$ time steps, at most one resource is used. The same holds for the last $c_{0}$ time steps. We show that this is a feasible scheduling by contradiction. Suppose that there are two resources needed at some time step $\beta$, and that $\beta$ is in the interval $I=I_{v_{i},v_{j},C_{a},C_{b}}$ for checking whether the chains $C_{a}$ and $C_{b}$ do not select the edge $v_{i}v_{j}$. Write $I=[\ell(I),r(I)]$. Notice that job of $C_{a}$ in the interval $I$ starts at time $\ell(I)+c_{0}-s_{i}+s_{i_{a}}$. And the job of $C_{b}$ in the interval $I$ starts at time $\ell(I)+c_{0}-s_{j}+s_{i_{b}}$. Thus $\ell(I)+c_{0}-s_{i}+s_{i_{a}}=\beta=\ell(I)+c_{0}-s_{j}+s_{i_{b}}$. Equivalently, $s_{i_{a}}-s_{i}=s_{i_{b}}-s_{j}$. Since $S^{n}$ is a Golomb ruler, it follows that either $s_{i_{a}}=s_{i}$ and $s_{i_{b}}=s_{j}$ or $s_{i_{a}}=s_{i_{b}}$ and $s_{i}=s_{j}$.

In the first case, $s_{i_{a}}=s_{i}$ and $s_{i_{b}}=s_{j}$, we see that $v_{i}=v_{i_{a}}$ and $v_{j}=v_{i_{b}}$. But there is no edge $v_{i_{a}}v_{i_{b}}$, since $v_{i_{a}}$ and $v_{i_{b}}$ are in an independent set. This yields a contradiction.

In the second case, $s_{i_{a}}=s_{i_{b}}$ and $s_{i}=s_{j}$, we see that $v_{i_{a}}=v_{i_{b}}$, but this yields a contradiction with the fact that the vertices of the independent set are distinct.

We conclude that this schedule always uses at most one machine.

Suppose that there is a feasible schedule. Notice that the time of the first step of a vertex selection chains must be an element of $S^{n}$, otherwise the job conflicts with the start time forcing chain. Now, set $W=\{v_{i}\mid\textrm{there is a vertex selection chain that starts at time }s_{i}\}$. Notice that $|W|=k$, as otherwise two vertex selection chains start at the same time, and conflict with each other for their first job.

We prove that $W$ is an independent set by contradiction. Suppose that there exists an edge $v_{i}v_{j}$, with $v_{i},v_{j}\in W$. Let $C_{a}$ be the chain that starts at $s_{i}$ and $C_{b}$ the chain that starts at $s_{j}$. Now consider the interval $I=I_{v_{i},v_{j},j_{a},j_{b}}$, and write $I=[\ell(I),r(I)]$. By the construction of the chains, it follows that $C_{a}$ starts its job of the interval $I$ at time $\ell(I)+c_{0}-s_{i}+s_{i}=\ell(I)+c_{0}$. The job of $C_{b}$ in the interval starts at time $\ell(I)+c_{0}$ as well. This yields a contradiction. We conclude that $W$ is an independent set. This shows that the Chain Scheduling with Exact Delays problem with a single machine, parametrized by the number of chains, is $W[1]$-hard.

We show that the problem belongs to $W[1]$ by a transformation to Independent Set. Suppose we are given an instance of the Chain Scheduling with Exact Delays problem with a single machine. We now build a graph as follows: for each chain $C$ and each possible starting time $t$ of this chain, we take one vertex $v_{C,t}$. We now add edges as follows: each pair of vertices that represent the same chain with different starting times is adjacent (i.e., for each chain, its vertices form a clique). For each pair of vertices $v_{C,t}$, $v_{C^{\prime},t^{\prime}}$, $C\neq C^{\prime}$, we add an edge if and only if starting chain $C$ on time $t$ and starting chain $C^{\prime}$ on time $t^{\prime}$ will cause that there is a time step where a job of both chains is scheduled.

It is not hard to see that the given instance of the Chain Scheduling with Exact Delays problem has a solution, if and only if the constructed graph has an independent set of size at least $k$ (with $k$ the number of chains). Indeed, if there is an independent set of size at least $k$, then for every chain $C$ there is a vertex $v_{C,t}$ in the independent set, since for every chain $C$ we can have at most one vertex $v_{C,t}$ in an independent set. Scheduling chain $C$ at time $t$ gives a feasible schedule. On the other hand, if we have a feasible schedule, then the set $\{v_{C,t}\mid C\textrm{ a chain },t\textrm{ its starting time in the schedule}\}$ is an independent set of size $k$.

Let $x_{C_{1},t_{1}},x_{C_{2},t_{2}},\ldots,x_{C_{k},t_{k}}$ be a solution for the CNF-SAT instance. The chain clauses garantee that for every chain $C$ there is at least one variable $x_{C,t}$ true. Since $k$ equals the number of chains, we have that for every chain $C$ there is exactly one variable $x_{C,t}$ true. Let chain $C_{i}$ start at time $t_{i}$ for all $i=1,2,\ldots k$.

We show that this is a feasible schedule by contradiction. Suppose that at time $t$ more than $m$ jobs are scheduled to start. Let $X$ be the set of chains that those jobs are in. Let $X^{\prime}\subseteq X$ be a subset of size $m+1$. Then the time clause $\bigvee_{C_{i}\in X^{\prime}}\neg x_{C_{i},t_{i}}$ is not satisfied. This yields a contradiction.

Suppose we have a feasible schedule. For each chain $C$ set the variable $x_{C,t}$ to true, where $t$ is the starting time of $C$ in the schedule. Notice that there are exactly $k$ variables set to true.

For every chain $C$ we set one variable $x_{C,t}$ to true, so the chain clauses are satisfied.

For every time $t$, at most $m$ jobs start at $t$. Consider the set $S_{t}$. At most $m$ variables in this set are set to true. So, for every subset $X\subseteq S_{t}$ of $m+1$ variables, there is at least one variable false. Hence, the clause is satisfied.

Let $G=(V,E)$, $k$ be an instance of Dominating Set. Assume that $V=\{v_{1},v_{2},\ldots,v_{n}\}$. We will make an instance of the scheduling problem with $k$ machines and $k+1$ chains.

The first chain will have release date $2n-1$ and deadline $n(n+1)$. It will consist of $n$ jobs, with delay $n-1$ between each pair of consecutive jobs. Notice that this chain has to start at time $2n-1$. We will call this chain the check chain.

The other $k$ chains will be identical, we call them the vertex selection chains. They have release date $0$ and deadline $(n+1)n+n-1$. They will have total execution time $(n+1)n$, so they can start at times $0,1,\ldots,n-1$. Start each chain with a job and then a delay of $n-1$.

Consider the interval $[n,(n+1)n]$ from left to right. For each integer $in+j$ with $0\leq j\leq n-1$, add the following to the vertex selection chains:

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  a delay of $1$ if $i=n-j$,

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  a delay of $1$ is $v_{i}\sim v_{n-j}$,

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  a job otherwise.

Notice that for each interval $[in,(i+1)n]$, we made a gadget the represents the adjacencies of vertex $v_{i}$: there is no job at time $(i+1)n-j$ of the execution of the chain if $i=j$ or $v_{i}\sim v_{j}$ for $j=1,2,\ldots,n$. See Figure 5.

Let $v_{j_{1}},v_{j_{2}},\ldots,v_{j_{k}}$ be a dominating set. Let the vertex selection chains start at times $j_{1}-1$, $j_{2}-1$, $\ldots$, $j_{k}-1$. We will show that this is a feasible schedule.

Notice that at each time that is not of the form $in-1$ at most $k$ jobs are scheduled to start, since the check chain has no job starting at this time.

Now consider the time $in-1$, for some $i=1,2,\ldots,n$. The check chain has a job scheduled to start at this time. We will show that there is a vertex selection chain that has no job starting at this time. Let $v_{j_{a}}$ be the vertex that dominates $v_{i}$. Then we know that $i=j_{a}$ or $v_{i}\sim v_{j_{a}}$. So the vertex selection chains have a delay of $1$ starting at time $(i+1)n-j_{a}$ of their execution time. Let $C$ be the chain that starts at time $j_{a}-1$. It follows that $C$ has a delay of $1$ starting at time $(i+1)n-j_{a}+(j_{a}-1)=(i+1)n-1$.

Let $j_{1},j_{2},\ldots,j_{k}$ be the starting times of the vertex selection chains in a feasible schedule. Consider the set $U=\{v_{j_{1}+1},v_{j_{2}+1},\ldots,v_{j_{k}+1}\}$. We will show that this is a dominating set.

Let $v_{i}$ be a vertex. Consider the time $(i+1)n-1$. Notice that a job of the check chain is scheduled to start at this time, so one of the vertex selection chains $C$ has no job starting at this time. Let $j_{a}$ be the starting time of the chain $C$. It follows that $C$ has no job at time $(i+1)n-1-j_{a}$ of its execution time. Hence $j_{a}+1=i$ or $v_{j_{a}+1}\sim v_{i}$. We conclude that $v_{i}$ is dominated. We conclude that the Chain Scheduling with Exact Delays problem with a variable number of machines is $W[2]$-hard, when parametrized by the number of chains.

We use a reduction to Threshold Dominating Set. In the Threshold Dominating Set problem, we are given a graph $G$ and integers $k$ and $r$, and ask for a set of at most $k$ vertices, such that each vertex in $G$ is dominated at least $r$ times.

Downey and Fellows [8] showed that Threshold Dominating Set, parameterized by $k$ and $r$ is $W[2]$-complete.

Suppose we have an instance of the Chain Scheduling with Exact Delays scheduling problem with $m$ machines and $c$ chains. Let $t_{\max}$ the largest deadline in this instance. We will distinguish three cases: $c-m\leq 0$, $c-m=1$, and $c-m>1$. If $c\leq m$, any schedule is feasible, so we reduce to a trivial dominating set instance.

Suppose that $c-m=1$. We will construct an instance of Threshold Dominating Set where we look for a set of $c$ vertices that dominates each vertex at least once.