Task assignment algorithms for two-type heterogeneous multiprocessors

Abstract

Consider the problem of assigning implicit-deadline sporadic tasks on a heterogeneous multiprocessor platform comprising two different types of processors—such a platform is referred to as two-type platform. We present two low degree polynomial time-complexity algorithms, SA and SA-P, each providing the following guarantee. For a given two-type platform and a task set, if there exists a task assignment such that tasks can be scheduled to meet deadlines by allowing them to migrate only between processors of the same type (intra-migrative), then (i) using SA, it is guaranteed to find such an assignment where the same restriction on task migration applies but given a platform in which processors are $1+\frac{\alpha}{2}$ times faster and (ii) SA-P succeeds in finding a task assignment where tasks are not allowed to migrate between processors (non-migrative) but given a platform in which processors are 1+α times faster. The parameter 0<α≤1 is a property of the task set; it is the maximum of all the task utilizations that are no greater than 1.

We evaluate average-case performance of both the algorithms by generating task sets randomly and measuring how much faster processors the algorithms need (which is upper bounded by $1+\frac{\alpha}{2}$ for SA and 1+α for SA-P) in order to output a feasible task assignment (intra-migrative for SA and non-migrative for SA-P). In our evaluations, for the vast majority of task sets, these algorithms require significantly smaller processor speedup than indicated by their theoretical bounds.

Finally, we consider a special case where no task utilization in the given task set can exceed one and for this case, we (re-)prove the performance guarantees of SA and SA-P.

We show, for both of the algorithms, that changing the adversary from intra-migrative to a more powerful one, namely fully-migrative, in which tasks can migrate between processors of any type, does not deteriorate the performance guarantees. For this special case, we compare the average-case performance of SA-P and a state-of-the-art algorithm by generating task sets randomly. In our evaluations, SA-P outperforms the state-of-the-art by requiring much smaller processor speedup and by running orders of magnitude faster.

Appendix: The hardness of the task assignment problems

In this section, we show that both intra-migrative and non-migrative task assignment problems on a two-type heterogeneous multiprocessors are intractable.

1.1 A.1 Intra-migrative task assignment problem is NP-Complete

In this section, we show that the problem of intra-migrative task assignment on two-type heterogeneous multiprocessor platform is NP-Complete. We denote this problem as HET2-INTRA-ASSIGN and is stated in Fig. 10. In order to show this, we will first consider a restricted version of this problem which is denoted as HET2-INTRA-ASSIGN-SPEC-CASE—see Fig. 11. We will show that this problem is NP-complete. It then follows that the HET2-INTRA-ASSIGN problem is NP-complete as well.

Fig. 10

The intra-migrative task assignment problem on a two-type heterogeneous multiprocessor platform

Fig. 11

A restricted version of the intra-migrative task assignment problem on a two-type heterogeneous multiprocessor platform

For showing that the HET2-INTRA-ASSIGN-SPEC-CASE problem is NP-Complete, we make use of the PARTITION problem. The PARTITION problem is shown in Fig. 12 and it is well-known that this problem is NP-Complete (Corollary 15.28 in Korte and Vygen 2006, p. 365).

Fig. 12

The partitioning problem, which is known to be NP-Complete (Korte and Vygen 2006)

Lemma 6

The HET2-INTRA-ASSIGN-SPEC-CASE problem is NP-Complete.

Proof

In order to show that a problem is NP-Complete, we need to: (1) show that the problem is in NP, (2) transform an NP-Complete problem to the problem under consideration and (3) show that the transformation (of Step (2)) can be done in polynomial time. We now show these for HET2-INTRA-ASSIGN-SPEC-CASE problem.

• 1. It is straightforward to see that the problem belongs to NP. To show that the problem is in NP, we should be able to verify, in polynomial time, the given certificate for a yes-instance of the problem. As a certificate, we take the assignment on each processor type. To check whether the given assignment in fact satisfies, for all t∈{1,2}: $(\sum_{i:f(i)=t}{u_{i}^{t}} \leq m_{t} ) \wedge (\forall i \in \{1, 2, \ldots, n \} \mbox{ such that } f(i)=t: u_{i}^{t} \leq1 )$, is obviously possible in polynomial time; specifically, the time complexity of this step is O(n).

• 2. We now transform the PARTITION problem (which is NP-Complete) to the above decision problem. Given an instance $c_{1}, c_{2}, \ldots, c_{n} \in\mathbb{N}$ of the PARTITION problem, transform it into an instance of HET2-INTRA-ASSIGN-SPEC-CASE problem with n tasks and compute utilizations of tasks as follows:

  $$ \forall\tau_i \in\tau, \forall t \in \{1, 2 \}:\quad u_i^t = \frac{2c_i}{\sum_{k=1}^{n}{c_k}} \in(0, 1] $$

  (44)

  We now show that (intra-migrative) assignment of these n tasks on two processor types is possible if and only if there is a set S⊆{1,2,…,n} such that ∑ _j∈S c _j =∑ _{j∈({1,2,…,n}∖S)} c _j . We do so by first showing, in (a), some results we will use and then showing, in (b), the implication in one direction and finally showing, in (c), the implication in the other direction.

  • (a) Results we will use:

    • (a.1) It is trivial to see that $(a = b ) \Rightarrow (a = b = \frac{a+b}{2} )$. This gives us:

      $$\begin{aligned} & \biggl(\sum_{j \in S}{c_j} = \sum _{j \in ( \{1, 2, \ldots , n \} \setminus S )}{c_j} \biggr) \\ &\quad\Rightarrow\quad \biggl(\sum_{j \in S}{c_j} = \sum_{j \in ( \{1, 2, \ldots , n \} \setminus S )}{c_j} = \frac{\sum_{j \in \{1, 2, \ldots, n \}}{c_j}}{2} \biggr) \end{aligned}$$

    • (a.2) It is also trivial to see that $( (a \leq \frac{a+b}{2} ) \wedge (b \leq\frac{a+b}{2} ) ) \Rightarrow (a = b )$. This gives us:

      $$\begin{aligned} & \biggl( \biggl(\sum_{j \in S}{c_j} \leq \frac{\sum_{k=1}^{n}{c_k}}{2} \biggr) \wedge \biggl(\sum_{j \in ( \{1, 2, \ldots, n \} \setminus S )}{c_j} \leq\frac{\sum_{k=1}^{n}{c_k}}{2} \biggr) \biggr) \\ &\quad\Rightarrow\quad \biggl(\sum_{j \in S}{c_j} = \sum_{j \in ( \{1, 2, \ldots , n \} \setminus S )}{c_j} \biggr) \end{aligned}$$

    • (a.3) Let us introduce g that maps an element in {1,2,…,n} to a processor type. It is defined as follows:

      $$\begin{aligned} &i \in S \quad \Leftrightarrow\quad g(i) = 1 \\ &i \in \bigl( \{1, 2, \ldots, n \} \setminus S \bigr)\quad \Leftrightarrow\quad g(i) = 2 \end{aligned}$$

  • (b) Implication in one direction: We now show (using g) that if there is a set S⊆{1,2,…,n} such that ∑ _j∈S c _j =∑ _{j∈({1,2,…,n}∖S)} c _j then intra-migrative assignment of these n tasks on two processor types is possible.

    We will do so by assuming that the if-condition of (b) is true and then show that this implies that the then-condition of (b) must also be true. We know that ∑ _j∈S c _j =∑ _{j∈({1,2,…,n}∖S)} c _j . Using (a.1) on this gives us:

    $$\begin{aligned} \sum_{j \in S}{c_j} =& \frac{\sum_{j \in \{1, 2, \ldots, n \}}{c_j}}{2} \\ \sum_{j \in ( \{1, 2, \ldots, n \} \setminus S )}{c_j} =& \frac{\sum_{j \in \{1, 2, \ldots, n \}}{c_j}}{2} \end{aligned}$$

    Multiplying each side by $\frac{2}{\sum_{k=1}^{n} {c_{k}}}$ and applying the definition of $u_{i}^{t}$ on the left hand side and using the definition of g gives us:

    $$\begin{aligned} \sum_{j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1}{u_j^1} =& 1 \\ \sum_{j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1}{u_j^2} =& 1 \\ \sum_{j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2}{u_j^1} =& 1 \\ \sum_{j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2}{u_j^2} =& 1 \end{aligned}$$

    It obviously holds that, for a set of non-negative numbers, each element cannot be greater than the sum of all numbers in the set. Using this observation on the above gives us:

    $$\begin{aligned} &\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1:\quad u_j^1 \leq1 \\ &\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1:\quad u_j^2 \leq1 \\ &\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2:\quad u_j^1 \leq1 \\ &\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2:\quad u_j^2 \leq1 \end{aligned}$$

    Hence, we have shown that g is an assignment of tasks to processor types that satisfies the constraints stated in HET2-INTRA-ASSIGN-SPEC-CASE problem.

  • (c) Implication in the other direction: We now show (using g) that if intra-migrative assignment of these n tasks on two processor types is possible then there is a set S⊆{1,2,…,n} such that ∑ _j∈S c _j =∑ _{j∈({1,2,…,n}∖S)} c _j .

    We will do so by assuming that the if-condition of (c) is true and then show that this implies that the then-condition of (c) must also be true. We know that an intra-migrative assignment of these n tasks is possible. Using the function g to express this gives us:

    $$\begin{aligned} & \biggl(\sum_{\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1} {u_j^1} \leq1 \biggr) \\ &\quad{}\wedge \biggl(\sum_{\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2} {u_j^2} \leq1 \biggr) \\ &\quad{}\wedge \bigl(\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 1:\quad u_j^1 \leq1 \bigr) \\ &\quad{}\wedge \bigl(\forall j \in \{1, 2, \ldots, n \} \text{ such that } g(j) = 2:\quad u_j^2 \leq1 \bigr) \end{aligned}$$

    Using the definition of $u_{i}^{t}$ and the mapping g and multiplying each side by $\frac{\sum_{k=1}^{n} {c_{k}}}{2}$ gives us:

    $$\begin{aligned} & \biggl(\sum_{\forall j \in S} c_j \leq \frac{\sum_{k=1}^{n} {c_k}}{2} \biggr) \\ &\quad{}\wedge \biggl(\sum_{\forall j \in ( \{1, 2, \ldots, n \} \setminus S )} c_j \leq\frac{\sum_{k=1}^{n} {c_k}}{2} \biggr) \\ &\quad{}\wedge \biggl(\forall j \in S: c_j \leq\frac{\sum_{k=1}^{n} {c_k}}{2} \biggr) \\ &\quad{}\wedge \biggl(\forall j \in \bigl( \{1, 2, \ldots, n \} \setminus S \bigr): c_j \leq\frac{\sum_{k=1}^{n} {c_k}}{2} \biggr) \end{aligned}$$

    Observing the first two expressions and using (a.2) gives us:

    $$ \sum_{j \in S}{c_j} = \sum _{j \in ( \{1, 2, \ldots, n \} \setminus S )}{c_j} $$

    This satisfies the constraints of the PARTITION problem.

• 3. Finally, it can be easily seen that the transformation from PARTITION to HET2-INTRA-ASSIGN-SPEC-CASE using Expression (44) is possible in polynomial time; specifically, the time complexity is O(n).

Hence the proof. □

Theorem 10

The HET2-INTRA-ASSIGN problem is NP-Complete.

Proof

Follows from Lemma 6 and the fact that HET2-INTRA-ASSIGN-SPEC-CASE problem is a restricted form of HET2-INTRA-ASSIGN problem. □

1.2 A.2 Non-migrative task assignment problem is NP-Complete in the strong sense

In this section, we show that the problem of non-migrative task assignment on a two-type heterogeneous multiprocessor platform is NP-Complete in the strong sense. We denote this problem as HET2-NON-ASSIGN and is stated in Fig. 13. In order to show this, we will first consider a restricted version of this problem which is denoted as HET2-NON-ASSIGN-SPEC-CASE—see Fig. 14. We will show that this problem is NP-complete in the strong sense. It then follows that the HET2-NON-ASSIGN problem is NP-complete in the strong sense as well.

Fig. 13

The non-migrative task assignment problem on a two-type heterogeneous multiprocessor platform

Fig. 14

A restricted version of the non-migrative task assignment problem on a two-type heterogeneous multiprocessor platform

For showing that the HET2-NON-ASSIGN-SPEC-CASE problem is NP-Complete in the strong sense, we make use of the 3-PARTITION problem. The 3-PARTITION problem is shown in Fig. 15 and it is well-known that this problem is NP-Complete in the strong sense (see, e.g., Garey and Johnson 1978).

Fig. 15

The 3-partitioning problem, which is known to be NP-Complete in the strong sense (Garey and Johnson 1978)

Lemma 7

The HET2-NON-ASSIGN-SPEC-CASE problem is NP-Complete in the strong sense.

Proof

In order to show that a problem is NP-Complete in the strong sense, we need to: (1) show that the problem is in NP, (2) transform a problem which is NP-Complete in the strong sense to the problem under consideration and (3) show that the transformation (of Step (2)) can be done in polynomial time. We now show these for HET2-NON-ASSIGN-SPEC-CASE problem.

• 1. It is straightforward to see that the problem belongs to NP. As a certificate, we take the assignment on each processor. To check whether the given assignment in fact satisfies $\sum_{i:f(i)=j}{u_{i}^{t}} \leq1$ for every processor j∈ type-t of π (where t∈{1,2}) is obviously possible in polynomial time; specifically the time complexity is O(n).

• 2. We now transform the 3-PARTITION problem (which is NP-Complete in the strong sense) to the above decision problem. Given an instance c ₁,c ₂,…,c _n=3m and B of the 3-PARTITION problem, transform it into an instance of HET2-NON-ASSIGN-SPEC-CASE problem with n=3m tasks by computing utilizations of tasks as follows:

  $$ \forall\tau_i \in\tau, \forall t \in \{1, 2 \}:\quad u_i^t = \frac{c_i}{B} $$

  (45)

  We now show that (non-migrative) assignment of these 3m tasks on m processors is possible if and only if c ₁,c ₂,…,c _n=3m can be partitioned into m subsets I ₁,I ₂,…,I _m such that $\forall j \in \{1, 2, \ldots, m \}: \sum_{i \in I_{j}}{c_{i}} = B$. We do so by first showing, in (a), some results we will use and then showing, in (b), the implication in one direction and finally showing, in (c), the implication in the other direction.

  • (a) Results we will use:

    • (a.1) Let us introduce g that maps an element in {1,2,…,3m} to a processor. It is defined as follows:

      $$ i \in I_j \quad\Leftrightarrow\quad g(i) = j $$

  • (b) Implication in one direction: We now show (using g) that if c ₁,c ₂,…,c _3m can be partitioned into m subsets I ₁,I ₂,…,I _m such that $\forall j \in \{1, 2, \ldots, m \}: \sum_{i \in I_{j}}{c_{i}} = B$ then there is an assignment of these 3m tasks on m processors.

    We will do so by assuming that the if-condition of (b) is true and then show that this implies that the then-condition of (b) must also be true. We know that c ₁,c ₂,…,c _3m can be partitioned into m subsets I ₁,I ₂,…,I _m such that $\forall j \in \{1, 2, \ldots, m \}: \sum_{i \in I_{j}}{c_{i}} = B$. Multiplying each side by $\frac{1}{B}$ and applying the definition of $u_{i}^{t}$ on the left hand side and using the definition of g gives us:

    $$\begin{aligned} &\forall j \in \{1, 2, \ldots, m \}:\quad \sum_{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^1 = 1 \\ &\forall j \in \{1, 2, \ldots, m \}:\quad \sum_{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^2 = 1 \end{aligned}$$

    Hence, we have shown that g is an assignment of tasks to processors that satisfies the constraints stated in HET2-NON-ASSIGN-SPEC-CASE problem.

  • (c) Implication in the other direction: We now show (using g) that if non-migrative assignment of these n tasks on m processors is possible then c ₁,c ₂,…,c _3m can be partitioned into m subsets I ₁,I ₂,…,I _m such that $\forall j \in \{1, 2, \ldots, m \}: \sum_{i \in I_{j}}{c_{i}} = B$.

    We will do so by assuming that the if-condition of (c) is true and then show that this implies that the then-condition of (c) must also be true. We know that a non-migrative assignment of these n tasks is possible. Using the function g to express this gives us:

    $$\begin{aligned} &\forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^1 \leq1 \biggr) \\ &\quad{}\wedge \forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^2 \leq1 \biggr) \end{aligned}$$

    Since it is a non-migrative assignment, it also holds that (from one of the assumptions of HET2-NON-ASSIGN-SPEC-CASE problem):

    $$ \frac{1}{m} \times \biggl(\sum_{i \in \{1, 2, \ldots, n \}} u_i^1 \biggr) = \frac{1}{m} \times \biggl(\sum _{i \in \{1, 2, \ldots, n \}} u_i^2 \biggr) = 1 $$

    Applying this on the earlier expression gives:

    $$\begin{aligned} &\forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^1 = 1 \biggr) \\ &\quad{}\wedge \forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} u_i^2 = 1 \biggr) \end{aligned}$$

    Multiply both sides by B and using the definition of $u_{i}^{t}$ gives us:

    $$\begin{aligned} &\forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} c_i = B \biggr) \\ &\quad{}\wedge \forall j \in \{1, 2, \ldots, m \}:\quad \biggl(\sum _{\forall i \in \{1, 2, \ldots, n \} \text{ such that } g(i)=j} c_i = B \biggr) \end{aligned}$$

    Note that these two expressions state the same thing so only one is needed. Also, we form the partitioning as follows. Let I _j be the set of all integers such that i∈{1,2,…,n} and g(i)=j. This gives us:

    $$ \forall j \in \{1, 2, \ldots, m \}:\quad \sum_{\forall i \in I_j} c_i = B $$

    This satisfies the constraints of the 3-PARTITION problem.

• 3. Finally, it can be easily seen that the transformation from 3-PARTITION to HET2-NON-ASSIGN-SPEC-CASE using Expression (45) is possible in polynomial time; specifically, the time complexity is O(n).

Hence the proof. □

Theorem 11

The HET2-NON-ASSIGN problem is NP-Complete in the strong sense.

Proof

Follows from Lemma 7 and the fact that HET2-NON-ASSIGN-SPEC-CASE problem is a restricted form of HET2-NON-ASSIGN problem. □