Model approximation for batch flow shop scheduling with fixed batch sizes

Abstract

Batch flow shops model systems that process a variety of job types using a fixed infrastructure. This model has applications in several areas including chemical manufacturing, building construction, and assembly lines. Since the throughput of such systems depends, often strongly, on the sequence in which they produce various products, scheduling these systems becomes a problem with very practical consequences. Nevertheless, optimally scheduling these systems is NP-complete. This paper demonstrates that batch flow shops can be represented as a particular kind of heap model in the max-plus algebra. These models are shown to belong to a special class of linear systems that are globally stable over finite input sequences, indicating that information about past states is forgotten in finite time. This fact motivates a new solution method to the scheduling problem by optimally solving scheduling problems on finite-memory approximations of the original system. Error in solutions for these “t-step” approximations is bounded and monotonically improving with increasing model complexity, eventually becoming zero when the complexity of the approximation reaches the complexity of the original system.

Appendix: Proof of Theorem 1

Lemma 11

Given $B \in \mathbb {R}_{\max }^{n \times n}$ that satisfies Eq. 7 and ( c, τ )(α), the matrix resulting from finitely many left multiplications of P _i (k) and R _i (k) for any i with B satisfies Eq. 7.

Proof

Let $B \in \mathbb {R}_{\max }^{n \times n}$ that satisfies Eq. 7, (c, τ )(α), and i≤n be given. After a left multiply of P _i (k), Eq. 7 is trivially satisfied. Suppose that i<n. Consider the product A=R _i (k)⊗B. Because the only difference between A and B is in rows i and i+1, we want to show that a _{i j} ≥a _i,j+1 and a _i+1,j ≥a _i+1,j+1 for all j<n. We will consider two cases.

Suppose b _{i j} ≥b _i+1,j . Then a _{i j} =b _{i j} =a _i+1,j . Because B satisfies (7) it follows that b _{i j} ≥b _i,j+1 and b _{i j} ≥b _i+1,j ≥b _i+1,j+1. So a _{i j} ≥a _i,j+1 and a _i+1,j ≥a _i+1,j+1.

Suppose b _i+1,j ≥b _{i j} . We achieve the same result as the previous case in the same manner.

Thus, because B satisfies Eq. 7 after a left multiply of an arbitrary P _i or R _i , after a finite number of left multiplies, the resulting matrix will satisfy Eq. 7.

Lemma 12

Given $B \in \mathbb {R}_{\max }^{n \times n}$ that satisfies Eq. 8 and ( c, τ )(α), the matrix resulting from finitely many left multiplications of P _i (k) and R _i (k) for any i with B satisfies Eq. 8.

Proof

Let $B \in \mathbb {R}_{\max }^{n \times n}$ that satisfies Eq. 8, (c, τ )(α), and i≤n be given. Left multiplying B by P _i adds the same number to every element of row i. As this does not change ξ _{i j} for any j, the resulting matrix still satisfies Eq. 8. Now suppose that we left multiply B by R _i for some i<n. The resulting matrix, A, has rows i and i+1 equal. We now have

$$\begin{array}{*{20}l} \xi_{ij}(A) & = a_{ij} - a_{i,j+1} \\ & = \max(b_{ij}, b_{i+1,j}) - \max(b_{i,j+1}, b_{i+1,j+1}) \\ & = a_{i+1,j} - a_{i+1,j+1} \\ & = \xi_{i+1,j}(A). \end{array} $$

We need now show that ξ _{i −1,j} (A)≥ξ _{i j} (A) and that ξ _i+1,j (A)≥ξ _i+2,j (A). We will handle each of four cases separately.

• Suppose b _{i j} ≥b _i+1,j and b _i,j+1≥b _i+1,j+1. Then

  $$\begin{array}{rcl} a_{ij} - a_{i,j+1} & = & b_{ij} - b_{i,j+1} \end{array}$$

  and

  $$\begin{array}{rcl} a_{i+1,j} - a_{i+1,j+1} & = & b_{ij} - b_{i,j+1} \\ & \geq & b_{i+1,j} - b_{i+1,j+1}. \end{array}$$

• Suppose b _{i j} ≥b _i+1,j and b _i,j+1≤b _i+1,j+1. Then

  $$\begin{array}{rcl} a_{ij} - a_{i,j+1} & = &b_{ij} - b_{i+1,j+1} \\ & \leq & b_{ij} - b_{i,j+1} \\ \end{array} $$

  and

  $$\begin{array}{rcl} a_{i+1,j} - a_{i+1,j+1} & = & b_{ij} - b_{i+1,j+1} \\ & \geq & b_{i+1,j} - b_{i+1,j+1}. \end{array}$$

• Suppose b _{i j} ≤b _i+1,j and b _i,j+1≥b _i+1,j+1. But then

  $$\begin{array}{rcl} b_{ij} - b_{i,j+1} & \leq & b_{i+1,j} - b_{i+1,j+1} \end{array} $$

  which violates the assumption on B, so this case is not possible.

• Finally, suppose b _{i j} ≤b _i+1,j and b _i,j+1≤b _i+1,j+1. Then

  $$\begin{array}{rcl} a_{ij} - a_{i,j+1}& = &b_{i+1,j} - b_{i+1,j+1} \\ & \leq & b_{i,j} - b_{i+1,j} \end{array} $$

  and

  $$\begin{array}{rcl} a_{i+1,j} - a_{i+1,j+1}& = &b_{i+1,j} - b_{i+1,j+1}. \end{array} $$

We have shown that in any case, ξ _{i j} (A)≤ξ _{i j} (B) and ξ _i+1,j (A)≥ξ _i+1,j (B). For any l≠i and l≠i+1, ξ _{l j} (A)=ξ _{l j} (B) for all j. Thus, ξ _{i −1,j} (A)≥ξ _{i j} (A)=ξ _i+1,j (A)≥ξ _i+2,j (A), so Eq. 8 is satisfied after a left multiply of R _i .

Because we have shown that the resulting matrix after a left multiply of arbitrary P _i or arbitrary R _i satisfies Eq. 8, it follows that Eq. 8 is satisfied after a finite number of left multiplies.

We are now ready to prove the main theorem.

Proof of Theorem 1

Let (c, τ )(α) be given. The algorithm for constructing A(α) begins with I _{m a x} . The block of I _{m a x} consisting of i ₁₁ is trivially in $\mathcal {M}^n$. Thus we know by the Lemmata 11 and 12 that this block satisfies Eq. 7 and Eq. 8 throughout the algorithm. We will show by induction that by the end of the algorithm the entire matrix satisfies both of these. Suppose that at the p ^th stage of the algorithm, the block from b ₁₁ to b _{k k} of the current matrix B satisfies Eq. 7 and Eq. 8 and the remainder of the matrix consists of 𝜖 off the diagonal and e on the diagonal. Consider now C=R _k ⊗B, we can suppose without loss of generality that this is the next operation the algorithm performs. Row k+1 of C is equal to row k of C, with the only difference between row k of C and row k of B being that c _k,k+1=e and b _k,k+1=𝜖. Therefore ξ _{i k} (C)=∞ for i<k and ξ _{k k} (C)<∞, so the block from c ₁₁ to c _k+1,k+1 satisfies (8). Also, because we must multiply by P _k before R _k , c _{k k} >c _k,k+1=0, so the block from c ₁₁ to c _k+1,k+1 satisfies Eq. 7. The remaining rows and columns of C consist of 𝜖 off the diagonal and e on the diagonal. Therefore, by induction, A must satisfy Eq. 7 and Eq. 8.

It remains to be shown that A satisfies Eq. 6 and Eq. 9. By the algorithm, we can write,

$$A = R_{n-1} \otimes B_{n-1} \otimes R_{n-2} \otimes B_{n-2} \otimes \ldots \otimes R_1\otimes P_1 \otimes I_{max} $$

where B _{n −j} is some intermediate matrix sum. This sequence of R _i ’s shows that the 0 entries on the diagonal or I _{m a x} will be cascaded down through the columns of the matrix, it will also ensure that the entries just above the diagonal are greater than −∞, and we see that A satisfies Eq. 9.

For Eq. 6, consider the last multiplication of P _i for some i. This is necessarily followed by a multiplication of R _i , at this point the resulting matrix, C, has c _{i j} =c _i+1,j . This matrix is necessarily multiplied by P _i+1 before another R _i or P _i ; the resulting matrix, which we will call D, has d _{i j} ≤d _i+1,j , at this point we know a _{i j} =d _{i j} and a _i+1,j ≥d _i+1,j , so A must satisfy Eq. 6.