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Machine assignment in a nonlinear multi-product flowline

Abstract

A group of machines for processing a set of jobs in a manufacturing system is often located in a serial line. An efficient strategy for locating these machines such that the total travel distance or the cost of transporting the jobs is minimized is desired. In this research, the assumption of a linear line with equally spaced machine location is relaxed. This research addressed problems of locating unique machines. It is found that the machine distances possess unique properties in this type of a problem. Utilizing these properties, heuristic strategies are proposed to obtain efficient solution where optimal methods are expected to be computationally prohibitive. A lower bound for the optimum solution is also proposed. Results are encouraging.

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Correspondence to B R Sarker.

Appendices

Appendix: A Proof of properties

Proof of Property 1 If machine i is assigned to location k and machine j at location h, x ik = 1 and x jh = 1. Also x il = 0 (for lk) and x jm = 0 for (for mh). From (4), d ij = δ kh and d ji = δ hk . □

Proof of Property 2 From Property 1, d ij = δ kh if machine i is assigned to location k and machine j at location h. The row-sum is given as and the column-sum is given as . Therefore, .

Proof of Property 3 In the first assignment (a 1), let x mk = 1, x nh = 1 and x lj = 1 for lm, n and jk, h. From Equation (4), d mn 1 = δ kh , d nm 1 = δ hk , d ml 1 = δ kj and d lm 1 = δ jk . In the second assignment (a 2), x nk = 1, x mh = 1 and x lj = 1 for lm,n and jk h. From (4), d nm 2 = δ kh , d mn 2 = δ hk , d nl 2 = δ kj and d ln 2 = δ jk . Therefore, (a)-(c) follow. □

Appendix: B Illustration of SH

Step 0: Given R and Δ in (6) and (3), respectively. Set k←1: m←1: λ←0. Assume an initial solution a 0 =(4, 1, 2, 5, 3) and set a k * = a 1 * = a 0. The matrices D(a 0) and Q(a 0)are

The corresponding total flow distance is TD(a 0)=206 and the non-increasing ordered array of is ={1094, 843, 792, 755, 651}. Set Φ ← Φ0.

Step 1: Iteration: Produce Mm =5−1=4 new assignments a 1, …, a j , …, a Mm : Swap the machine that corresponds to , which is 4, with other machines that correspond to for j=1,…, 4, which are 3, 2, 5 and 1:

Step 2: Since min l {TD a l } =TD(a 2) =200<TD(a 1 *) =206 then Set m←1: λ←1: a 1 *a 2=(2, 1, 4, 5, 3): TD(a 1 *) ← 200. The D(a 1 *) and Q(a 1 *) matrices can be computed accordingly (but are not shown here for compactness). Hence, = {1112, 853, 705, 694, 651} and return to Step 1.

Step 1: Produce Mm=5−1=4 assignments: Swap the machine that corresponds to , which is 2, with all other machines that correspond to for j=1,…, 4: a 1=(3, 1, 4, 5, 2), a 2=(5, 1, 4, 2, 3), a 3=(4, 1, 2, 5, 3), a 4=(1, 2, 4, 5, 3) with TD(a 1)=229, TD(a 2)=246, TD(a 3)=206 and TD(a 4)=198.

Step 2: Since min l {TD(a l ) =TD(a 4)=198<TD(a 1 *) = 200 set m←1: λ←1: a 1 *a 4 =(1, 2, 4, 5, 3): TD(a 1 *) ← 198. Thus, φ ={963, 951, 752, 654, 655}, go to Step 1.

Note from this point forward, we use symbol ‘∼’ to mean ‘corresponds to’ for brevity.

Step 1: Swap 3 with those , j=1,…, 4: a 1=(3, 2, 4, 5, 1), a 2=(1, 3, 4, 5, 2), a 3=(1, 2, 3, 5, 4) and a 4=(1, 2, 4, 3, 5) with TD(a 1)=259, TD(a 2)=242, TD(a 3)=218 and TD(a 4)=203.

Step 2: Since min l { TD(a l ) } =203>TD(a 1 *)=198, go to Step 3.

Step 3: Since m=1<M−1=4 and λ=1, then mm+1=2 and go to Step 1.

Step 1: Swap 1 with machines , j=1, 2, 3: a 1=(2, 1, 4, 5, 3), a 2=(4, 2, 1, 5, 3) and a 3=(5, 2, 4, 1, 3) with TD(a 1)=200, TD(a 2)=199 and TD(a 3)=224.

Step 2: Since min l {TD(a l )} =199>TD(a 1 *)=198, go to Step 3.

Step 3: Since m=2<M−1=4 and λ=1, then mm+1=3 and go to Step 1.

Step 1: Swap 2 with machines , j=1, 2: a 1=(1, 4, 2, 5, 3) and a 2=(1, 5, 4, 2, 3) with TD(a 1)=211 and TD(a 2)=254.

Step 2: Since min l {TD(a l )} =211>TD(a 1 *)=198, go to Step 3.

Step 3: Since m=3<M−1=4 and λ=1, then mm+1=4 and go to Step 1.

Step 1: Swap 4 with those for j=1: a 1=(1, 2, 5, 4, 3) with TD(a 1)=220.

Step 2: Since min l {TD(a l )} =220>TD(a 1 *)=198, go to Step 3.

Step 3: Since m=4 and since k=1<M−1=4 (searching the first branch is completed) then set kk+1=2: mk: λ←0: a 2 *a 0 =(4, 1, 2, 5, 3): Φ ← Φ0 = {1094, 843, 792, 755, 651}: TD(a 2 *)←TD(a 0)=206 and go to Step 1.

Step 1: Swap 3 with those for j=1, 2, 3: a 1=(4, 1, 3, 5, 2), a 2=(4, 1, 2, 3, 5) and a 3=(4, 3, 2, 5, 1) with TD(a 1)=211 TD(a 2)=243 and TD(a 3)=260.

Step 2: Since min l { TD(a l ) } =211>TD(a 2 *)=206, go to Step 3.

Step 3: Since λ≠1 and since k=2<M−1=4, then set kk+1=3: mk: λ←0: a 3 *a 0: Φ ← Φ0: TD(a 3 *)←TD(a 0) and go to Step 1.

Step 1: Swap 2 with those for j=1, 2: a 1=(4, 1, 5, 2, 3) and a 2=(4, 2, 1, 5, 3) with TD(a 1)=206 and TD(a 1)=199.

Step 2: Since min l { TD(a l ) } =199<TD(a 3 *)=206, then set m←1: λ←1: a 3 *a 1 =(4, 2, 1, 5, 3): TD(a 3 *)←199. Thus, Φ ={1014, 835, 793, 772, 581} and return to Step 1.

Step 1: Swap 4 with those , j=1,…, 4: a 1=(5, 2, 1, 4, 3), a 2=(3, 2, 1, 5, 4), a 3=(2, 4, 1, 5, 3) and a 4=(1, 2, 4, 5, 3) with TD(a 1)=238, TD(a 2)=240, TD(a 3)=206 and TD(a 4)=198.

Step 2: Since min l {TD(a l )} =TD(a 4)=198<TD(a 3 *) =199, then set m←1: λ←1: a 3 *a 4=(1, 2, 4, 5, 3): TD(a 3 *)←199. Find Φ ={963, 951, 752, 654, 655}, go to Step 1.

Step 1: Swap 3 with those , j=1,…, 4: a 1=(3, 2, 4, 5, 1), a 2=(1, 3, 4, 5, 2), a 3=(1, 2, 3, 5, 4) and a 4=(1, 2, 4, 3, 5) with TD(a 1)=259, TD(a 2)=242, TD(a 3)=218 and TD(a 4)=203.

Step 2: Since min l { TD(a l ) } =203>TD(a 3 *)=198, go to Step 3.

Step 3: Since m=1<M−1=4 and λ=1, mm+1=2 and go to Step 1.

Step 1: Swap 1 with those , j=1, 2, 3: a 1=(2, 1, 4, 5, 3), a 2=(4, 2, 1, 5, 3) and a 3=(5, 2, 4, 1, 3) with TD(a 1)=200, TD(a 2)=199 and TD(a 3)=224.

Step 2: Since min l { TD(a l ) } =199>TD(a 3 *)=198, go to Step 3.

Step 3: Since m=2<M−1=4 and λ=1, then mm+1=3 and go to Step 1.

Step 1: Swap 2 with those for j=1, 2: a 1=(1, 4, 2, 5, 3) and a 2=(1, 5, 4, 2, 3) with TD(a 1)=211 and TD(a 2)=254.

Step 2: Since min l { TD(a l ) } =211>TD(a 3 *)=198, go to Step 3.

Step 3: Since m=3<M−1 and λ=1, mm+1=4 and go to Step 1.

Step 1: Swap 4 with those for j=1: a 1=(1, 2, 5, 4, 3) with TD(a 1)=220.

Step 2: Since min l { TD(a l ) } =220>TD(a 3 *)=198, go to Step 3.

Step 3: Since m=4 and since k=3<M−1=4, set kk+1=4: mk: λ←0: a 4 *a 0: Φ ← Φ0: TD(a 4 *) ←TD(a 0)=206 and go to Step 1.

Step 1: Swap 4 (∼ P [4] Q(a 4 *) with those ∼ P [4 + j] Q(a 4 *) for j = 1: a 1 = (4, 5, 2, 1, 3) with TD(a 1) = 215.

Step 2: Since min l { TD(a l ) } =215>TD(a 4 *)=206, go to Step 3.

Step 3: Since λ≠1 and since k=4, go to Step 4.

Step 4: The best total distance is TD *= min s {TD(a s *)} = TD(a 1 *) = 198 with a 1 * =(2, 1, 4, 5, 3).□

Appendix C: Solution of problems for empirical tests

The solution of problems for empirical tests are shown in Tables 8,9,10,11.

Table 8 Solution of problem P-m 2
Table 9 Solution of problem O-m 16
Table 10 Solution of problem S-m 2
Table 11 Solution of asymmetric and unequally spaced machine location*

Appendix D: Location distance for problem A-m

The location distance for problem A−m is shown in Table 12

Table 12 Distance between adjacent locations for asymmetric problem

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Diponegoro, A., Sarker, B. Machine assignment in a nonlinear multi-product flowline. J Oper Res Soc 54, 472–489 (2003). https://doi.org/10.1057/palgrave.jors.2601488

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Keywords

  • multi-product flowline
  • machine location problem
  • heuristic solution