Ind. Eng. Chem. Res. 2003, 42, 811-824
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Efficient Solution Approaches for the Multifloor Process Plant Layout Problem Dimitrios I. Patsiatzis and Lazaros G. Papageorgiou* Centre for Process Systems Engineering, Department of Chemical Engineering, UCL (University College London), London WC1E 7JE, U.K.
In this paper, two novel efficient solution approaches for the multifloor process plant layout problem are presented. The first one is a rigorous decomposition approach, and the second one is based on an iterative solution scheme. Both approaches, which are based on integer linear programming models, determine the number of floors, land area, floor allocation of each equipment item, and detailed layout for each floor. Their applicability is finally demonstrated by five illustrative examples. 1. Introduction Plant layout is an important part of the design or retrofit of a chemical plant as it involves decisions concerning the spatial allocation of equipment items and the required connections among them.1 Significant attention has been received on the plant layout problem though since chemical companies started looking for potential savings at every stage of the design process. Equipment items are allocated to one floor (single-floor case) or many floors (multifloor case) considering a number of cost and management/engineering drivers (e.g., connectivity, operational, land area, safety, construction, retrofit, maintenance, and production organization) within the same framework. To resolve various tradeoffs at an optimal manner, we need to develop computer-aided methods to support engineers in the selection of optimal layouts. A first extensive approach of the problem was given by industrial engineers studying the facility layout and location problem in ref 2, where a given number of departments with unequal areas are located in the plane, minimizing the material handling costs subject to location restrictions and department and floor area requirements. Two approaches have been developed for the solution of the single-floor problem: the quadratic assignment problem (QAP)3 and the graph theory approach.4 The first approach is a special case of the facility layout problem because it assumes equal area departments and a priori fixed and known locations5 and it is solved through branch and bound6 and heuristic7,8 techniques. The second approach maximizes an objective function of weights of the adjacencies (arcs) between department pairs (nodes).5 A limited number of procedures have been developed to solve the multifloor facility layout problem. They can be divided into single-stage and two-stage methods. In single-stage procedures, departments are allowed to occupy any floor during execution.9,10 In the two-stage procedures11,12 though, each department is permanently assigned to each floor in the first stage. In the second stage, the layout is determined for each floor. In this work, we focus on the process plant layout problem where a given number of equipment items are * To whom all correspondence should be addressed. Tel.: +44 20 7679 2563. Fax: +44 20 7383 2348. E-mail:
[email protected].
located in the plane, minimizing the total plant layout cost. In the single-floor case, initial approaches were based on heuristic techniques,13,14 which are efficient from the computational point of view but do not guarantee optimality of the solution obtained. Graph theory approaches were also applied to the problem of organizing items into sections created by aisles or corridors.15 Furthermore, stochastic optimization techniques utilizing either genetic algorithms16 or simulated annealing17 were demonstrated. Finally, a number of mathematical programming approaches have been suggested including a mixed-integer nonlinear programming (MINLP) approach presented in ref 18 integrating safety and economics, as well as a number of continuous-domain integer linear programming (MILP) approaches presenting simultaneous models to determine the optimal location (i.e., coordinates) and orientation for each equipment item,19 alternative formulations for equipment allocation, utilizing a piecewise linear function representation for absolute value functionals,20 and formulations that consider irregular equipment shapes and different equipment connectivity inputs and outputs.21 In the multifloor case, the assignment of items to different floors, by taking into account vertical pumping and land costs and satisfying a number of preferences, was considered in ref 22. A combination of a graphical heuristic approach and a mathematical programming formulation in order to allocate the units in different floors with no consideration of the detailed layout within each floor was proposed in ref 23. Grid-based MILP models have been described in refs 24 and 25, where land area is discretized in a number of candidate locations, with equipment items occupying one and more than one location, respectively. Reference 26 suggested a general mathematical programming MILP formulation which determines simultaneously the number of floors, land area, floor allocation of each equipment item, and detailed layout for each floor. Solving the multifloor process plant layout problem with a continuous-domain formulation is restricted to examples with a small number of units (see, for example, ref 26). The aim of this work is to investigate alternative solution schemes capable of tackling larger flowsheets without comprising significantly the solution quality.
10.1021/ie020586t CCC: $25.00 © 2003 American Chemical Society Published on Web 01/17/2003
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In this work, two novel solution approaches for the multifloor process plant layout problem are presented. The first one is a rigorous decomposition approach, and the second one is based on an iterative solution scheme. The rest of the paper is structured as follows: in the next section, the multifloor process plant layout problem is stated. Section 3 presents two novel solution approaches, while their applicability is demonstrated in section 4 through five examples. Finally, some concluding remarks are made in section 5. 2. Problem Description The multifloor plant layout problem can be stated as follows: Given (i) a set of N equipment items and their dimensions, (ii) a set of K potential floors, (iii) connectivity network, (iv) cost data (connection, pumping, land, and construction), (v) floor height, (vi) space and equipment allocation limitations, and (vii) minimum safety distances between equipment items, determine (i) the number of floors, (ii) land area, (iii) equipment-floor allocation, and (iv) detailed layout (i.e., orientation and coordinates) of each floor, so as to minimize the total plant layout cost. To solve the problem, a number of assumptions are made. Equipment items are described by rectangular shapes, and they are connected through their geometrical centers with rectilinear distances. Units can be allocated only to one floor (i.e., they cannot split into two or more different floors), but a 90° rotation is allowed in the floor plane. In ref 26, a simultaneous approach for the multifloor process plant layout problem was presented using a general mathematical programming formulation which determines simultaneously the number of floors, land area, floor allocation of each equipment item, and detailed layout for each floor. The overall problem was formulated as an MILP model based on a continuousdomain representation. The model minimizes the total plant layout cost (connection, pumping, land area, and floor construction costs) subject to floor constraints (allocate each equipment item to one floor and calculate the number of floors), equipment orientation constraints (determine the length and depth of each item), nonoverlapping constraints (avoid situations where two equipment items occupy the same physical location when allocated to the same floor), distance constraints (calculate the absolute distances between equipment items), and additional layout design constraint (calculate the land and floor area). The complete model [P] is presented in the appendix. This approach has proven capable of solving cases up to 11 units (see ref 26). However, in the case of larger flowsheets, the size of the model (number of equations and integer and continuous variables) increases and the problem is difficult to solve. Therefore, the need for developing efficient solution procedures for the multifloor process plant layout is evident. 3. Solution Approaches In this section, two novel solution procedures for the multifloor process plant layout problem are proposed.
Both approaches comprise a master problem and a subproblem. The master problem provides a lower bound to the optimal solution of the original problem. The number of floors and the allocation of units to floors are determined by the master problem. Then, the solution of the subproblem provides an upper bound to the optimal solution of the original problem and determines the detailed layout for every floor. The master problem and the subproblem are solved iteratively until convergence (crossover of upper and lower bounds), with an acceptable tolerance. As will be explained later in this section, both approaches presented have the same master problem but incorporate different solution schemes for the resulting subproblems. 3.1. Decomposition Approach. 3.1.1. Master Problem. The master problem determines the optimal number of floors as well as equipment-floor assignment. An approximation of the original objective function is used including an approximation of the connection and the pumping costs based on minimum rectilinear distances and the land and floor construction costs. The feasible region of the master problem essentially constitutes a relaxation of that of the original problem [P] because some constraints are left out or approximated. In particular, the orientation (constraints 20 and 21), nonoverlapping (constraints 22-25), horizontal distance (constraints 26 and 27), and additional layout constraints (constraints 38 and 39) are omitted while an approximation of the area constraints (constraints 3035) is used. Next, the mathematical model of the master problem is described in detail. Floor Constraints. The same constraints as the ones of the simultaneous model are used [see the appendix; constraints (15)-(19)]. New integer constraints are included in the model to relate the number of floors, NF, with the new binary variables, Wk.
NF )
∑k Wk
(1)
A floor is occupied only if the previous floor is occupied:
Wk e Wk-1
∀ k ) 2, ..., K
(2)
A unit can only be allocated to an existing floor:
Vik e Wk
∀ i, k
(3)
It should be noted that if one floor does not exist (i.e., Wk ) 0), then constraints (3) do not allow any equipment item to be allocated to floor k by forcing the associated Vik variables to zero. If floor k exists (i.e., Wk ) 1), then at least one equipment item should be allocated to it:
∑i Vik g Wk
∀k
(4)
In addition, each floor may comprise up to a maximum number of units (Λ), to simplify the operational complexity of each floor:
∑i Vik e ΛWk
∀k
(5)
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It should be mentioned that the above constraint can further be generalized to take into account the limited availability of resources as follows:
∑i λrikVik e ΛrkWk
∀ k, r
defined by
∀k
WFAk e FAmaxWk
WFAk g FA - FAmax(1 - Wk)
(6)
(8) ∀k
(9)
where FAmax is given by Vertical Distance Constraints. Vertical distances between equipment items are determined by constraints (28) (see the appendix). Also, minimum horizontal distances are used for approximating horizontal connection and pumping costs in the objective function. Area Constraints. The total floor area is always larger than the summation of the footprint areas of the units allocated to that floor:
FA g
∑i SiVik
∀k
(7)
∑i j*i|f∑)1δijZij + CcijUij + (Ccij + Cvij)Dij ij
where parameter δij is given by
min(ai,bi) + min(aj,bj) 2
δij ) (Ccij + Chij)
(b) floor construction cost
∑k WkFA
and (c) land area cost
LC × FA Overall, the master problem can be summarized as follows:
[Problem MO] min
∑i j*i|f∑)1δijZij +
∑i Si
Finally, the linearized problem corresponds to the following MILP model:
[Problem MR] min
∑i j*i|f∑)1δijZij + CvijDij + CcijUij + (Ccij + Cvij)Dij + ij
FC1 × NF + FC2
In addition, constraints (30) and (31) (see the appendix) are used. Objective Function. The objective function to be minimized includes (a) an approximation of the connection and (horizontal and vertical) pumping costs by using the same cost factors as those in the simultaneous model but for the minimum possible horizontal distances
FC1 × NF + FC2
FAmax )
∑k WFAk + LC × FA
subject to constraints (1)-(9) and part from the appendix [constraints (15)-(19), (28), (30), and (31)]. It is clear that the solution of [MR] provides a lower bound to the solution of the original problem [P] because the feasible region of [M] is simply a relaxation of [P] while the objective function constitutes an underestimator for the original objective function. 3.1.2. Subproblem Model. A reduced simultaneous model is used as the subproblem. In our case, the number of floors and the allocation of units to floor are given by the solution of the master problem. The number of floors, NF, will now be treated as a parameter according to the solution of the master problem. Because the assignment of equipment items to floors is part of the master problem, all floor constraints [see the appendix; constraints (15)-(19)] can be omitted from the model. Similarly, the Zij variables are not necessary and can be treated as parameters, while the nonoverlapping constraints are only defined for units assigned to the same floor. Moreover, vertical distances between units are now known from the master problem and can be treated as parameters in the subproblem model. Note that, because the number of floors is now a parameter, construction cost is no longer a nonlinear term in the objective function and therefore linearization constraints (34) and (35) are not required in the model anymore. In summary, the complete subproblem can be stated as
[Problem S] CcijUij
+
(Ccij
ij
NF + FC2
+
Cvij)Dij
∑k
+ FC1 ×
WkFA + LC × FA
subject to constraints (1)-(7) and part from the appendix [constraints (15)-(19), (28), (30), and (31)]. All continuous variables in the formulation are defined as nonnegative. The above problem is a MINLP model because of the bilinearities involved in the penultimate term of the objective function, which can easily be linearized by introducing new continuous variables, WFAk;
WFAk ≡ WkFA
∀k
min
∑i j*i|f∑)1[CcijDij + Cvij Dzij + Chij(Rij + Lij + Aij + ij
Bij)] + FC1 × NF + FC2 × NF × FA + LC × FA subject to (a) part of the appendix such as (i) equipment orientation constraints (20) and (21); (ii) distance constraints (26), (27), and (29); (iii) area constraints (30)(33); and (iv) additional layout constraints (36)-(39) and (b) nonoverlapping constraints written only for equipment items assigned by the master problem to the same floor (i.e., Zij ) 1)
li + lj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (10)
xi - xj + M(E1ij + E2ij) g
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li + lj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (11)
xj - xi + M(1 - E1ij + E2ij) g
di + dj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (12)
yi - yj + M(1 + E1ij - E2ij) g
di + dj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (13)
yj - yi + M(2 - E1ij - E2ij) g
All continuous variables are defined as nonnegative. The above problem [S] corresponds to an MILP model. Because problem [S] constitutes a reduced version of the original problem [P], the solution of [S] will provide a valid upper bound on the solution of [P]. 3.1.3. Decomposition Solution Procedure. Here, a decomposition procedure is described by solving iteratively master problem [MR] and subproblem [S] until convergence. At every iteration, all previous solutions of the master problem are excluded. This can be achieved by introducing the following constraints in the master problem at iteration m:
∑ Wk + ∑
k∈Um w
(i,k)∈Um v
Vik -
∑ Wk - ∑
k∈Lm w
Vik e |Um w| +
(i,k)∈Lm v
|Um v | - 1 (14) m where Um w and Lw denote the subsets of Wk variables that were at their upper and lower bounds, respectively, at the solution of the master problem at iteration m. In m a similar way, Um v and Lv denote the subsets of Vik variables at their upper and lower bounds, respectively, at the solution of the master problem at iteration m. The algorithm will terminate when the current lower ΦL and the best upper ΦU,min bounds of the optimal solution Φ of the original problem objective function converge with a prespecified tolerance or maximum number of iterations mmax has been exceeded. The optimal solution corresponds to the subproblem solution with the lowest objective function value. The decomposition solution algorithm comprises the following steps:
[Algorithm D] 1. Set best feasible solution ΦU,min ) +∞. Initialize iterations counter; m :) 0. 2. Set m :) m + 1. If m > mmax, STOP. 3. Solve master problem [MR] [including cuts (14)] and get ΦL. If [MR] is infeasible, STOP. 4. If (ΦU,min - ΦL)/ΦU,min e , STOP. 5. Solve subproblem [S] according to the solution of master problem [MR] and get a valid upper bound ΦU. If at any stage the subproblem is infeasible, go to step 2. 6. If ΦU e ΦU,min, keep the current solution as the best feasible solution and set ΦU,min :) ΦU. 7. If (ΦU,min - ΦL)/ΦU,min e , STOP. Otherwise, go to step 2.
3.2. Iterative Approach. In this approach, the same models as those above are used for the master problem and the subproblem. The main difference between the two approaches is in the solution procedure used for the subproblem. Depending on the size of the example, the subproblem is sometimes difficult or time-expensive to solve. For this reason, an alternative solution algorithm is suggested by solving a sequence of smaller versions
of the subproblem with a prefixed number of binary variables before we solve the complete subproblem and get ΦU. The equipment-floor assignment has already been determined by the master problem. The subproblem solution procedure first selects a floor, usually the top or the bottom floor. Then, for the chosen floor, an initial set of equipment items are selected and the resulting reduced subproblem [S] is solved for these items only. Then, the corresponding nonoverlapping binary variables [i.e., variables E1ij and E2ij in constraints (10)(13)] are fixed to the solution of the current subproblem. The next subproblem to be tackled augments the previous set of items by inserting new ones according to some rules. Here, we simply insert new items that are connected to items of the previous iterations. A maximum number of new inserted items may be imposed in the solution procedure. If none of the remaining items on the floor are connected to the previous ones, then we solve the subproblem for all equipment items of the chosen floor. The resulting subproblem will only involve the nonoverlapping binary variables of the new items with respect to existing (i.e., items considered in all previous iterations) and new items, thus reducing the combinatorial nature of the problem. It should be added that, although the nonoverlapping binary variables among existing items are fixed, all remaining variables are determined by the optimization algorithm each time a reduced subproblem is solved. The above equipment insertion scheme is repeated until all equipment items of the chosen floor are considered. The same procedure is then applied to the neighboring floor while including all previously examined floors in the same model with all nonoverlapping binary variables fixed. The above procedure is repeated until all occupied floors are considered. The last subproblem solved, which involves all plant equipment items, gives a valid upper bound, ΦU. The following sets are defined for the description of the iterative algorithm: Sets I ) set of equipment items in the plant Ik ) set of equipment items in floor k determined by the master problem ∆ ) set of equipment items considered by the subproblem Θ ) set of new equipment items inserted
Next, the iterative algorithm used for the solution of the subproblem is outlined: [Algorithm U] 1. Initialize ∆ ) Ø. 2. Select floor, k. 3. Select the initial set of equipment items of floor k, Θ, and set ∆ ) Θ. 4. Solve reduced subproblem [S] for ∆. If ∆ ) I, STOP. 5. Fix nonoverlapping binary variables E1ij and E2ij for i, j ∈ ∆. If all equipment items of floor k are considered (i.e., ∆ ∩ Ik ) Ik), then go to step 7. 6. Insert new equipment items, Θ. Update ∆; ∆ ) ∆ ∪ Θ. go to step 4. 7. Consider the next floor and update k. go to step 6.
Overall, the iterative solution approach uses the same steps as those in algorithm [D], but step 5 of algorithm
Ind. Eng. Chem. Res., Vol. 42, No. 4, 2003 815 Table 1. Dimensions of Equipment Items for Examples 1-5 example 1 Ri item [m] 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
15.8 3.2 15.8 6.3 9.5
example 2
example 3 example 4 example 5
Ri [m]
Ri [m]
βi [m]
Ri [m]
βi [m]
Ri [m]
βi [m]
3.2 5.22 5.22 6.0 3.2 11.42 11.42 7.2 3.2 7.68 7.68 6.0 6.3 8.48 8.48 4.8 3.2 7.68 7.68 4.8 2.60 2.60 6.0 2.40 2.40 4.8 7.2 4.8 7.2 6.0 7.2
4.8 6.0 7.2 6.0 6.0 4.8 6.0 4.8 6.0 6.0 4.8 4.8
3.0 2.5 4.0 3.0 2.5 4.0 4.5 3.5 3.0 3.0 2.0 4.0 3.5 2.5
4.5 3.5 3.0 4.5 4.0 4.0 3.5 3.0 4.5 3.0 3.5 2.0 3.0 3.5
2.5 2.5 3.0 2.5 2.5 3.0 3.5 3.0 3.5 3.0 2.0 3.5 4.0 3.5 3.5 2.0
2.0 3.0 4.0 2.0 2.5 3.0 3.5 2.0 2.5 3.0 2.0 1.5 1.5 3.0 2.5 1.5
βi [m]
βi [m]
Table 2. Connection and Pumping Costs for Example 1 connection
Ccij [rmu/m]
Chij [rmu/m]
Cvij [rmu/m]
(1, 2) (1, 3) (2, 3) (2, 4) (4, 5)
600 800 350 400 500
2525 3783 631 1879 1420
25250 37830 6310 18790 14200
Figure 1. Flowsheet for example 1. Table 3. Connection and Pumping Costs for Example 2
[D] is now replaced by algorithm [U] for the subproblem solution to obtain a valid upper bound, ΦU. As in the previous approach, the optimal solution corresponds to the subproblem solution with the lowest objective function value. Unlike the decomposition approach, the iterative approach cannot guarantee global optimality because of the iterative insertion scheme used in algorithm [U]. However, as shown later, the solution quality obtained by the iterative approach compares favorably with that of the decomposition approach while reducing significantly the computational requirements. 4. Computational Results In this section, the proposed solution approaches are applied to five examples of process plant layout optimization. All examples were modeled using the GAMS modeling system27 coupled with the CPLEX V6.5 MILP optimization package. All of the computational experiments were performed on an IBM RS6000 with 1% margin of optimality for the master problem and the subproblem. The algorithms terminate when the current lower bound (current master problem solution) and the best upper bound (best subproblem solution) differ by less than 5%. Examples 1 and 2 were also solved with the simultaneous solution approach,26 using a 5% margin of optimality, to provide a measure of the solution quality obtained by both proposed approaches. It should be added that the last three examples could not be solved to optimality after a computational limit of 10 000 s. The dimensions of equipment items for all examples are given in Table 1 while the connection and pumping costs are given in Tables 2-6 for examples 1-5, respectively (rmu stands for relative money units). Floor and land cost parameters for every example are given in Table 7. There are variations in the floor- and landrelated coefficients in order to capture differences in construction and land costs. As mentioned earlier, the
connection
Ccij [rmu/m]
Chij [rmu/m]
Cvij [rmu/m]
(1, 2) (2, 3) (3, 4) (4, 5) (5, 1) (5, 6) (6, 7) (7, 5)
200 200 200 200 200 200 200 200
400 400 300 300 100 200 150 150
4000 4000 3000 3000 1000 2000 1500 1500
Table 4. Connection and Pumping Costs for Example 3 connection
Ccij [rmu/m]
Chij [rmu/m]
Cvij [rmu/m]
(1, 2) (2, 3) (2, 4) (3, 2) (4, 1) (4, 5) (4, 6) (5, 1) (6, 7) (7, 8) (8, 9) (8, 11) (9, 10) (10, 11) (11, 7) (11, 12) (12, 1) (12, 4)
120.0 195.0 135.0 195.0 12.0 45.0 135.0 42.0 135.0 165.0 90.0 24.0 90.0 24.0 13.5 36.0 12.0 27.0
939 789 909 789 42 210 669 168 669 540 570 72 420 72 33 108 42 66
9390 7890 9090 7890 420 2100 6690 1680 6690 5400 5700 720 4200 720 330 1080 420 660
land area is chosen from a given set of alternative rectangular area sizes. Here, we use five alternative sizes for the x and y directions, thus resulting in 25 candidate area sizes, each of them being 10 m × 10 m. The first example studied is the five-unit instant coffee process (see Figure 1), introduced in ref 23. The example was first solved with the simultaneous solution approach in ref 26 given three potential floors. The decomposition and iterative approaches give identical objective function values for the master problem and subproblem for each iteration as shown in Figure 2. Both algorithms require 12 iterations to converge while
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Figure 2. Convergence for the decomposition and iterative solution approach for example 1.
Figure 3. Optimal layout for example 1: decomposition solution approach.
Table 5. Connection and Pumping Costs for Example 4 connection
Ccij [rmu/m]
Chij [rmu/m]
Cvij [rmu/m]
(1, 3) (2, 3) (3, 4) (3, 5) (4, 3) (5, 6) (6, 7) (6, 13) (7, 8) (8, 9) (8, 14) (9, 10) (10, 11) (10, 12) (10, 7) (11, 12) (12, 8) (13, 14) (14, 8)
150 150 50 300 50 300 250 50 280 250 80 250 150 100 30 150 30 50 20
300 300 100 600 100 600 500 100 560 500 160 500 300 200 60 300 60 100 40
3000 3000 1000 6000 1000 6000 5000 1000 5600 5000 1600 5000 3000 2000 600 3000 600 1000 400
Table 6. Connection and Pumping Costs for Example 5 connection
Ccij [rmu/m]
Chij [rmu/m]
Cvij [rmu/m]
(1, 2) (2, 3) (3, 4) (4, 5) (5, 3) (5, 6) (6, 7) (7, 8) (8, 9) (9, 8) (9, 10) (10, 11) (11, 3) (8, 12) (12, 13) (13, 14) (14, 15) (15, 16)
175 160 205 205 15 190 190 190 45 15 30 35 25 140 140 140 140 140
525 480 615 615 45 570 570 570 285 45 90 105 75 420 420 420 420 420
5250 4800 6150 6150 450 5700 5700 5700 2850 450 900 1050 750 4200 4200 4200 4200 4200
the optimal solution was obtained in iteration 9. The optimal objective function values and the CPU times for each approach are presented in Tables 16 and 17. The optimal two-floor layouts determined by decomposition and iterative approaches are shown in Figures 3 and 4, respectively. It is worth noting that both optimal layouts are equivalent because each of them can easily be derived by shifting appropriately all equipment items
Figure 4. Optimal layout for example 1: iterative solution approach. Table 7. Floor Construction and Land Area Cost Parameters example
FC1 [rmu]
FC2 [rmu/m2]
LC [rmu/m2]
1 2 3 4 5
3330 3330 2331 3330 3330
66.7 6.7 9.9 6.6 6.6
66.7 26.7 133.2 66.7 66.7
Table 8. Optimal Solution for Example 1: Decomposition Solution Approach orientation
location
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5
15.8 3.2 15.8 6.3 9.5
3.2 3.2 3.2 6.3 3.2
10.50 10.50 10.50 15.25 15.25
2.10 5.30 2.10 6.85 6.85
2 2 1 2 1
along the x direction. The above degeneracy stems mainly from the discretization of the land area. The optimal equipment orientation and location are given in Tables 8 and 9 for decomposition and iterative approaches, respectively.
Ind. Eng. Chem. Res., Vol. 42, No. 4, 2003 817
Figure 5. Flowsheet for example 2.
Figure 6. Optimal layout for example 2: decomposition solution approach.
Figure 7. Optimal layout for example 2: iterative solution approach.
Table 9. Optimal Solution for Example 1: Iterative Solution Approach
Table 10. Optimal Solution for Example 2: Decomposition Solution Approach
orientation
orientation
location
location
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5
15.8 3.2 15.8 6.3 9.5
3.2 3.2 3.2 6.3 3.2
7.9 7.9 7.9 12.65 12.65
2.1 5.3 2.1 6.85 6.85
2 2 1 2 1
1 2 3 4 5 6 7
5.22 11.42 7.68 8.48 7.68 2.60 2.40
5.22 11.42 7.68 8.48 7.68 2.60 2.40
12.32 12.32 12.32 4.24 4.24 9.38 9.38
16.4 8.08 8.08 8.08 16.16 16.16 18.66
2 2 1 1 1 1 1
The second example considers the ethylene oxide plant, derived from the case study presented in ref 18. The plant flowsheet includes seven units, as shown in Figure 5. As can be seen in Table 16, both proposed approaches obtain the same objective function value as that of the simultaneous approach. However, it is evident from Tables 10 and 11 and Figures 6 and 7 that degenerate optimal solutions were obtained with the same item-floor assignment but different locations of equipment items. In all cases, two of the three initial floors are occupied. Note that the iterative approach solves the problem in a significantly smaller CPU time
than the decomposition and simultaneous approaches (see Table 17), thus illustrating the efficiency of the iterative approach. Both approaches require nine iterations in total to converge with the second iteration, obtaining the best solution as is clearly seen in Figure 8. It should be mentioned that the same subproblem solutions are obtained by both approaches apart from the very first iteration. The third example was first presented in ref 15 and considers the layout design for a 12-unit plant (see
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Ind. Eng. Chem. Res., Vol. 42, No. 4, 2003 Table 14. Optimal Solution for Example 4: Iterative Solution Approach orientation
location
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5 6 7 8 9 10 11 12 13 14
4.5 2.5 3.0 4.5 4.0 4.0 3.5 3.5 3.0 3.0 2.0 2.0 3.0 3.5
3.0 3.5 4.0 3.0 2.5 4.0 4.5 3.0 4.5 3.0 3.5 4.0 3.5 2.5
4.50 1.75 4.50 3.75 8.00 8.00 8.00 8.00 4.75 4.75 2.25 2.25 8.25 8.00
1.50 5.25 5.00 8.50 4.75 8.00 7.75 4.00 4.00 7.75 8.00 4.25 1.75 1.25
2 2 2 2 2 2 1 1 1 1 1 1 2 1
Table 15. Optimal Solution for Example 5: Iterative Solution Approach Figure 8. Convergence for the decomposition and iterative solution approach for example 2.
orientation li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
2.5 2.5 3.0 2.0 2.5 3.0 3.5 2.0 3.5 3.0 2.0 3.5 4.0 3.5 3.5 2.0
2.0 3.0 4.0 2.5 2.5 3.0 3.5 3.0 2.5 3.0 2.0 1.5 1.5 3.0 2.5 1.5
8.75 5.75 6.00 3.50 1.25 1.75 1.75 4.50 7.25 8.50 8.50 4.50 4.50 4.50 4.50 4.50
6.50 4.50 8.00 8.00 8.00 5.00 1.75 1.50 1.25 4.00 8.50 0.75 2.25 4.50 7.25 9.25
2 2 2 2 2 2 2 2 2 2 2 1 1 1 1 1
Table 11. Optimal Solution for Example 2: Iterative Solution Approach orientation
location
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5 6 7
5.22 11.42 7.68 8.48 7.68 2.60 2.40
5.22 11.42 7.68 8.48 7.68 2.60 2.40
8.08 8.08 8.08 8.08 16.16 16.16 18.66
4.00 12.32 12.32 4.24 4.24 9.38 9.38
2 2 1 1 1 1 1
Table 12. Optimal Solution for Example 3: Decompostion Solution Approach orientation equipment 1 2 3 4 5 6 7 8 9 10 11 12
li [m] 4.8 6.0 6.0 4.8 6.0 4.8 6.0 7.2 6.0 6.0 6.0 7.2
di [m] 6.0 7.2 7.2 6.0 4.8 6.0 4.8 4.8 4.8 7.2 4.8 4.8
Table 16. Objective Function Values (in rmu)
location xi [m] 5.4 10.8 16.8 10.8 5.4 15.6 15.6 15.6 9.0 3.0 9.0 4.8
yi [m] 3.6 3.6 3.6 3.6 2.8 3.0 2.8 7.6 7.6 6.4 2.8 7.6
approach
allocation floor 3 3 3 2 2 2 1 1 1 1 1 2
example
simultaneous
decomposition
iterative
1 2 3 4 5
82 366 50 817 111 074 102 335 88 767
82 366 50 817 101 100 53 919 51 239
82 366 50 817 102 086 42 147 40 602
Table 17. CPU Times (in s) approach example
simultaneous
decomposition
iterative
1 2 3 4 5
3.1 174.0 10 000.0a 10 000.0a 10 000.0a
1.5 37.0 4457.6 10 000.0a 10 000.0a
1.5 7.0 413.7 110.0 20.7
Table 13. Optimal Solution for Example 3: Iterative Solution Approach orientation
location
equipment
location
equipment
li [m]
di [m]
xi [m]
yi [m]
allocation floor
1 2 3 4 5 6 7 8 9 10 11 12
4.8 6.0 6.0 6.0 4.8 6.0 6.0 7.2 4.8 6.0 6.0 7.2
6.0 7.2 7.2 4.8 6.0 4.8 4.8 4.8 6.0 7.2 4.8 4.8
2.4 7.8 13.8 7.8 2.4 7.8 6.2 5.6 11.6 17.0 14.4 14.4
6.4 6.4 6.4 7.2 7.0 2.4 2.4 7.2 7.0 6.4 2.4 7.2
3 3 3 2 2 2 1 1 1 1 2 2
Figure 9) manufacturing cosmetic-grade isopropyl alcohol. As shown in Tables 16 and 17, the simultaneous approach cannot solve this example to optimality, obtaining an integer feasible solution with an objective function value of 111 074 rmu (gap of 25.9%) after
a
A computational limit of 10 000 s is used.
10 000 s. The decomposition approach requires 40 iterations to converge as shown in Figure 10 while the 21st iteration provides the optimal solution (101 100 rmu). The iterative approach, which converges in 93 iterations (see Figure 11), solves the problem with a modest CPU time of 413.7 s vs 4457.6 s for the decomposition approach. The optimal solution of the iterative approach (102 086 rmu) is less than 1% higher than the respective one from the decomposition approach. The optimal layout details for both approaches are shown in Figures 12 and 13 and Tables 12 and 13. The two optimal structures differ only in the floor allocation of item 11: the first floor for the decomposition approach and the second floor for the iterative
Ind. Eng. Chem. Res., Vol. 42, No. 4, 2003 819
Figure 9. Flowsheet for example 3.
Figure 10. Convergence for the decomposition solution approach for example 3. Figure 12. Optimal layout for example 3: decomposition solution approach.
Figure 11. Convergence for the iterative solution approach for example 3.
approach. Both optimal solutions comprise three constructive floors out of four potential ones. Example 4 considers a 14-unit maleic anhydride process, based on ref 28. The simplified flowsheet is shown in Figure 14. The simultaneous and decomposition approaches cannot solve the problem within the prespecified time limit of 10 000 s. For the simultaneous approach, we get an objective function of 102 335 rmu with a gap of 82.6%. For the decomposition approach, the subproblem of the first iteration cannot be solved for an optimality margin of 1%. However, an integer feasible solution of 53 919 rmu exhibiting a gap of 32.4% has been obtained. On the other hand, the iterative approach solves the problem within 17 iterations while the third iteration obtains the optimal solution (42 147 rmu) as shown in Figure 15. The optimal layout is presented in Figure 16 and Table 14. Two floors out of the three initially available are finally occupied.
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better computational performance has been achieved by starting on the top floor for all examples except example 5. It is worth mentioning that the optimal solution was obtained in just a few iterations (usually less than 10 iterations) for most of the examples presented, which can be adopted for larger and/or more complicated flowsheets. In this way, the CPU time can significantly be reduced (i.e., for example 4 from 110 to 64 s). 5. Concluding Remarks
Figure 13. Optimal layout for example 3: iterative solution approach.
Example 5 considers the layout design for a 16-unit cis-polybutadiene process (ref 28). Figure 17 shows a simplified flowsheet of the process. As shown in Table 17, only the iterative approach tackles the problem within the time limit of 10 000 s while the corresponding optimal solution is presented in Figure 18 and Table 15. The simultaneous approach results in an objective function value of 88 767 rmu (gap of 81.62%) in 10 000 s. The decomposition approach cannot provide a solution for the subproblem of the first iteration with 1% margin of optimality (objective function value: 51 239 rmu with a gap of 30.3%). For the iterative approach, an optimal solution of 40 602 rmu requires 20.7 s, thus clearly illustrating the efficiency of this approach. The convergence between solutions of the master problem and subproblem is shown in Figure 19. As is evident from section 3.2, algorithm [U] of the iterative approach is based on two user-defined parameters: selection of initial floor (step 2) and equipment insertion strategy (step 6). It has been found that a
Figure 14. Flowsheet for example 4.
In this paper, two efficient solution approaches for the optimal multifloor process plant layout problem have been considered based on decomposition and iterative solution approaches. The proposed approaches have been shown to compare favorably with the simultaneous approach.26 Both approaches determine the number of floors, land area, optimal equipment-floor allocation, and equipment location (i.e., coordinates and orientation) so as to minimize the total plant layout cost. The applicability of the proposed approaches has been demonstrated by five illustrative examples. Finally, it should be mentioned that the iterative approach, which does not compromise solution quality, proved particularly successful for tackling larger flowsheets with modest computational requirements. Acknowledgment The authors gratefully acknowledge the financial support from the Centre for Process Systems Engineering and the U.K. Engineering and Physical Sciences Research Council (EPSRC). Appendix The simultaneous approach is presented in ref 26 as follows: A1. Floor Constraints. Each equipment item should be assigned to one floor:
∑k Vik ) 1
∀i
(15)
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mined by
∀i
li ) RiOi + βi(1 - Oi)
∀i
di ) Ri + βi - li
(20) (21)
A3. Nonoverlapping Constraints. To avoid situations where two equipment items i and j occupy the same physical location when allocated to the same floor (i.e., Zij ) 1), appropriate constraints should be included in the model that prohibit overlapping of their equipment footprint projections, either in the x or y direction.
li + lj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (22)
xi - xj + M(1 - Zij + E1ij + E2ij) g Figure 15. Convergence for example 4: iterative solution approach.
li + lj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (23)
xj - xi + M(2 - Zij - E1ij + E2ij) g
di + dj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (24)
yi - yj + M(2 - Zij + E1ij - E2ij) g
di + dj 2 ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N (25)
yj - yi + M(3 - Zij - E1ij - E2ij) g
Each pair of values (0 or 1) to variables E1ij and E2ij determines which constraint from (22) to (25) is active. A4. Distance Constraints. Distance constraints calculate the relative distances in x, y, and z coordinates:
Figure 16. Optimal layout for example 4: iterative solution approach.
If units i and j are allocated to the same floor, then binary variable Zij is equal to 1:
Zij g Vik + Vjk - 1 Zij e 1 - Vik + Vjk Zij e 1 + Vik - Vjk
∀ i ) 1, ..., N - 1, j ) i + 1, ..., N, k ) 1, ..., K (16) ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N, k ) 1, ..., K (17) ∀ i ) 1, ..., N - 1, j ) i + 1, ..., N, k ) 1, ..., K (18)
∑k kVik
∀i
∀ (i, j): fij ) 1
(26)
Aij - Bij ) yi - yj
∀ (i, j): fij ) 1
(27)
Uij - Dij ) H
∑k k(Vik - Vjk)
∀ (i, j): fij ) 1
(28)
Thus, the total rectilinear distance between items i and j is given by
TDij ) Rij + Lij + Aij + Bij + Uij + Dij ∀ (i, j): fij ) 1 (29) A5. Area Constraints. The value of land area, FA, is chosen from a set of S candidate rectangular area h s and Y h s dimensions: sizes, ARs, with X
The number of floors is determined by
NF g
Rij - Lij ) xi - xj
(19)
A2. Equipment Orientation Constraints. The length and the depth of equipment item i are deter-
FA )
∑s ARsQs
∑s Qs ) 1
(30) (31)
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Figure 17. Flowsheet for example 5.
Figure 19. Convergence for example 5: iterative solution approach. Figure 18. Optimal layout for example 5: iterative solution approach.
The values of Xmax and Ymax variables are forced to coincide with the dimensions of the selected area size:
Xmax ) Y
max
)
∑s Xh sQs
(32)
∑s Yh sQs
(33)
Some extra linearization constraints are included to avoid bilinearities in the objective function:
NQs e KQs NF )
∀s
∑s NQs
(34) (35)
A6. Additional Layout Design Constraints. Lower bound constraints on the coordinates of the geometrical center are included in order to avoid intersection of items with the origin of axes:
xi g
li 2
∀i
(36)
yi g
di 2
∀i
(37)
A rectangular shape of land area is assumed to be used, and its dimensions are determined by
xi +
li e Xmax 2
∀i
(38)
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yi +
di e Ymax 2
∀i
(39)
A7. Objective Function. The overall objective function minimizes connection, (horizontal and vertical) pumping, land area, and construction cost:
min
∑i j*i|∑f )1[CcijTDij + CvijDij + Chij(Rij + Lij + Aij + ij
∑s ARsNQs + LC × FA
Bij)] + FC1 × NF + FC2 Nomenclature Indices
i, j ) equipment item k ) floor s ) candidate rectangular area size Parameters Ri, βi ) dimensions of item i (m) Si ) footprint area of item i (m2) H ) floor height (m) fij ) 1 if flow is from item i to item j; 0 otherwise Ccij ) connection cost between items i and j (rmu/m) Cvij ) vertical pumping cost between items i and j (rmu/ m) Chij ) horizontal pumping cost between items i and j (rmu/m) FC1 ) fixed floor construction cost (rmu) FC2 ) area-dependent floor construction cost (rmu/m2) LC ) land cost (rmu/m2) Λ ) maximum number of units per floor Λrk ) availability level on resource r for floor k λrik ) relevant equipment resource utilization Integer Variable NF ) number of floors Binary Variables Vik ) 1 if item i is assigned to floor k; 0 otherwise Zij ) 1 if equipment items i and j are allocated to the same floor; 0 otherwise Oi ) 1 if length of item i is equal to Ri (i.e., parallel to the x axis); 0 otherwise E1ij, E2ij ) nonoverlapping binary variables (as used in ref 19) Qs ) 1 if candidate area s is selected; 0 otherwise Wk ) 1 if floor k is occupied; 0 otherwise Continuous Variables li ) length of item i (m) di ) depth of item i (m) xi, yi ) coordinates of the geometrical center of item i (m) Rij ) relative distance in x coordinates between items i and j, if i is to the right of j (m) Lij ) relative distance in x coordinates between items i and j, if i is to the left of j (m) Aij ) relative distance in y coordinates between items i and j, if i is above j (m) Bij ) relative distance in y coordinates between items i and j, if i is below j (m) Uij ) relative distance in z coordinates between items i and j, if i is higher than j (m) Dij ) relative distance in z coordinates between items i and j, if i is lower than j (m)
TDij ) total rectilinear distance between items i and j (m) FA ) floor land area (m2) Xmax, Ymax ) dimensions of floor land area (m) NQs ) linearization variable expressing the product of NF and Qs WFAk ) linearization variable expressing the product of Wk and FA (m2)
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Received for review July 31, 2002 Revised manuscript received October 28, 2002 Accepted December 4, 2002 IE020586T