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HENS       S EQUENTIAL F RAMEWORK           M IN U NITS SUB - PROBLEM   S UMMARY




       T HE S EQUENTIAL F RAMEWORK FOR H EAT
          E XCHANGER N ETWORK S YNTHESIS

                          Rahul Anantharaman
                    rahul.anantharaman@ntnu.no

                Department of Energy & Process Engineering
               Norwegian University of Science and Technology


                               IIT Madras
                           Chennai, 18.12.2009
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS                  S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



H EAT E XCHANGER N ETWORK S YNTHESIS


       For a given set of hot and cold process streams as well as
       external utilities, design a heat exchanger network that
       minimizes Total Annualized Cost (TAC).
       TAC = Capital Cost + Energy Cost




        Sequential Framework Engine
HENS           S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
S OLUTION METHODS




        Evolutionary methods such as Pinch Design Method
        Sequential synthesis methods
        Simultaneous synthesis methods
        Stochastic optimization methods
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS        S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


H EAT E XCHANGER N ETWORK S YNTHESIS
T IMELINE
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS             S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW




       224 references published from 2000-2008
           216 journal papers
               48 jounals
               43 countries
           4 conference proceedings
           10 Ph.D. theses
           4 texts
HENS                  S EQUENTIAL F RAMEWORK                 M IN U NITS SUB - PROBLEM           S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW


          45



          40



          35



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          25



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          15



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          5



          0
               2000        2001     2002       2003   2004    2005       2006      2007   2008
HENS                  S EQUENTIAL F RAMEWORK         M IN U NITS SUB - PROBLEM                          S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW




                                                                         Computers & Chem Eng
                                                43
                              77                                         Applied Thermal Eng

                                                                         Industrial & Eng Chem 
                                                                         Research
                                                                         Chem Eng Research & Design

                                                      39                 Latin American Appl Research

                                                                         Heat Transfer Eng
          7
                          7                                              Chem Eng Science
                              7                23
              7                    7     10                              Chem Eng and Processing
                                                                         Ch   E     dP       i
                  7
HENS           S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW



          50


          45


          40


          35


          30


          25


          20


          15


          10


          5


          0
HENS      S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW
HENS             S EQUENTIAL F RAMEWORK    M IN U NITS SUB - PROBLEM   S UMMARY


HENS IN THE 21 ST CENTURY
R EVIEW

           HENS still an active area of research interest
           Over 25% of references devoted to case studies
               Pinch Analysis based evolutionary methods dominate
           Sustained interest in simultaneous MINLP methods
               Yee and Grossmann (1990) superstructure
               Pressure drop and detailed HX design considerations
               Small test problems
           Number of references related to genetic programming and
           other meta-heuristic methods increasing in frequency

       Though there has been significant developments in HENS
       using mathematical programming methods, synthesis of large
       scale HENS problems without simplifications and heuristics
       have been lacking. This is an area that requires more research
       for mathematical programming based approaches to be used in
       the industry
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS         S EQUENTIAL F RAMEWORK      M IN U NITS SUB - PROBLEM   S UMMARY



M OTIVATION FOR THE S EQUENTIAL F RAMEWORK


       Pinch based methods for network design
           Improper trade-off handling
           Time consuming
           Several topological traps
       MINLP methods for network design
           Severe numerical problems
           Difficult user interaction
           Fail to solve large scale problems
       Stochastic optimization methods for network design
           Non-rigorous algorithms
           Quality of solution depends on time spent on search
HENS              S EQUENTIAL F RAMEWORK    M IN U NITS SUB - PROBLEM   S UMMARY



M OTIVATION FOR THE S EQUENTIAL F RAMEWORK


       HENS TECHNIQUES DECOMPOSE THE MAIN PROBLEM
           Pinch Design Method is sequential and evolutionary
           Simultaneous MINLP methods let math considerations
           define the decomposition
           The Sequential Framework decomposes the problem into
           subproblems based on knowledge of the HENS problem

       Engineer acts as optimizer at the top level
       Quantitative and qualitative considerations included
HENS         S EQUENTIAL F RAMEWORK      M IN U NITS SUB - PROBLEM   S UMMARY



U LTIMATE G OAL

       Solve Industrial Size Problems
           Defined to involve 30 or more streams
       Include Industrial Realism
           Multiple and ``Complex´´Utilities
           Constraints in Heat Utilization (Forbidden matches)
           Heat exchanger models beyond pure countercurrent
       Avoid Heuristics and Simplifications
           No global or fixed ∆ Tmin
           No Pinch Decomposition
       Develop a Semi-Automatic Design Tool
           EXCEL/VBA (preprocessing and front end)
           MATLAB (mathematical processing)
           GAMS (core optimization engine)
           Allow significant user interaction and control
           Identify near optimal and practical networks
HENS                      S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UR ENGINE




        3 way trade-off


       Compromise between Pinch Design and MINLP methods
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS     S EQUENTIAL F RAMEWORK              M IN U NITS SUB - PROBLEM   S UMMARY



E XAMPLE 1



             Stream    Tin    Tout    mCp       ∆H             h
                       K       K      kW/K      kW         kW/m2 K
               H1     626     586    9.802     392.08        1.25
               H2     620     519    2.931     296.03        0.05
               H3     528     353    6.161    1078.18        3.20
               C1     497     613    7.179     832.76        0.65
               C2     389     576    0.641     119.87        0.25
               C3     326     386    7.627     457.62        0.33
               C4     313     566     1.69     427.57        3.20
               ST     650     650       -        -           3.50
              CW      293     308       -        -           3.50
             Exchanger cost ($) = 8,600 + 670A0.83 (A is in m2 )
HENS                S EQUENTIAL F RAMEWORK              M IN U NITS SUB - PROBLEM   S UMMARY


E XAMPLE 1
L OOPING TO THE SOLUTION



       HRAT fixed at 20K (Qh,min = 244.1 kW & Qc,min = 172.6)
       Umin = 8 units


                       Soln. No       U      EMAT (K)    HLD         TAC ($)
                           1          8        2.5        A          199,914
                           2          8         5         A          199,914
                           3          8        7.5        -          No Soln
                           4          9        2.5        A          147,861
                           5          9        2.5        B          151,477
                           6          9         5         A          147,867
                           7          9         5         B          151,508
                           8          9        7.5        A          149,025
                           9          9        7.5        B          149,224
                          10          10       2.5        A          164,381
                          11          10        5         A          167,111
                          12          10       7.5        A          164,764
HENS             S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


E XAMPLE 1
B EST SOLUTION
HENS           S EQUENTIAL F RAMEWORK           M IN U NITS SUB - PROBLEM              S UMMARY


E XAMPLE 1
C OMPARISON




                                        No. of units    Area (m2 )          Cost ($)
         Colberg and Morari (1990)          22            173.6
         Colberg and Morari (1990)          12            188.9             177,385
        Yee and Grossmann (1990)             9            217.8             150,998
           Sequential Framework              9            189.7             147, 861
HENS     S EQUENTIAL F RAMEWORK            M IN U NITS SUB - PROBLEM   S UMMARY



E XAMPLE 2

         Stream    Tin    Tout      mCp      ∆H               h
                  (°C)    (°C)    (kW/°C)    (kW)       (kW/m2 °C)
           H1     180      75        30      3150             2
           H2     280     120        60      9600             1
           H3     180      75        30      3150             2
           H4     140      40        30      3000             1
           H5     220     120        50      5000             1
           H6     180      55        35      4375             2
           H7     200      60        30      4200            0.4
           H8     120      40       100      8000            0.5
           C1      40     230        20      3800             1
           C2     100     220        60      7200             1
           C3      40     290        35      8750             2
           C4      50     290        30      7200             2
           C5      50     250        60     12000             2
           C6      90     190        50      5000             1
           C7     160     250        60      5400             3
           ST     325     325                                 1
          CW       25      40                                 2
         Exchanger cost ($) = 8,000 + 500A0.75 (A is in m2 )
HENS                S EQUENTIAL F RAMEWORK              M IN U NITS SUB - PROBLEM   S UMMARY


E XAMPLE 2
L OOPING TO THE SOLUTION




       HRAT fixed at 20.35°C (Qh,min = 11539.25 kW & Qc,min = 9164.25 kW)
       Umin = 14 units


                      Soln. No        U      EMAT (C)   HLD          TAC ($)
                          1           14       2.5       A          1,545,375
                          2           15       2.5       A          1,532,148
                          3           15       2.5       B          1,536,900
                          4           15        5        A          1,529,968
                          5           15        5        B          1,533,261
                          6           16       2.5       A          1,547,353
HENS             S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


E XAMPLE 2
B EST SOLUTION
HENS           S EQUENTIAL F RAMEWORK    M IN U NITS SUB - PROBLEM   S UMMARY


E XAMPLE 2
C OMPARISON




        The solution given here with a TAC of $1,529,968, about
        the same cost as the solution presented in the original
        paper by Björk and Nordman (2005) (TAC $1,530,063)
        When only one match was allowed between a pair of
        streams the TAC reported by Björk & Nordman (2005) was
        $1,568,745
              The Sequential Framework allows only 1 match between a
              pair of streams
        Unable to compare the solutions apart from cost as the
        paper did not present the networks in their work
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS               S EQUENTIAL F RAMEWORK          M IN U NITS SUB - PROBLEM      S UMMARY



C HALLENGES

       C OMBINATORIAL E XPLOSION
            Reason: Binary Variables in MILP models
            Physical and engineering insights will mitigate, not remove, the problem
            MILP models are the bottlenecks that limit problem size due to
            computational time

       L OCAL OPTIMA
            Reason: Non-convexities in the NLP model
            Convex estimators developed for MINLP models are computationally
            intensive
            Time to solve the basic NLP is not a problem


       Sequence of MILP and NLP problems considerably easier to solve than
       MINLP formulations
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS                 S EQUENTIAL F RAMEWORK          M IN U NITS SUB - PROBLEM      S UMMARY


M INIMUM NUMBER OF UNITS SUB - PROBLEM
D EFINITION



       Given:
            a set H of hot process streams to be cooled,
            a set C of cold process streams to be heated,
            start and target temperatures, heat capacities and flow rates of the hot
            and cold process streams,
            a set of utilities,
            temperatures or temperature ranges and minimum requirement of the
            utilities and
            Exchanger Minimum Approach Temperature (EMAT) set to zero,
       calculate the minimum number of matches between hot process streams and
       utilities and cold process streams and utilities such that the heating and
       cooling requirements for each stream are met.
HENS                          S EQUENTIAL F RAMEWORK                               M IN U NITS SUB - PROBLEM   S UMMARY


 M INIMUM NUMBER OF UNITS SUB - PROBLEM
 F ORMULATION



                                  X X
                        min z =                  yij         (P1)
                                  i∈H j∈C



                                          s.t.

                                      H
                    X
Ri,k − Ri,k −1 +           Qijk = Qik                  ∀ i ∈ Hk , k ∈ TI
                    j∈Ck

                                      C
                    X
                           Qijk = Qjk                  ∀ j ∈ Ck , k ∈ TI
                    i∈Hk
         X
                 Qijk − Uij yij ≤ 0                    ∀ i ∈ H, j ∈ C
         k ∈TI

                           Rik ≥ 0                     ∀ i ∈ Hk , k ∈ TI
                  Ri0 = RiK = 0                        ∀i ∈H
                           Qijk ≥ 0                    ∀ i ∈ Hk , j ∈ Ck , k ∈ TI
                            yij = {0, 1}               ∀ i ∈ H, j ∈ C
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS           S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


M INIMUM NUMBER OF UNITS SUB - PROBLEM
C HALLENGES




       Combinatorial explosion
       Problem is proved to be
       N P-hard in the strong sense
HENS           S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY


M INIMUM NUMBER OF UNITS SUB - PROBLEM
C HALLENGES




       Combinatorial explosion
       Problem is proved to be
       N P-hard in the strong sense
HENS               S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



A LLEVIATING COMBINATORIAL EXPLOSION




       The three major ways to improve the model solution time are:
         1   Pre-processing to reduce model size using insight and
             heuristics
         2   Model modification/reformulation
         3   Improving efficiency of the B&B method
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS             S EQUENTIAL F RAMEWORK    M IN U NITS SUB - PROBLEM    S UMMARY



S HARPENING LP RELAXATION BY DECREASING BIG M




       The gap, i.e. the difference between the LP relaxation and the
       actual binary solution, is dependent on the value of Uij , the
       upper limit on the amount of heat transfer between streams i
       and j.
       A smaller value of Uij corresponds to a smaller gap and thus
       reduced computing times.
HENS                    S EQUENTIAL F RAMEWORK                                      M IN U NITS SUB - PROBLEM   S UMMARY



S HARPENING LP RELAXATION BY DECREASING BIG M

       O RIGINAL U
                                                                8                          9
                                                                <X                         =
                                                                          H            C
                                                                                X
                                                    Uij = min           Qik ,        Qjk
                                                                :                          ;
                                                                    k           k




       M ODIFIED U
                             8                                                                       9
                             <X                           h   “              ” “                ” i=
                                       H          C               H        C      H     C
                                             X
                 Uij = min           Qik ,       Qjk , max min mCpi , mCpj    · Tsi − Tsj − EMAT , 0
                             :                                                                       ;
                                 k           k
HENS                  S EQUENTIAL F RAMEWORK               M IN U NITS SUB - PROBLEM           S UMMARY


S HARPENING LP RELAXATION BY DECREASING BIG M
L OCAL U




       L OCAL U: Define maximum amount of heat exchanged between streams on a
       tempertaure interval basis

                                                  80       1     9
                                                  < X            =
                                                       HA     C
                                     Uijk   = min  @  Qi k , Qjk
                                                         ¯
                                                  :              ;
                                                    ¯
                                                    k ≤k


       Logical constraint utilizing the local U


                             Qijk − Uijk yij ≤ 0   ∀ i ∈ Hk , j ∈ Ck , k ∈ TI

       This constraint will reduce the feasible region as compared to the earlier one - thus
       leading to a tighter formulation.
HENS          S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



I NTEGER C UTS



       The gap between the LP relaxation based lower bound and
       the optimal integer solution. This gap can be reduced by
       employing integer cuts to the model.
       A potential drawback of adding such cuts is the increase in
       model size and hence computation time.
       2 types of integer cuts applied to the model
         1   Compulsory matches
         2   Minimum number of matches per stream
HENS                   S EQUENTIAL F RAMEWORK            M IN U NITS SUB - PROBLEM               S UMMARY


M ODEL MODIFICATION
T EST PROBLEM 22TP1




       ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 22TP1 WITH IP SOLUTION 23

         U definition       No integer      Compulsory   Minimum matches              Both cuts
                              cuts          matches        per stream
            Eq 1             12.21             -                -                        -
         Global Eq 2          15.17             16.27         16.82                   18.62
        Local Eqs 3,4         15.78             16.56         17.63                   18.62
HENS                   S EQUENTIAL F RAMEWORK            M IN U NITS SUB - PROBLEM               S UMMARY


M ODEL MODIFICATION
T EST PROBLEM 21TP1




       ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 21TP1 WITH IP SOLUTION 22

         U definition       No integer      Compulsory   Minimum matches              Both cuts
                              cuts          matches        per stream
            Eq 1             11.93             -                -                        -
         Global Eq 2          14.30             15.14         14.49                   15.21
        Local Eqs 3,4         14.87             15.39         14.95                   15.40
HENS                   S EQUENTIAL F RAMEWORK            M IN U NITS SUB - PROBLEM               S UMMARY


M ODEL MODIFICATION
T EST PROBLEM 21TP2




       ROOT NODE LP RELAXATION VALUES AND TOTAL SOLUTION TIMES FOR TEST PROBLEM
       21TP2 WITH IP SOLUTION 22

         U definition       No integer      Compulsory   Minimum matches              Both cuts
                             cuts           matches        per stream
            Eq 1               16              -                -                        -
                             20 s              -                -                        -
         Global Eq 2          17.67             18.80         18.13                   18.83
                               19 s              27 s          23 s                    21 s
        Local Eqs 3,4         18.80             18.80         18.81                   18.83
                               36 s              46 s          45 s                    42 s
HENS                S EQUENTIAL F RAMEWORK          M IN U NITS SUB - PROBLEM      S UMMARY


M ODEL MODIFICATION
D ISCUSSION




            Modified U and Local U definitions give tighter lower bounds than
            original U definition
            Compulsory matches integer cuts always improve the lower bound
            Minimum number of matches integer cuts have varying results
            21TP1 and 22TP1 still do not solve in less than 12 hours


       Gap is not the only measure of the complexity of an integer problem. Two of
       the main issues in this particular integer problem are the number of feasible
       solutions and the number of multiple optima.
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS                 S EQUENTIAL F RAMEWORK                     M IN U NITS SUB - PROBLEM   S UMMARY



S ET PARTITIONING FORMULATION

       SSis be defined to represent all feasible sets of matches, s ∈ Si , between a hot stream
       i and all cold streams:

                  SSis = { j | j ∈ C if stream j is in set of matches s for stream i}

                                      s    1      2   3    4     5     6     7
                                     C1    1      0   0    1     1     0     1
                                     C2    0      1   0    1     0     1     1
                                    CW     0      0   1    0     1     1     1
                         (
                             1,   if set of matches s is chosen for hot process stream i
                 λis =
                             0,   otherwise


       S ET PARTITIONING CONSTRAINT
                                           X
                                                 λis = 1       ∀i∈H
                                          s∈Si
HENS         S EQUENTIAL F RAMEWORK     M IN U NITS SUB - PROBLEM   S UMMARY



S ET PARTITIONING FORMULATION



       Maximum number of set of matches for a hot stream =
       2nc − 1
       Potentially large number of binary variables will be
       introduced
       Use thermodynamics and physical insight to reduced the
       number of binary variables
       The set partitioning constraint is expected to have more
       efficient branching characteristics
HENS                  S EQUENTIAL F RAMEWORK              M IN U NITS SUB - PROBLEM              S UMMARY



S ET PARTITIONING FORMULATION

       The constraints for defining the set of feasible matches are:
          1   At least one cold process stream or utility in the set of matches must have a
              supply temperature lower than or equal to the hot process stream’s target
              temperature and satisfy the hot process stream’s duty in this temperature range.
          2   The total heat demand for the set of cold process streams or utilities below the
              hot process stream’s supply temperature must be greater than or equal to the
              hot process stream’s total heat duty.
          3   The total number of cold process streams and utilities in a set of matches should
              not exceed a user-specified maximum value.
          4   For cases with streams having large duties, the number of streams in a set of
              matches for a utility must be larger than one.


       Constraints 1 and 2 are thermodynamically based while constraints 3 and 4 are
       heuristics based on insights gained from testing various problems.

       Constraint 3 is user specified to allow the user to impart their knowledge about the
       problem on hand to reduce the problem size.
HENS         S EQUENTIAL F RAMEWORK                                  M IN U NITS SUB - PROBLEM            S UMMARY



S ET PARTITIONING FORMULATION
                                                X X
                                    min z =                cis λis                                       (P2)
                                                i∈H s∈Si


                                   s.t.
                                                    H
                              X
       Ri,k − Ri,k −1 +               Qijk = Qik                            ∀ i ∈ Hk , k ∈ TI
                             j∈Ck

                                                    C
                              X
                                      Qijk = Qjk                            ∀ j ∈ Ck , k ∈ TI
                             i∈Hk
          X                X
                  Qijk −           Uij λis ≤ 0                              ∀ i ∈ H, j ∈ C
          k ∈TI            s∈Pij
                              X
                                          λis = 1                           ∀i ∈H
                              s∈Si
                       X X
                                          λis ≥ 1                           ∀j ∈C
                       i∈H s∈Pij
                       X X
                                          λis ≤ max value                   ∀j ∈C
                       i∈H s∈Pij

                                          cis = card (SSis )                ∀ i ∈ H, s ∈ Si
                                          Rik ≥ 0                           ∀ i ∈ Hk , k ∈ TI
                            Ri0 = RiK = 0                                   ∀i ∈H
                                  Qijk ≥ 0                                  ∀ i ∈ Hk , j ∈ Ck , k ∈ TI
                                          λis = {0, 1}                      ∀ i ∈ H, s ∈ Si
HENS       S EQUENTIAL F RAMEWORK                             M IN U NITS SUB - PROBLEM          S UMMARY



S ET PARTITIONING FORMULATION
                                           XX
                                min z =                 yij                                     (P3)
                                              i     j

                             s.t.
                                                H
                            X
        Ri,k − Ri,k −1 +            Qijk = Qik                     ∀ i ∈ Hk , k ∈ TI
                            j∈Ck

                                                C
                            X
                                    Qijk = Qjk                     ∀ j ∈ Ck , k ∈ TI
                            i∈Hk
                 X
                         Qijk − Uij yij ≤ 0                        ∀ i ∈ H, j ∈ C
                 k ∈TI
                             X
                                    λis = 1                        ∀i ∈H
                             s∈Si
                     X X
                                    λis ≥ 1                        ∀j ∈C
                     i∈H s∈Pij
                     X X
                                    λis ≤ max value                ∀j ∈C
                     i∈H s∈Pij
                             X
                                    λis = yij                      ∀ i ∈ H, j ∈ C
                            s∈Pij

                                    Rik ≥ 0                        ∀ i ∈ Hk , k ∈ TI
                          Ri0 = RiK = 0                            ∀i ∈H
                                    Qijk ≥ 0                       ∀ i ∈ Hk , j ∈ Ck , k ∈ TI
HENS                     S EQUENTIAL F RAMEWORK             M IN U NITS SUB - PROBLEM      S UMMARY


M ODEL REFORMULATION
T EST PROBLEM 22TP1




       ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 22TP1 WITH IP SOLUTION 23

           Model             Binary          U model      Additional       LP relaxation
                            variables                     constraints          value
             P1               143          Global Eq-2     Eqs 5,6            18.62
                                          Local Eqs 3,4    Eqs 5,6            18.62
            P3                2500         Global Eq-2     Eqs 5,6              18.62
          with λis                        Local Eqs 3,4    Eqs 5,6              18.62
             P4               2625         Global Eq-2     Eqs 5,6              18.67
           with µjt                       Local Eqs 3,4    Eqs 5,6              18.67
             P5               4999         Global Eq-2      None                18.67
        with λis & µjt                    Local Eqs 3,4                         18.67
HENS                     S EQUENTIAL F RAMEWORK             M IN U NITS SUB - PROBLEM      S UMMARY


M ODEL REFORMULATION
T EST PROBLEM 21TP1




       ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 21TP1 WITH IP SOLUTION 22

           Model             Binary          U model      Additional       LP relaxation
                            variables                     constraints          value
             P1               131          Global Eq-2     Eqs 5,6            15.21
                                          Local Eqs 3,4    Eqs 5,6            15.40
            P3                4645         Global Eq-2     Eqs 5,6              15.74
          with λis                        Local Eqs 3,4    Eqs 5,6              15.82
             P4               5435         Global Eq-2     Eqs 5,6              16.33
           with µjt                       Local Eqs 3,4    Eqs 5,6              16.45
             P5               9879         Global Eq-2      None                16.86
        with λis & µjt                    Local Eqs 3,4                         16.93
HENS           S EQUENTIAL F RAMEWORK     M IN U NITS SUB - PROBLEM   S UMMARY


M ODEL REFORMULATION
D ISCUSSION




         Reformulated models may results in strengthening the LP
         relaxation
              Contain more information regarding thermodynamics of the
              matches than the basic model
         Reformulated models for 21TP1 and 22TP2 cannot be
         solved within 12 hours
         These models are larger and with more binary variables
         counteracting any potential benefit
         Reduction in gap not significant enough
HENS              S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



O UTLINE

       1   HENS
             Background
             HENS in the 21st century
       2   S EQUENTIAL F RAMEWORK
             Introduction
             Examples
             Challenges
       3   M IN U NITS SUB - PROBLEM
             Background
             Challenges
             Model modification
             Model reformulation
             Further work
       4   S UMMARY
HENS         S EQUENTIAL F RAMEWORK    M IN U NITS SUB - PROBLEM    S UMMARY



F URTHER WORK



       Optimum value is reached early in the solution process
       and most of the effort is expended in proving optimality.
       Develop heuristics to stop the search after an appropriate
       solution time.
       Identifying subnetworks by relaxing stream temperatures
       and flow rates it may be possible to get a good initial bound
       on the minimum number of units using U = N + L − S.
       This value could be used by CPLEX as the initial lower
       bound thus tightening the gap.
HENS         S EQUENTIAL F RAMEWORK   M IN U NITS SUB - PROBLEM   S UMMARY



F URTHER WORK

       All NP-complete problems have a phase separation i.e.
       both hard and easy regions, with a sharp boundary
       between them. On crossing that frontier, the problem
       undergoes a phase transition, analogous to the boiling or
       freezing of water. Identifying the phase transition of the
       minimum units problem would enable the user to decide on
       using deterministic methods for solving it for other
       non-deterministic methods depending on which phase the
       problem happens to be in.
HENS         S EQUENTIAL F RAMEWORK          M IN U NITS SUB - PROBLEM     S UMMARY



S UMMARY
       Sequential Framework has many advantages
           Automates the design process
           Allows significant User interaction
       Progress
           EMAT identified as an optimizing ‘area variable´
           Improved HLDs from Stream match generator subproblem
           Significantly better and automated starting values for NLP
           subproblem
           Global optimum for NLP ensured using a modified GBD
           scheme
       Limiting elements
           Stream match generator Transportation Model for
           promising HLDs
                  Significant improvements required to fight combinatorial
                  explosion
           MILP Transhipment model for minimum number of units
                  Similar combinatorial problems as the Transportation model
           Reliability of NLP solutions is no longer limiting

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Sequential Framework for HENS @ IITM

  • 1. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY T HE S EQUENTIAL F RAMEWORK FOR H EAT E XCHANGER N ETWORK S YNTHESIS Rahul Anantharaman rahul.anantharaman@ntnu.no Department of Energy & Process Engineering Norwegian University of Science and Technology IIT Madras Chennai, 18.12.2009
  • 2. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 3. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 4. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS For a given set of hot and cold process streams as well as external utilities, design a heat exchanger network that minimizes Total Annualized Cost (TAC). TAC = Capital Cost + Energy Cost Sequential Framework Engine
  • 5. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS S OLUTION METHODS Evolutionary methods such as Pinch Design Method Sequential synthesis methods Simultaneous synthesis methods Stochastic optimization methods
  • 6. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 7. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 8. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 9. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 10. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 11. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 12. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 13. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY H EAT E XCHANGER N ETWORK S YNTHESIS T IMELINE
  • 14. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 15. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW 224 references published from 2000-2008 216 journal papers 48 jounals 43 countries 4 conference proceedings 10 Ph.D. theses 4 texts
  • 16. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW 45 40 35 30 25 20 15 10 5 0 2000 2001 2002 2003 2004 2005 2006 2007 2008
  • 17. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW Computers & Chem Eng 43 77 Applied Thermal Eng Industrial & Eng Chem  Research Chem Eng Research & Design 39 Latin American Appl Research Heat Transfer Eng 7 7 Chem Eng Science 7 23 7 7 10 Chem Eng and Processing Ch E dP i 7
  • 18. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW 50 45 40 35 30 25 20 15 10 5 0
  • 19. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW
  • 20. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY HENS IN THE 21 ST CENTURY R EVIEW HENS still an active area of research interest Over 25% of references devoted to case studies Pinch Analysis based evolutionary methods dominate Sustained interest in simultaneous MINLP methods Yee and Grossmann (1990) superstructure Pressure drop and detailed HX design considerations Small test problems Number of references related to genetic programming and other meta-heuristic methods increasing in frequency Though there has been significant developments in HENS using mathematical programming methods, synthesis of large scale HENS problems without simplifications and heuristics have been lacking. This is an area that requires more research for mathematical programming based approaches to be used in the industry
  • 21. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 22. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M OTIVATION FOR THE S EQUENTIAL F RAMEWORK Pinch based methods for network design Improper trade-off handling Time consuming Several topological traps MINLP methods for network design Severe numerical problems Difficult user interaction Fail to solve large scale problems Stochastic optimization methods for network design Non-rigorous algorithms Quality of solution depends on time spent on search
  • 23. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M OTIVATION FOR THE S EQUENTIAL F RAMEWORK HENS TECHNIQUES DECOMPOSE THE MAIN PROBLEM Pinch Design Method is sequential and evolutionary Simultaneous MINLP methods let math considerations define the decomposition The Sequential Framework decomposes the problem into subproblems based on knowledge of the HENS problem Engineer acts as optimizer at the top level Quantitative and qualitative considerations included
  • 24. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY U LTIMATE G OAL Solve Industrial Size Problems Defined to involve 30 or more streams Include Industrial Realism Multiple and ``Complex´´Utilities Constraints in Heat Utilization (Forbidden matches) Heat exchanger models beyond pure countercurrent Avoid Heuristics and Simplifications No global or fixed ∆ Tmin No Pinch Decomposition Develop a Semi-Automatic Design Tool EXCEL/VBA (preprocessing and front end) MATLAB (mathematical processing) GAMS (core optimization engine) Allow significant user interaction and control Identify near optimal and practical networks
  • 25. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UR ENGINE 3 way trade-off Compromise between Pinch Design and MINLP methods
  • 26. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 27. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 1 Stream Tin Tout mCp ∆H h K K kW/K kW kW/m2 K H1 626 586 9.802 392.08 1.25 H2 620 519 2.931 296.03 0.05 H3 528 353 6.161 1078.18 3.20 C1 497 613 7.179 832.76 0.65 C2 389 576 0.641 119.87 0.25 C3 326 386 7.627 457.62 0.33 C4 313 566 1.69 427.57 3.20 ST 650 650 - - 3.50 CW 293 308 - - 3.50 Exchanger cost ($) = 8,600 + 670A0.83 (A is in m2 )
  • 28. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 1 L OOPING TO THE SOLUTION HRAT fixed at 20K (Qh,min = 244.1 kW & Qc,min = 172.6) Umin = 8 units Soln. No U EMAT (K) HLD TAC ($) 1 8 2.5 A 199,914 2 8 5 A 199,914 3 8 7.5 - No Soln 4 9 2.5 A 147,861 5 9 2.5 B 151,477 6 9 5 A 147,867 7 9 5 B 151,508 8 9 7.5 A 149,025 9 9 7.5 B 149,224 10 10 2.5 A 164,381 11 10 5 A 167,111 12 10 7.5 A 164,764
  • 29. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 1 B EST SOLUTION
  • 30. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 1 C OMPARISON No. of units Area (m2 ) Cost ($) Colberg and Morari (1990) 22 173.6 Colberg and Morari (1990) 12 188.9 177,385 Yee and Grossmann (1990) 9 217.8 150,998 Sequential Framework 9 189.7 147, 861
  • 31. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 2 Stream Tin Tout mCp ∆H h (°C) (°C) (kW/°C) (kW) (kW/m2 °C) H1 180 75 30 3150 2 H2 280 120 60 9600 1 H3 180 75 30 3150 2 H4 140 40 30 3000 1 H5 220 120 50 5000 1 H6 180 55 35 4375 2 H7 200 60 30 4200 0.4 H8 120 40 100 8000 0.5 C1 40 230 20 3800 1 C2 100 220 60 7200 1 C3 40 290 35 8750 2 C4 50 290 30 7200 2 C5 50 250 60 12000 2 C6 90 190 50 5000 1 C7 160 250 60 5400 3 ST 325 325 1 CW 25 40 2 Exchanger cost ($) = 8,000 + 500A0.75 (A is in m2 )
  • 32. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 2 L OOPING TO THE SOLUTION HRAT fixed at 20.35°C (Qh,min = 11539.25 kW & Qc,min = 9164.25 kW) Umin = 14 units Soln. No U EMAT (C) HLD TAC ($) 1 14 2.5 A 1,545,375 2 15 2.5 A 1,532,148 3 15 2.5 B 1,536,900 4 15 5 A 1,529,968 5 15 5 B 1,533,261 6 16 2.5 A 1,547,353
  • 33. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 2 B EST SOLUTION
  • 34. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY E XAMPLE 2 C OMPARISON The solution given here with a TAC of $1,529,968, about the same cost as the solution presented in the original paper by Björk and Nordman (2005) (TAC $1,530,063) When only one match was allowed between a pair of streams the TAC reported by Björk & Nordman (2005) was $1,568,745 The Sequential Framework allows only 1 match between a pair of streams Unable to compare the solutions apart from cost as the paper did not present the networks in their work
  • 35. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 36. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY C HALLENGES C OMBINATORIAL E XPLOSION Reason: Binary Variables in MILP models Physical and engineering insights will mitigate, not remove, the problem MILP models are the bottlenecks that limit problem size due to computational time L OCAL OPTIMA Reason: Non-convexities in the NLP model Convex estimators developed for MINLP models are computationally intensive Time to solve the basic NLP is not a problem Sequence of MILP and NLP problems considerably easier to solve than MINLP formulations
  • 37. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 38. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M INIMUM NUMBER OF UNITS SUB - PROBLEM D EFINITION Given: a set H of hot process streams to be cooled, a set C of cold process streams to be heated, start and target temperatures, heat capacities and flow rates of the hot and cold process streams, a set of utilities, temperatures or temperature ranges and minimum requirement of the utilities and Exchanger Minimum Approach Temperature (EMAT) set to zero, calculate the minimum number of matches between hot process streams and utilities and cold process streams and utilities such that the heating and cooling requirements for each stream are met.
  • 39. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M INIMUM NUMBER OF UNITS SUB - PROBLEM F ORMULATION X X min z = yij (P1) i∈H j∈C s.t. H X Ri,k − Ri,k −1 + Qijk = Qik ∀ i ∈ Hk , k ∈ TI j∈Ck C X Qijk = Qjk ∀ j ∈ Ck , k ∈ TI i∈Hk X Qijk − Uij yij ≤ 0 ∀ i ∈ H, j ∈ C k ∈TI Rik ≥ 0 ∀ i ∈ Hk , k ∈ TI Ri0 = RiK = 0 ∀i ∈H Qijk ≥ 0 ∀ i ∈ Hk , j ∈ Ck , k ∈ TI yij = {0, 1} ∀ i ∈ H, j ∈ C
  • 40. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 41. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M INIMUM NUMBER OF UNITS SUB - PROBLEM C HALLENGES Combinatorial explosion Problem is proved to be N P-hard in the strong sense
  • 42. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M INIMUM NUMBER OF UNITS SUB - PROBLEM C HALLENGES Combinatorial explosion Problem is proved to be N P-hard in the strong sense
  • 43. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY A LLEVIATING COMBINATORIAL EXPLOSION The three major ways to improve the model solution time are: 1 Pre-processing to reduce model size using insight and heuristics 2 Model modification/reformulation 3 Improving efficiency of the B&B method
  • 44. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 45. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S HARPENING LP RELAXATION BY DECREASING BIG M The gap, i.e. the difference between the LP relaxation and the actual binary solution, is dependent on the value of Uij , the upper limit on the amount of heat transfer between streams i and j. A smaller value of Uij corresponds to a smaller gap and thus reduced computing times.
  • 46. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S HARPENING LP RELAXATION BY DECREASING BIG M O RIGINAL U 8 9 <X = H C X Uij = min Qik , Qjk : ; k k M ODIFIED U 8 9 <X h “ ” “ ” i= H C H C H C X Uij = min Qik , Qjk , max min mCpi , mCpj · Tsi − Tsj − EMAT , 0 : ; k k
  • 47. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S HARPENING LP RELAXATION BY DECREASING BIG M L OCAL U L OCAL U: Define maximum amount of heat exchanged between streams on a tempertaure interval basis 80 1 9 < X = HA C Uijk = min @ Qi k , Qjk ¯ : ; ¯ k ≤k Logical constraint utilizing the local U Qijk − Uijk yij ≤ 0 ∀ i ∈ Hk , j ∈ Ck , k ∈ TI This constraint will reduce the feasible region as compared to the earlier one - thus leading to a tighter formulation.
  • 48. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY I NTEGER C UTS The gap between the LP relaxation based lower bound and the optimal integer solution. This gap can be reduced by employing integer cuts to the model. A potential drawback of adding such cuts is the increase in model size and hence computation time. 2 types of integer cuts applied to the model 1 Compulsory matches 2 Minimum number of matches per stream
  • 49. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL MODIFICATION T EST PROBLEM 22TP1 ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 22TP1 WITH IP SOLUTION 23 U definition No integer Compulsory Minimum matches Both cuts cuts matches per stream Eq 1 12.21 - - - Global Eq 2 15.17 16.27 16.82 18.62 Local Eqs 3,4 15.78 16.56 17.63 18.62
  • 50. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL MODIFICATION T EST PROBLEM 21TP1 ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 21TP1 WITH IP SOLUTION 22 U definition No integer Compulsory Minimum matches Both cuts cuts matches per stream Eq 1 11.93 - - - Global Eq 2 14.30 15.14 14.49 15.21 Local Eqs 3,4 14.87 15.39 14.95 15.40
  • 51. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL MODIFICATION T EST PROBLEM 21TP2 ROOT NODE LP RELAXATION VALUES AND TOTAL SOLUTION TIMES FOR TEST PROBLEM 21TP2 WITH IP SOLUTION 22 U definition No integer Compulsory Minimum matches Both cuts cuts matches per stream Eq 1 16 - - - 20 s - - - Global Eq 2 17.67 18.80 18.13 18.83 19 s 27 s 23 s 21 s Local Eqs 3,4 18.80 18.80 18.81 18.83 36 s 46 s 45 s 42 s
  • 52. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL MODIFICATION D ISCUSSION Modified U and Local U definitions give tighter lower bounds than original U definition Compulsory matches integer cuts always improve the lower bound Minimum number of matches integer cuts have varying results 21TP1 and 22TP1 still do not solve in less than 12 hours Gap is not the only measure of the complexity of an integer problem. Two of the main issues in this particular integer problem are the number of feasible solutions and the number of multiple optima.
  • 53. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 54. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S ET PARTITIONING FORMULATION SSis be defined to represent all feasible sets of matches, s ∈ Si , between a hot stream i and all cold streams: SSis = { j | j ∈ C if stream j is in set of matches s for stream i} s 1 2 3 4 5 6 7 C1 1 0 0 1 1 0 1 C2 0 1 0 1 0 1 1 CW 0 0 1 0 1 1 1 ( 1, if set of matches s is chosen for hot process stream i λis = 0, otherwise S ET PARTITIONING CONSTRAINT X λis = 1 ∀i∈H s∈Si
  • 55. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S ET PARTITIONING FORMULATION Maximum number of set of matches for a hot stream = 2nc − 1 Potentially large number of binary variables will be introduced Use thermodynamics and physical insight to reduced the number of binary variables The set partitioning constraint is expected to have more efficient branching characteristics
  • 56. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S ET PARTITIONING FORMULATION The constraints for defining the set of feasible matches are: 1 At least one cold process stream or utility in the set of matches must have a supply temperature lower than or equal to the hot process stream’s target temperature and satisfy the hot process stream’s duty in this temperature range. 2 The total heat demand for the set of cold process streams or utilities below the hot process stream’s supply temperature must be greater than or equal to the hot process stream’s total heat duty. 3 The total number of cold process streams and utilities in a set of matches should not exceed a user-specified maximum value. 4 For cases with streams having large duties, the number of streams in a set of matches for a utility must be larger than one. Constraints 1 and 2 are thermodynamically based while constraints 3 and 4 are heuristics based on insights gained from testing various problems. Constraint 3 is user specified to allow the user to impart their knowledge about the problem on hand to reduce the problem size.
  • 57. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S ET PARTITIONING FORMULATION X X min z = cis λis (P2) i∈H s∈Si s.t. H X Ri,k − Ri,k −1 + Qijk = Qik ∀ i ∈ Hk , k ∈ TI j∈Ck C X Qijk = Qjk ∀ j ∈ Ck , k ∈ TI i∈Hk X X Qijk − Uij λis ≤ 0 ∀ i ∈ H, j ∈ C k ∈TI s∈Pij X λis = 1 ∀i ∈H s∈Si X X λis ≥ 1 ∀j ∈C i∈H s∈Pij X X λis ≤ max value ∀j ∈C i∈H s∈Pij cis = card (SSis ) ∀ i ∈ H, s ∈ Si Rik ≥ 0 ∀ i ∈ Hk , k ∈ TI Ri0 = RiK = 0 ∀i ∈H Qijk ≥ 0 ∀ i ∈ Hk , j ∈ Ck , k ∈ TI λis = {0, 1} ∀ i ∈ H, s ∈ Si
  • 58. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S ET PARTITIONING FORMULATION XX min z = yij (P3) i j s.t. H X Ri,k − Ri,k −1 + Qijk = Qik ∀ i ∈ Hk , k ∈ TI j∈Ck C X Qijk = Qjk ∀ j ∈ Ck , k ∈ TI i∈Hk X Qijk − Uij yij ≤ 0 ∀ i ∈ H, j ∈ C k ∈TI X λis = 1 ∀i ∈H s∈Si X X λis ≥ 1 ∀j ∈C i∈H s∈Pij X X λis ≤ max value ∀j ∈C i∈H s∈Pij X λis = yij ∀ i ∈ H, j ∈ C s∈Pij Rik ≥ 0 ∀ i ∈ Hk , k ∈ TI Ri0 = RiK = 0 ∀i ∈H Qijk ≥ 0 ∀ i ∈ Hk , j ∈ Ck , k ∈ TI
  • 59. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL REFORMULATION T EST PROBLEM 22TP1 ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 22TP1 WITH IP SOLUTION 23 Model Binary U model Additional LP relaxation variables constraints value P1 143 Global Eq-2 Eqs 5,6 18.62 Local Eqs 3,4 Eqs 5,6 18.62 P3 2500 Global Eq-2 Eqs 5,6 18.62 with λis Local Eqs 3,4 Eqs 5,6 18.62 P4 2625 Global Eq-2 Eqs 5,6 18.67 with µjt Local Eqs 3,4 Eqs 5,6 18.67 P5 4999 Global Eq-2 None 18.67 with λis & µjt Local Eqs 3,4 18.67
  • 60. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL REFORMULATION T EST PROBLEM 21TP1 ROOT NODE LP RELAXATION VALUES FOR TEST PROBLEM 21TP1 WITH IP SOLUTION 22 Model Binary U model Additional LP relaxation variables constraints value P1 131 Global Eq-2 Eqs 5,6 15.21 Local Eqs 3,4 Eqs 5,6 15.40 P3 4645 Global Eq-2 Eqs 5,6 15.74 with λis Local Eqs 3,4 Eqs 5,6 15.82 P4 5435 Global Eq-2 Eqs 5,6 16.33 with µjt Local Eqs 3,4 Eqs 5,6 16.45 P5 9879 Global Eq-2 None 16.86 with λis & µjt Local Eqs 3,4 16.93
  • 61. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY M ODEL REFORMULATION D ISCUSSION Reformulated models may results in strengthening the LP relaxation Contain more information regarding thermodynamics of the matches than the basic model Reformulated models for 21TP1 and 22TP2 cannot be solved within 12 hours These models are larger and with more binary variables counteracting any potential benefit Reduction in gap not significant enough
  • 62. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY O UTLINE 1 HENS Background HENS in the 21st century 2 S EQUENTIAL F RAMEWORK Introduction Examples Challenges 3 M IN U NITS SUB - PROBLEM Background Challenges Model modification Model reformulation Further work 4 S UMMARY
  • 63. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY F URTHER WORK Optimum value is reached early in the solution process and most of the effort is expended in proving optimality. Develop heuristics to stop the search after an appropriate solution time. Identifying subnetworks by relaxing stream temperatures and flow rates it may be possible to get a good initial bound on the minimum number of units using U = N + L − S. This value could be used by CPLEX as the initial lower bound thus tightening the gap.
  • 64. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY F URTHER WORK All NP-complete problems have a phase separation i.e. both hard and easy regions, with a sharp boundary between them. On crossing that frontier, the problem undergoes a phase transition, analogous to the boiling or freezing of water. Identifying the phase transition of the minimum units problem would enable the user to decide on using deterministic methods for solving it for other non-deterministic methods depending on which phase the problem happens to be in.
  • 65. HENS S EQUENTIAL F RAMEWORK M IN U NITS SUB - PROBLEM S UMMARY S UMMARY Sequential Framework has many advantages Automates the design process Allows significant User interaction Progress EMAT identified as an optimizing ‘area variable´ Improved HLDs from Stream match generator subproblem Significantly better and automated starting values for NLP subproblem Global optimum for NLP ensured using a modified GBD scheme Limiting elements Stream match generator Transportation Model for promising HLDs Significant improvements required to fight combinatorial explosion MILP Transhipment model for minimum number of units Similar combinatorial problems as the Transportation model Reliability of NLP solutions is no longer limiting