# Structural Optimization of Process Flowsheets

TomSym implementation of GAMS Example (PROCSEL,SEQ=116)

The goal is the profitable production of chemical C, which can be produced from chemical B where B may be the raw material that can be purchased from the external market or an intermediate that is produced from raw material A. There are two alternative paths of producing B from A. A mixed-integer nonlinear formulation is presented to solve the optimal production and capacity expansion problem.

Kocis, G R, and Grossmann, I E, Relaxation Strategy for the Structural Optimization of Process Flow Sheets. Independent Engineering Chemical Research 26, 9 (1987), 1869-1880.

Morari, M, and Grossmann, I E, Eds, Chemical Engineering Optimization Models with GAMS. Computer Aids for Chemical Engineering Corporation, 1991.

Process flowsheet:

A2 +-----+ B2 BP +----->| 2 |----->+ | A | +-----+ | | B1 +-----+ C1 ---->| +----+------->| 1 |--------> | +-----+ | +-----+ +----->| 3 |----->+ A3 +-----+ B3

a2: consumption of chemical a in process 2 a3: consumption of chemical a in process 3 b2: production capacity of chemical b in process 2 b3: production capacity of chemical b in process 3 bp: amount of chemical b purchased in external market b1: consumption of chemical b in process 1 c1: production capacity of chemical c in process 1

toms a2 a3 b2 b3 bp b1 c1 posvbls = [a2 a3 b2 b3 bp b1 c1]'; % Variables are positive cbnd1 = {posvbls >= 0}; % Binary variables, existence of process 1 2 and 3 toms integer y1 y2 y3 eq1 = {c1 == 0.9*b1 % Input-output for process 1 exp(b2) - 1 == a2 % Input-output for process 2 exp(b3/1.2) - 1 == a3 % Input-output for process 3 b1 == b2 + b3 + bp % Mass balance for chemical b c1 <= 2*y1 % Logical constraint for process 1 b2 <= 4*y2 % Logical constraint for process 2 b3 <= 5*y3}; % Logical constraint for process 3 % total profit in million dollars per year % sales revenue, fixed investment cost, operating cost and purchases pr = 11*c1; pr = pr - 3.5*y1 - y2 - 1.5*y3; pr = pr - b2 - 1.2*b3; pr = pr - 1.8*(a2+a3) - 7*bp; % c1 is less than 1 cbnd2 = {c1 <= 1}; solution = ezsolve(-pr,{cbnd1,cbnd2,eq1});

Problem type appears to be: minlp ===== * * * =================================================================== * * * TOMLAB - Tomlab Optimization Inc. Development license 999001. Valid to 2010-02-05 ===================================================================================== Problem: --- 1: f_k -1.923098737768011500 f(x_0) -0.000000000000000000 Solver: minlpBB. EXIT=0. INFORM=0. Dense Branch and Bound MINLP Optimal integer solution found FuncEv 31 GradEv 36 HessEv 29 ConstrEv 36 ConJacEv 34 ConHessEv 29 Iter 9 CPU time: 0.046875 sec. Elapsed time: 0.094000 sec.