\$title Haverly's Pooling Problem Example (HAVERLY,SEQ=214) \$onText Haverly's pooling problem example. This is a non-convex problem. Setting initial levels for the nonlinear variables is a good approach to find the global optimum. Haverly, C A, Studies of the Behavior of Recursion for the Pooling Problem. ACM SIGMAP Bull 25 (1978), 29-32. Adhya, N, Tawaralani, M, and Sahinidis, N, A Lagrangian Approach to the Pooling Problem. Independent Engineering Chemical Research 38 (1999), 1956-1972. ----- crudeA ------/--- pool --| / |--- finalX ----- crudeB ----/ | |--- finalY ----- crudeC ------------------| Keywords: nonlinear programming, chemical engineering, pooling problem \$offText Set s 'supplies (crudes)' / crudeA, crudeB, crudeC / f 'final products' / finalX, finalY / i 'intermediate sources for final products' / Pool, CrudeC / poolin(s) 'crudes going into pool tank' / crudeA, crudeB /; Table data_S(s,*) 'supply data summary' price sulfur crudeA 6 3 crudeB 16 1 crudeC 10 2; Table data_f(f,*) 'final product data' price sulfur demand finalX 9 2.5 100 finalY 15 1.5 200; Parameter sulfur_content(s) 'supply quality in (percent)' req_sulfur(f) 'required max sulfur content (percentage)' demand(f) 'final product demand'; sulfur_content(s) = data_S(s,'sulfur'); req_sulfur(f) = data_F(f,'sulfur'); demand(f) = data_F(f,'demand'); Equation costdef 'cost equation' incomedef 'income equation' blend(f) 'blending of final products' poolbal 'pool tank balance' crudeCbal 'balance for crudeC' poolqualbal 'pool quality balance' blendqualbal 'quality balance for blending' profitdef 'profit equation'; Positive Variable crude(s) 'amount of crudes being used' stream(i,f) 'streams' q 'pool quality'; Variable profit 'total profit' cost 'total costs' income 'total income' final(f) 'amount of final products sold'; profitdef.. profit =e= income - cost; costdef.. cost =e= sum(s, data_S(s,'price')*crude(s)); incomedef.. income =e= sum(f, data_F(f,'price')*final(f)); blend(f).. final(f) =e= sum(i, stream(i,f)); poolbal.. sum(poolin, crude(poolin)) =e= sum(f, stream('pool',f)); crudeCbal.. crude('crudeC') =e= sum(f, stream('crudeC',f)); poolqualbal.. q*sum(f, stream('pool', f)) =e= sum(poolin, sulfur_content(poolin)*crude(poolin)); blendqualbal(f).. q*stream('pool',f) + sulfur_content('CrudeC')*stream('CrudeC',f) =l= req_sulfur(f)*sum(i,stream(i,f)); final.up(f) = demand(f); Model m / all /; * Because of the product terms, some local solver may get * trapped at 0*0, we therefore set an initial value for q. q.l = 1; solve m maximizing profit using nlp;