$title Aluminum Alloy Smelter Sample Problem (IBM1,SEQ=79)

$onText
This simple alloy smelter blending problem is used as an introductory
example in several mpsx manuals.

IBM, MPSX/370 Primer. Tech. rep., IBM, 1979.

Keywords: linear programming, blending problem, scenario analysis, chemical engineering
$offText

Set
   s     'scrap metals for blending' / bin-1*bin-5, aluminum, silicon /
   sl(s) 'locally available blends'  / bin-1*bin-5 /
   e     'chemical elements'         / iron, copper, manganese, magnesium
                                       aluminum, silicon /

Table bspec(e,*) 'blending specs (lb for 2000 lb ingot)'
                  maximum  minimum
   iron                60
   copper             100
   manganese           40
   magnesium           30
   aluminum           inf     1500
   silicon            300      250;

Table prop(e,s)  'chemical properties     (proportions)'
                  bin-1  bin-2  bin-3  bin-4  bin-5  aluminum  silicon
   iron             .15    .04    .02    .04    .02       .01      .03
   copper           .03    .05    .08    .02    .06       .01
   manganese        .02    .04    .01    .02    .02
   magnesium        .02    .03                  .01
   aluminum         .70    .75    .80    .75    .80       .97
   silicon          .02    .06    .08    .12    .02       .01      .97;

Parameter dcheck(s) 'other elements in prop      (prop)';
dcheck(s) = 1 - sum(e, prop(e,s));
display dcheck;

Table sup(s,*) 'supply and cost data'
                inventory  min-use   cost
*                    (lb)     (lb) ($/lb)
   bin-1              200             .03
   bin-2              750             .08
   bin-3              800      400    .17
   bin-4              700      100    .12
   bin-5             1500             .15
   aluminum           inf             .21
   silicon            inf             .38;

Variable
   x(s)    'blending components      (lb)'
   bc(e)   'elements in blend        (lb)'
   cost    'total material cost       ($)';

Equation
   yield   'final blend requirements (lb)'
   ebal(e) 'element balance          (lb)'
   cdef    'cost definition           ($)';

yield..   sum(s, x(s)) =e= 2000;

ebal(e).. bc(e) =e= sum(s, prop(e,s)*x(s));

cdef..    cost  =e= sum(s, sup(s,"cost")*x(s));

Model alloy 'alloy blending model' / all /;

x.lo(s)  = sup(s,"min-use");
x.up(s)  = sup(s,"inventory");
bc.lo(e) = bspec(e,"minimum");
bc.up(e) = bspec(e,"maximum");

Parameter report(s,*) 'blending results';

solve alloy minimizing cost using lp;
report(s,"run-1") = x.l(s);

option solPrint = off, limCol = 0, limRow = 0;

prop("iron","silicon")    = .02;
prop("silicon","silicon") = .98;
solve alloy minimizing cost using lp;
report(s,"run-2") = x.l(s);

prop("iron","silicon")    = .01;
prop("silicon","silicon") = .99;
solve alloy minimizing cost using lp;
report(s,"run-3") = x.l(s);

prop("iron","silicon")    = 0;
prop("silicon","silicon") = 1.0;
option solPrint   = on;
solve alloy minimizing cost using lp;
report(s,"run-4") = x.l(s);

display report;

Parameter costrep 'example cost report for 2000lb batch';
costrep(s          ,"cost    ") = sup(s,"cost");
costrep(s          ,"quantity") = x.l(s);
costrep(s          ,"c-cost  ") = costrep(s,"cost")*costrep(s,"quantity");
costrep("**total**","c-cost  ") = sum(s, costrep(s,"c-cost"));

display costrep;

* demonstrates GAMS' rounding capabilities according to ecl manual
Parameter
   xr(s)  'rounded solution value'
   xr3(s) 'numerical error rounding at third decimal'
   num    'rounded solution counter';

xr(s)  = round(x.l(s));
xr3(s) = round(x.l(s),3);
num    = 0;

loop(s$(xr(s) <> xr3(s) and num < 2),
   x.lo(s)$(xr(s) > xr3(s)) = xr(s);
   x.up(s)$(xr(s) < xr3(s)) = xr(s);
   num = num + 1;
);

solve alloy minimizing cost using lp;