2.6 Limits on the Availability of Components

So far we have considered only quality constraints in this problem. We have effectively been imagining that we have adequate quantities of each component whatever blend we choose to make. This is not true. We will usually have limited availabilities of each component. We shall now incorporate these constraints into our problem.

Suppose that we want to make 200 units of HFO. (The choice of units is completely arbitrary; for the sake of argument let it be MBBL.) The availability of components is

Availability (MBBL)

LGO 20.0

HGO 50.0

WAXD 40.0

ARES 200.0

VRES 100.0

We shall use the objective function from COST2FO so, if we had unlimited quantities of each component, the optimum solution would be

Use (MBBL)

LGO 43.4

ARES 95.6

VRES 60.8

with these properties

ACTUALSG = 0.9578

ACTUALVI  = 37.0

ACTUALSU = 3.7

The matrix is shown in Fig 6.1 and the solution in Fig 6.2.

Fig 6.1: LP Matrix - Minimise Cost of Blend with Availability Limits

{ Simple Oil Blending Model - Version 6 }

{ In this model we have restricted the availability of each component }

{ Note that, rather than representing PROPORTIONS of the blend, the }
{ component variables now represent actual QUANTITIES. The total    }
{ quantity of the blend required is 200 units. We have adjusted the }
{ constraints to account for this }


TITLE Simple_Oil_Blending_Model_V6 ;

MIN  OBJ =   25.0 LGO + 22.0 HGO + 20.0 WAXD + 15.0 ARES + 12.0 VRES ;

SUBJECT TO

SGRAV :       0.83 LGO  + 0.88 HGO  + 0.92 WAXD
            + 0.97 ARES + 1.03 VRES - 200.0 ACTUALSG
            = 0.0;

VBI :         15.0 LGO + 26.0 HGO + 30.0 WAXD
            + 40.0 ARES + 48.0 VRES - 200.0 ACTUALVI
            = 0.0 ;

SULPHUR:      1.0 LGO + 2.2 HGO + 2.8 WAXD
            + 4.1 ARES + 5.0 VRES - 200.0 ACTUALSU
            = 0.0 ;


MBALANCE:     1.0 LGO  + 1.0 HGO + 1.0 WAXD
            + 1.0 ARES + 1.0 VRES
            = 200.0 ;

BOUNDS

            ACTUALSG <= 0.98 ;
            ACTUALVI <= 37.0 ;
            ACTUALSU <= 3.7 ;
            LGO <= 20.0 ;
            HGO <= 50.0 ;
            WAXD <= 40.0 ;
            ARES <= 200.0 ;   
            VRES <= 100.0 ;
END

Fig 6.2: LP Solution - Minimised Cost of Blend with Availability Limits

     ----------------------------------------------------------------------
       Problem Title  :  Simple_Oil_Blending_Mode Date  :     Apr-29-1995
                                                  Time  :           16:13

       Input File     :  SIMBLN6.SIM                              
       Output File    :  SIMBLN6.LIS                              
     ----------------------------------------------------------------------


     ----------------------------------------------------------------------

       Optimal Solution  : 


        MIN  OBJ               =          3266.3158

     ----------------------------------------------------------------------




            Decision variables            Values           Reduced Cost
     ======================================================================

        1)   LGO'              =            20.0000            1.4211
        2)   HGO'              =             9.4737             .   
        3)   WAXD              =              .                0.2105
        4)   ARES              =           170.5263             .   
        5)   VRES              =              .                0.3158
        6)   ACTUALSG          =             0.9517             .   
        7)   ACTUALVI'         =            36.8368             .   
        8)   ACTUALSU'         =             3.7000          736.8421

     ----------------------------------------------------------------------



            Constraints       Type         Slack          Shadow prices
     ======================================================================

        1)   SGRAV            '='             .                 .   
        2)   VBI              '='             .                 .   
        3)   SULPHUR          '='             .               -3.6842
        4)   MBALANCE         '='             .               30.1053

     ----------------------------------------------------------------------

The solution is

Use (MBBL)

LGO 20.0

HGO 9.5

ARES 170.5

with these properties

ACTUALSG = 0.9517

ACTUALVI  = 36.8

ACTUALSU = 3.7

You may be wondering why the variables LGO, HGO, etc, which used to be proportions, are now taking values greater than 1.0. This is because I have changed their definition in the matrix (Fig. 6.1). Recall that these variables were defined in the earlier matrices to be proportions by the material balance row:

MBALANCE : LGO + HGO + WAXD + ARES + VRES = 1.0

Now, I have redefined the right hand side of this row to be 200.0:

MBALANCE : LGO + HGO + WAXD + ARES + VRES = 200.0

This means that the variables LGO, HGO, etc now represent the quantities of material used in producing 200 units of HFO.

The limits on the availability of each component are expressed by bounds:

LGO <= 20.0

HGO <= 50.0

WAXD <= 40.0

ARES <= 200.0

VRES <= 100.0

I have had to make one further change to the matrix. Consider a quality constraint row, e.g.

SGRAV : ACTUALSG = 0.83 LGO + 0.88 HGO + 0.92 WAXD + 0.97 ARES + 1.03 VRES

We want ACTUALSG to be the actual specific gravity of the HFO which we make. When the variables LGO, HGO, etc were proportions, the actual specific gravity was as expressed in the equation. But when the variables LGO, MGO, etc represent quantities used to blend 200 MBBL or HFO, the ACTUALSG as given by the equation represents the quantity of the quality "specific gravity" in 200 MBBL of HFO. In other works, if we were to have an actual specific gravity of 0.95, ACTUALSG would be 190.0 (strictly 190 MBBL as specific gravity is dimensionless). We need to put a coefficient of 200 in front of ACTUALSG to restore its definition. We therefore have the row

SGRAV : 0.83 LGO + 0.88 HGO + 0.92 WAXD + 0.97 ARES + 1.03 VRES

-200.0 ACTUALSG = 0.0

Similarly, the coefficients of ACTUALVI and ACTUALSU in the VBI and SULPHUR rows are - 200.0.

We shall look further at the meaning of the different parts of the solution file in another lecture, but now let us consider only the value of the objective function. In all the problems with costs the objective function has been, e.g.

25.0 LGO + 22.0 HGO + 20.0 WAXD + 15.0 ARES + 12.0 VRES

That is, it has been the sum of the cost of the material multiplied by the proportion or quantity used.

When the variables LGO, HGO, etc were proportions, the objective thus represented the unit cost of the HFO which we produced. This was:

Unit Cost of HFO Blend

COST1FO 15.0

COST2FO 16.26

COST3FO 16.0

In this last problem the variables LGO, HGO, etc have represented quantities, so the value of the objective is the cost of making 200 MBBL of HFO. This is 3266.3. The cost of each MBBL is therefore 16.33 which should be compared with the unit cost of 16.26 for the comparable problem COST2FO without the capacity constraints. As we add more constraints, the cost increases.

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