Introduction

It is always preferred to use
uniform high grade ore for production of metallic alloy. But in real life
situation, it is not possible to get a homogeneous ore. Chrome ore is one of
the precious natural resources having limited availability. The chemical
composition of the ore varies horizontally and vertically across the ore body
in mines. The ore quality is not uniform in all the faces of the mines. The
production requirement being uniform grade of ore, blending of different grades
of ore is required at the mines head before using it in the furnaces for
smelting. The production requirement is to have a uniform grade metal
production to meet the customer specification with minimum or zero deviation.

 

Further, Chrome ore being one of the precious
natural resources need to be preserved and used properly; the low and or medium
grade ore should be used by suitably blending with the high grade ore instead
of misusing.

 

Present
Practice

The important chemical parameters of the
chrome ore are Cr2O3, Cr/Fe, FeO,
Al2O3, Phos, SiO2, CaO, MgO and
Cost ore. Each chemical parameter is having some lower and upper limit for
being suitable to be used for Ferro Chrome production. The requirement being uniform, higher grade of ore having uniform
chemical composition, this requirement forced to mine only high-grade ore, thereby
creating a huge stock of other medium and lower grade of ore at mines pit. This
not only affects the mines operation but also leads to the wastage of natural
resources.

 

Considering the practical problems
of reduction in cost of alloy production, preservation and use of high-grade
ore and enhanced value addition by using the medium and low-grade ores, a
mathematical model of non-linear nature has been developed. The model was
formulated for blending of chrome ore from different lots with three different
objective functions. The objective functions are

 

1.      Maximizing the quantity of ore
supply

2.      Minimizing the cost of Ore.

3.      Maximizing the satisfaction index

 

 

 

 

Maximize
the quantity of supply:

The problem formulated
with an objective of
maximizing the quantity of supply from the available lots of ore subject to
fulfilling the chemical parameters. The particulars of the model are as under.

 

Objective Function   Maximize
Z=  (Maximize the quantity of supply)

Qi = Quantity of ore from Lot (i)

 

The Constraints being

          (Quantity considered for blending
should be less than the availability)       

Qsi= Availability (stock) of ore in lot (i)

 

  and                                                          

Weighted average Cr2O3
% in ore should be less than the upper limit and more than the lower limit.

   and     

Weighted average FeO % in ore
should be less than the upper limit and more than the lower limit.

 

   and                                                                   

Weighted average SiO2 % in ore should be less than the upper
limit and more than the lower limit.

 

 and                                                                     

Weighted average Al2O3 % in ore should be less than
the upper limit and more than the lower limit.

 

    and                                                                    

Weighted average CaO % in ore should be less than the upper limit and
more than the lower limit.

 and                                                                              

Weighted average MgO % in ore should be less than the upper limit and
more than the lower limit.

    and                                                                                  

Weighted average Phos % in ore should be less than the upper limit and
more than the lower limit.

 and                                                                 

Weighted average Cr/Fe in ore should be less than the upper limit and more
than the lower limit.

 

Typical stocks of different lots
of chromium ore available with the upper and lower limit specification for
acceptance are as shown in table 1

 

Table 1 Different grade of chromium ore and its chemical composition

 

The problem formulation in case of maximizing the quantity available
considering the above available data can be as under.

Z= Maximize (Q1+Q2+Q3+……………. +Q10)

Where Q1, Q2,
Q3………Q10 are the quantity of material selected for
blending from lot-1, lot-2, Lot-3 and lot-10 respectively

Constraints for quantity availability (Quantity considered for blending should be
less than or equal with the availability)

Q1<= 1000 MT, Q2<=2000 MT, Q3<=3000 MT, Q4<=1500 MT, Q5<=7000 MT      Q6<=2000 MT , Q7<=3000 MT, Q8<=1500 MT, Q9<=7000 MT, Q10<=10000 MT   Constraints for Cr2O3 % in the Ore (0.427Q1+0.48Q2+0.46Q3+0.43Q4+0.46Q5+0.44Q6+0.48Q7+0.46Q8+0.43Q9+ 0.50Q10)  / (Q1+Q2+Q3…….+Q10) >=0.46

(0.427Q1+0.48Q2+0.46Q3+0.43Q4+0.46Q5+0.44Q6+0.48Q7+0.46Q8+0.43Q9
+0.50Q10)  / (Q1+Q2+Q3…….+Q10)
<=0.47 Weighted average Cr2O3 % in ore should be less than the upper limit and more than the lower limit. Constraints for FeO % in the Ore (0.188Q1+0.1523Q2+0.1802Q3+0.1717Q4+0.1681Q5+0.231Q6+0.1681Q7+0.231Q8+0.1142Q9+0.1257Q10)  / (Q1+Q2+Q3+……+Q10) >=0.16

(0.188Q1+0.1523Q2+0.1802Q3+0.1717Q4+0.1681Q5+0.231Q6+0.1681Q7+0.231Q8+0.1142Q9+0.1257Q10)
/(Q1+Q2+Q3+…….+Q10) <=0.18   Weighted average FeO % in ore should be less than the upper limit and more than the lower limit. Constraints for SiO2 % in the Ore (0.032Q1+0.032Q2+0.05Q3+0.032Q4+0.0368Q5+0.0368Q6+0.0368Q7+0.0368Q8+0.1515Q9+0.06Q10) / (Q1+Q2+Q3……+Q10) >=0.035

(0.032Q1+0.032Q2+0.05Q3+0.032Q4+0.0368Q5+0.0368Q6+0.0368Q7+0.0368Q8+0.1515Q9+0.06Q10)
/ (Q1+Q2+Q3……+Q10) <=0.04   Weighted average SiO2 % in ore should be less than the upper limit and more than the lower limit. Constraints for Al2O3% in the Ore   (0.1336Q1+0.1346Q2+0.139Q3+0.1223Q4+0.1704Q5+0.1487Q6+0.1704Q7+0.1487Q8+0.0892Q9+0.0892Q10) / (Q1+Q2+Q3…….+Q10) >=0.09

 

(0.1336Q1+0.1346Q2+0.139Q3+0.1223Q4+0.1704Q5+0.1487Q6+0.1704Q7+0.1487Q8+0.0892Q9+0.0892Q10)
/ (Q1+Q2+Q3…….+Q10) <=0.135   Weighted average Al2O3 % in ore should be less than the upper limit and more than the lower limit. Constraints for CaO% in the Ore   (0.0225Q1+0.0375Q2+0.015Q3+0.225Q4+0.0098Q5+0.0098Q6+0.0.0098Q7+0.0098Q8+0.0102Q9+0.0102Q10) / (Q1+Q2+Q3…….+Q10) >=0.01

 

(0.0225Q1+0.0375Q2+0.015Q3+0.225Q4+0.0098Q5+0.0098Q6+0.0.0098Q7+0.0098Q8+0.0102Q9+0.0102Q10)
/ (Q1+Q2+Q3…….+Q10) <=0.025   Weighted average CaO % in ore should be less than the upper limit and more than the lower limit. Constraints for MgO % in the Ore   (0.0802Q1+0.0725Q2+0.08Q3+0.08Q4+0.0958Q5+0.0958Q6+0.0.0958Q7+0.0958Q8+0.1881Q9+0.1881Q10) / (Q1+Q2+Q3…….+Q10) >=0.07

 

(0.0802Q1+0.0725Q2+0.08Q3+0.08Q4+0.0958Q5+0.0958Q6+0.0.0958Q7+0.0958Q8+0.1881Q9+0.1881Q10)
/ (Q1+Q2+Q3…….+Q10) <=0.14   Weighted average MgO % in ore should be less than the upper limit and more than the lower limit. Constraints for Phos % in the Ore   (0.00013Q1+0.00013Q2+0.0001Q3+0.00013Q4+0.0001Q5+0.0001Q6+0.0.0001Q7+0.0001Q8+0.00007Q9+0.00007Q10) / (Q1+Q2+Q3…….+Q10) >=0.00001

 

(0.00013Q1+0.00013Q2+0.0001Q3+0.00013Q4+0.0001Q5+0.0001Q6+0.0.0001Q7+0.0001Q8+0.00007Q9+0.00007Q10)
/ (Q1+Q2+Q3…….+Q10) <=0.00013   Weighted average Phos % in ore should be less than the upper limit and more than the lower limit.   Constraints for Cr/Fe in the Ore (2Q1+2.78Q2+2.25Q3+2.21Q4+2.41Q5+1.68Q6+2.52Q7+1.76Q8+3.32Q9+3.51Q10)/ (Q1+Q2+Q3…….+Q10) >=2.2

 

(2Q1+2.78Q2+2.25Q3+2.21Q4+2.41Q5+1.68Q6+2.52Q7+1.76Q8+3.32Q9+3.51Q10)/
(Q1+Q2+Q3…….+Q10) <=2.8 Weighted average Cr/Fe in ore should be less than the upper limit and more than the lower limit.   The problem was solved by using excel solver optimizer and the output of the optimization has been summarized in table 2 Table 2 Output of the optimization with objective function of maximizing the availability The table shows that the optimum quantities (Qi) of ore selected from different lots in order to maximize the availability are             Lot-1               Q1        1000 MT                     Lot-2               Q2          2000 MT             Lot-3               Q3          0       MT                     Lot-4               Q4          1500 MT             Lot-5               Q5          2874 MT                     Lot-6               Q6          200 MT             Lot-7               Q7          26 MT                         Lot-8               Q8          1500 MT             Lot-9               Q9          0 MT                           Lot-10             Q10         2824 MT   Average Chemical composition of ore against the upper and lower limits under.                                     Cr2O3   FeO     SiO2     Al2O3   CaO     MgO    Ph        Cr/Fe Upper limit                  47.00   18        4.00     13.50   2.50     14.00   0.013   2.80 Lower Limit                46.0     16.0     3.5       9.0       1.0       7.0       0.000   2.2 Actual with above       46.26   17.50   4.00     13.50   1.62     10.85   0.010   2.46 ore mix The average cost of the above blended ore is Rs 277 per metric tonne.   Minimize the cost of Ore The problem formulated with an objective of minimizing the cost of ore supply from the available lots of ore subject to fulfilling the minimum quantity and chemical parameters requirement. The particulars of the model are as under.   Objective Function = Minimize the cost of Ore Objective Function Minimize Z= Qi = Quantity of ore from Lot ( i) Ci is the cost of Mining of the ore from Lot ( i) The Constraints being          (Quantity considered for blending should be less than the availability)        Qsi= Availability (stock) of ore in lot (i)   Constraints for all quality parameters being same as in the same of maximizing the quantity of ore supply   The problem formulation in case of minimizing the cost of ore supply from the available lots of ore subject to fulfilling the minimum quantity and chemical parameters requirement. The problem formulation using the data available in Table-10 can be as shown below.   Z=Min (250Q1+275Q2+275Q3+250Q4+250Q5+250Q6+275Q7+275Q8+250Q9+350Q10) Where Q1, Q2, Q3……Q10 are the quantity of material from lot-1, lot-2, lot-3 and lot-10 respectively.   The constraints of the model being same as that of the maximizing the quantity problem. The output of the optimization has been summarized in table 3   Table 3 Output of the optimization with objective function of minimizing the average cost of ore.   The table shows that the optimum quantities (Qi) of ore selected from different lots in order to minimize the cost are               Lot-1               Q1        1000    MT                  Lot-2               Q2          2000    MT             Lot-3               Q3          0          MT                  Lot-4               Q4          1500    MT             Lot-5               Q5          2541    MT                 Lot-6               Q6          1893    MT             Lot-7               Q7          0          MT                  Lot-8               Q8          0          MT Lot-9               Q9          97        MT                  Lot-10             Q10         1969    MT   Average Chemical composition of ore against the upper and lower limits under.                                     Cr2O3   FeO     SiO2     Al2O3   CaO     MgO    Ph        Cr/Fe Upper limit                  47.00   18        4.00     13.50   2.50     14.00   0.013   2.80 Lower Limit                46.0     16.0     3.5       9.0       1.0       7.0       0.000   2.2 Actual with above       46.00   17.03   4.00     13.50   1.78     10.53   0.01     2.49 ore mix The average cost of the above blended ore is Rs 272 per metric tonne. Where the present cost of ore is about Rs.290   Maximizing the User satisfaction Index:   The user requirement is a homogeneity ore with higher metallic content and less of impurities available at low cost. There are few elements of ore, higher the value of which results in higher level of user satisfaction. Parameters Cr2O3, FeO, Cr/Fe, CaO, SiO2 and MgO fall in this category. The other category where lower the value, higher is the satisfaction are Al2O3, Phos and Cost of Ore.   Objective Function – Maximizing the User satisfaction Index. Objective Function Maximize Z= Dj – Satisfaction index of the parameter (j) Wj – Weighatage of the Parameter (j) The Constraints being            (Quantity considered for blending should be less than the availability)        Qsi= Availability (stock) of ore in lot (i) Qi = Quantity of ore from Lot ( i)   Constraints for all quality parameters being same as in the same of maximizing the quantity of ore supply   To decide on the user satisfaction index the parameters of the ore be divided in to two broad categories as   Category-1:     Higher the value of the parameter higher is the level of user satisfaction. Parameters falling in category- 1 are Cr2O3, FeO, Cr/Fe, CaO, SiO2 and MgO Category-2:     Higher the value of the parameter Lower is the level of user satisfaction. Parameters falling in category- 2 are Al2O3, Phos and Cost of Ore   Each parameter is having lower and upper limit. Let Rj be the range of the parameter, i.e., (difference of the maximum and minimum limit). The satisfaction index for each parameter has been divided in a five-point scale as follows.   Satisfaction index       Satisfaction index Category-1                  Category-2 Actual Parameter        >= Lower
Limit                                

<= Lower Limit + Rj/5             (Di)=1                                    (Di) =5   Actual Parameter        >= Lower
Limit + Rj/5                                   

<= Lower Limit + 2Rj/5          (Di)=2                         (Di) =4   Actual Parameter        >= Lower
Limit + 2Rj/5                                 

<= Lower Limit + 3Rj/5          Di) =3                          Di) =3   Actual Parameter        >= Lower
Limit + 3Rj/5

<= Lower Limit + 4Rj/5           (Di) =4                       (Di)=2   Actual Parameter        >= Lower
Limit + 4Rj/5          (Di) =5                        (Di)=1

 

Weightages of
the Parameters

 

Based on the feed back from the furnace operators the weightages for the
different parameters (Wj) for
comfortable operation has been decided as under.

 

                                    Parameters                 Weightage

                                               Cr2O3               0.30

                                               Cr/Fe               0.20

                                               FeO                 0.10

                                               Cost                 0.10

                                               Al2O3               0.075

                                               Ph                    0.075

                                               SiO2                 0.05

                                               CaO                 0.05

                                               MgO                0.05

The problem formulation in case of maximizing the satisfaction index can
be formulated as under

 

Z=Maximize (30% Satisfaction index
Cr2O3 +20% Satisfaction index Cr/Fe +10% Satisfaction index FeO+10% Satisfaction index Cost+7.5% Satisfaction index Al2O3 +7.5% Satisfaction index Ph+5% Satisfaction index SiO2+5%
Satisfaction index CaO)

 

The constraints are same as that
of the maximizing the quantity problem. The problem was solved and the output
of the model is presented in table 4

 

Table 4. Output of the model of
maximizing user satisfaction index.

(out of 5)

 

The table shows that the optimum quantity (Qi) of ore selected from different lots in order to
maximizing user satisfaction index.

            Lot-1               Q1          1000 MT                     Lot-2               Q2          1670 MT

            Lot-3               Q3          699   MT                     Lot-4               Q4          1500 MT

            Lot-5               Q5          921   MT                     Lot-6               Q6          639   MT

            Lot-7               Q7          1156 MT                     Lot-8               Q8          1433 MT

Lot-9               Q9          0       MT                     Lot-10             Q10         1982 MT

 

Average Chemical composition of ore against the upper and lower limits
under.

                                    Cr2O3  FeO     SiO2     Al2O3   CaO    MgO    Ph        Cr/Fe

Upper limit                  47.00   18        4.00     13.50   2.50
    14.00   0.013   2.80

Lower Limit                46.0     16.0     3.5       9.0       1.0       7.0       0.000   2.2

Actual with above       46.41   17.30   4.00     13.43   1.73     10.43   0.011   2.47

ore mix

 

The average cost of the above
blended ore is Rs 279 per metric tonne.

 

Summary and Observation

The output of the models having
different objective functions are summarized and presented in table 4.12

 

Table 5 Output comparison from different objective functions

 

There is a increase in cost of the
ore in the output of maximizing quantity and maximizing satisfaction index from
the out of the objective function of minimizing cost by Rs 5 and Rs 7 per MT
respectively, the impact of which is Rs 62477 and Rs 75237. The impact can be
derived as under

 

Impact of increasing in average
cost of ore from the objective function of minimizing cost to the objective
function of maximizing quantity = (272-277) Rs/MT x 13723MT = Rs (-)
62477/Month. Similarly the impact of increasing in average cost of ore from the
objective function of minimizing cost to the objective function of maximizing
satisfaction index =(272-279) Rs/MT x 11000MT 
= Rs (-) 75237/ Month

 

There is an increase in Cr2O3
% of the ore in the output of maximizing quantity and maximizing satisfaction
index from the out of the objective function of minimizing cost by 0.26% and
0.41% respectively, the impact of which is Rs 8.08 lacs and Rs 12.80 lacs .

 

The impact can be derived as under

                                                            Cr2O3
% in the Ore                 % Increase
from the                                                                                                                Output of Minimizing Cost

            Minimizing
Cost                                 46%                                         ——

            Maximizing
Quantity                          46.26%                                    0.26%

            Maximizing
Satisfaction index           46.41%                                    0.41%

 

The impact of additional chromium
input will lead to higher volume of production which can be derived as
(Additional chromium input x Chromium recovery) / (% chromium in the finished
goods)

Where Additional chromium input =
Difference in Cr2O3 % x Conversion factor (0.685) x
Quantity of ore in MT

Chromium Recovery   = 83% & percentage (%) of Chromium in the
finished good= 60%

 

In case of the output of
maximizing quantity (0.0026 x 0.685 x 13723 x 0.83) / 0.6 = 26.94 MT and in
case of the output of maximizing satisfaction index (0.0041 x 0.685 x 11000 x
0.83) / 0.6  = 42.66 MT respectively for
maximizing quantity or maximizing satisfaction index respectively.

 

By valuing the additional
production at the current market price @ Rs 30000 per MT the impact will be Rs
8.08 lacs and Rs 12.80 lacs. The net benefit by deducting the increasing in
cost of ore from the output minimizing cost will be about Rs 7.45 lacs and Rs
12.04 lacs.

 

Conclusion

It is suggested to use the model
with the objective function of maximizing the user satisfaction index.
The model suggests a
blend of available ore matching to all the elemental requirement of production.
The low grade ore which was otherwise idle will get utilized in due course of
time there by improving in productive use of natural resources without
affecting the quality of the output. In addition, there is a significant
improvement in the profitability of the organization.

Post Author: admin