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 Construction of Problem Objective Function:- Our Objective Function is to Maximize net profit. From Exhibit 3, we can derive the net profit as follows: S.No Alternatives Total Revenue Cost of Milk + Processing & Packing Cost Net Profit 1 FA Milk+Butter 7956 7453 503 2 FA Milk+Ghee 7952 7504 448 3 SMP+Butter 8782 8238 544 4 SMP+Ghee 8609 8254 355 5 WMP+Butter 9863 8953 910 6 WMP+Ghee 9698 8953 745 7 BabyFood+Butter 12296 11431 865 8 BabyFood+Ghee 12080 11447 633 9 Cheese+Butter 12650 10817 1833 10 Cheese+Ghee 12546 10936 1610 11 StdzdMilk+Butter 8764 8005 759 12 StdzdMilk+Ghee 8645 8013 632 Thus Objective function can be given as: Max z=(503x 1 + 448x 2 + 544x 3 + 355x 4 + 910x 5 + 745x 6 + 865x 7 + 633x 8 + 1833x 9 + 1610x 10 + 759x 11 + 632x 12 ) Constraints:- From the above two tables we can see that total product produced should be less than the production capacity. Thus, .062x 1 + .992 x 3 + .606x 5 + .560x 7 + .681x 9 <= 3650 .048x 2 + .756x 4 + .448x 6 + .529x 8 + .426x 10 +.519x 12 <= 912.5 S.No Alternatives Qty. of main product Qty of by-product Product Production capacity per year 1FA Milk+Butter 9874(L) = 1 unit 62= .062 units Butter 3650 2 FA Milk+Ghee 9874(L) = 1 unit 48 = .048 units Ghee 912.5 3SMP+Butter 813(K) 992 = .992 units Milk Powder(SMP+WMP) 2190000 4SMP+Ghee 813(K) 756 = .756 units Baby Food 1956.4 5 WMP+Butter 1179(K) 606 = .606 units Cheese 667.95 6WMP+Ghee 1179(K) 448 = .448 units FA Milk(min. to be supplied) 2772.4 7 BabyFood+Butter 1307(K) = 1 unit 696 = .696 units Raw Milk 6043 8 BabyFood+Ghee 1307(K) = 1 unit 529 = .529 units 9 Cheese+Butter 1094(K) = 1 unit 560 = .560 units 10 Cheese+Ghee 1094(K) = 1 unit 426 = .426 units 11 StdzdMilk+Bu tter 8675(L) = 1 unit 681 = .681 units 12 StdzdMilk+Ghee 8675(L) = 1 unit 519 =.519 units

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Construction of Problem

Objective Function:-

Our Objective Function is to Maximize net profit. From Exhibit 3, we can derive the net profit as follows:

S.No AlternativesTotalRevenue

Cost of Milk +Processing &Packing Cost Net Profit

1 FA Milk+Butter 7956 7453 503

2 FA Milk+Ghee 7952 7504 448

3 SMP+Butter 8782 8238 544

4 SMP+Ghee 8609 8254 355

5 WMP+Butter 9863 8953 910

6 WMP+Ghee 9698 8953 745

7 BabyFood+Butter 12296 11431 865

8 BabyFood+Ghee 12080 11447 633

9 Cheese+Butter 12650 10817 1833

10 Cheese+Ghee 12546 10936 1610

11 StdzdMilk+Butter 8764 8005 759

12 StdzdMilk+Ghee 8645 8013 632

Thus Objective function can be given as:

Max z=(503x1 + 448x2 + 544x3 + 355x4 + 910x5 + 745x6 + 865x7+ 633x8 + 1833x9 + 1610x10 + 759x11 + 632x12 )

Constraints:- 

From the above two tables we can see that total product produced should be less than the production

capacity. Thus,

.062x1 + .992 x3 + .606x5 + .560x7 + .681x9<= 3650

.048x2 + .756x4 + .448x6 + .529x8 + .426x10 +.519x12<= 912.5

S.No Alternatives Qty. of main product Qty of by-product ProductProductioncapacity per year 

1 FA Milk+Butter 9874(L) = 1 unit 62= .062 units Butter 3650

2 FA Milk+Ghee 9874(L) = 1 unit 48 = .048 units Ghee 912.5

3 SMP+Butter 813(K) 992 = .992 units Milk Powder(SMP+WMP) 2190000

4 SMP+Ghee 813(K) 756 = .756 units Baby Food 1956.4

5 WMP+Butter 1179(K) 606 = .606 units Cheese 667.95

6 WMP+Ghee 1179(K) 448 = .448 units

FA Milk(min. to be

supplied) 2772.4

7 BabyFood+Butter 1307(K) = 1 unit 696 = .696 units Raw Milk 6043

8 BabyFood+Ghee 1307(K) = 1 unit 529 = .529 units

9 Cheese+Butter 1094(K) = 1 unit 560 = .560 units

10 Cheese+Ghee 1094(K) = 1 unit 426 = .426 units

11 StdzdMilk+Butter 8675(L) = 1 unit 681 = .681 units

12 StdzdMilk+Ghee 8675(L) = 1 unit 519 =.519 units

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813x3+ 813x4 + 1179x5 + 1179x6<= 2190000

x7 + x8<= 1956.4

x9 + x10<= 667.95

x1 + x2>= 2772.40

x1 + x2 + x3 + x4 + x5 + x6 + x7 + x8 + x9 + x10 + x11 + x12<= 6043

Solutions using linear programming software:

a)  Annual

The formulation is shown in the next page. The solutions we get are:

ConstraintValueused inequality

max/Minvalue

Valueleft

Butter 1816.158 <= 3650 1833.84

ghee 284.5467 <= 912.5 627.953

milk powder 2190000 <= 2190000 0

baby food 745.1436 <= 1956.4 1211.26

cheese 667.95 <= 667.95 0

FA milk 2772.4 >= 2772.4 0

standardized

milk 0 <= 0

RM 6043 <= 6,043.00 0

So cheese should be produced as it utilizes whole of the constraint.

The final values for objective function are

Profit 4804797

Labor cost 950866.1

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cost 870000

Depriciation 810000

net margin 2173931

b)  May-Aug

The only difference lies in the procurement ofraw milk units per day, which is now 11.9 units/day

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Contraint

Value

used inequality

max/Min

value

Value

left

Butter 184.4356 <= 1230 1045.56

ghee 140.7329 <= 307.5 166.767

milk powder 358827.7 <= 738000 379172

baby food 0 <= 659.28 659.28

cheese 225.09 <= 225.09 0

FA milk 934.2608 >= 934.2608 0

standardized

milk 0 <= 0

RM 1463.7 <= 1,463.70 0

Again we see in spite of raw material constrain (equation 8 concerning RM) being changed, cheese

production is a f avoured function.

The final values for objective function are

Gross Profit 1057902

Labor cost 230313.2

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cost 290000

Depriciation 270000

net margin 267588.3

c)  Sep-April

Using similar logic as for May-Aug, the only changes are in milk supplied (19 units per day)

Contraint

Value

used inequality

max/Min

value

Value

left

Butter 1507.999 <= 2430 922.001

ghee 278.0332 <= 607.5 329.467

milk powder 1458000 <= 1458000 0

baby food 1089.934 <= 1302.48 212.546

cheese 444.69 <= 444.69 0

FA milk 1845.735 >= 1845.735 0

standardized

milk 0 <= 0

RM 4617 <= 4,617.00 0

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The final values for objective function are

Gross Profit 3610976

Labor cost 726485

Advertisementcost 580000

Depriciation 540000

net margin 1764492

  Howmuchmoney can the corporation invest in butter churning and powder drying?

Capacity of milk powder we can expand by 1884211 kg per year(5162 kg/day). Weshouldnt invest more

than 1884211*INR(0.13)= INR 244947 for expansion.

  Conclusion: Clearly, what made the production of cheese sound economically expensiveis high overhead

costs. The costs of processing and packaging are high for a perishable like cheese. As a result, the

calculations (without considering overhead costs) show cheese as a feasible product.

In the constraints table, constraints which are non-binding (non zero slack) can be considered as a

expensive weight function in the profit maximization equation. The sensitivity analysis can help us

realize how much constraints can be increased/decreased for the current optimal set. This can help us to

find out how much FA milk we can commit to supply to the state.