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Case study A blending problem, Ejercicios de Administración de Empresas

Asignatura: Operations Research, Profesor: Cecilio Mar Molinero, Carrera: Administració i Direcció d'Empreses - Anglès, Universidad: UAB

Tipo: Ejercicios

2013/2014

Subido el 18/01/2014

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A BLENDING PROBLEM
Blending was the first problem that was analysed using LP. The original problem,
formulated in the 1940s was concerned with optimal human diet. We know that we
need a minimum amount of calories, proteins, and various types of vitamins, and that
these are obtained from a variety of foods. The contribution of every type of food to the
diet is known, and so is the cost of the food. For example, we know that tomatoes do
not have many calories, that they have even less protein, but that they are rich in
vitamins. We also know that potatoes are rich in calories, and that they also contribute
vitamin C to the diet. There are various constraints that need to be satisfied. For
example, one can have an overdose of certain vitamins, and some ingredients need to be
balanced with others. The question is how we should feed ourselves if we want a good
diet at the cheapest possible cost. If you use LP to formulate and solve this problem
you will find that the optimal diet is surprisingly cheap but very boring. But optimal
diet formulation has found many industrial applications, from animal feed to recipes for
cheese. Here I have taken an example from the book “Linear Programming Models in
Business” by Derrick Smith (1973). The problem is a simplified version of a standard
industrial situation.
A company sells two types of fertiliser: Grade 1 and Grade 2.
Fertilisers need to contribute the soil with N (Nitrogen), Phosphorus (P), and Potassium
(K). Obviously, the company does not know where the fertiliser is going to be used;
otherwise it would make the mix appropriate to the characteristics of the soil. The
chemical components do not come as raw chemical elements. For example, N is a gas,
and P is toxic. The chemical elements come as part of chemical compounds:
Ammonium Nitrate (AN), Diammonium Phosphate (DAP), Single Superphosphate
(SS), Potash, and Mono-ammonium phosphate (MAP). The sacks of fertiliser also
include filler, which just makes bulk.
Grade 1 fertiliser contains 10% N, 10% P, and 10% K. 50 tons of grade 1 fertiliser need
to be made. Grade 2 fertiliser contains 22% N, 10% P, and 5% K. 100 tons of grade 2
fertiliser needs to be made. There is no problem finding AN, DAP, SS, Potash, or filler
in the market, but MAP is in short supply and the company has purchased 10 tons of
this product.
The table below gives the contribution (in percentage of weight) of each chemical
compound to each element of the fertiliser, as well as the unit cost per ton.
Raw material N% P% K% Cost per ton
AN 34 0 0 7
DAP 18 46 0 17
SS 0 20 0 8
Potash 0 0 60 6
Filler 0 0 0 2
MAP 13 52 0 19
You are required to formulate the problem so that the company can make its purchasing
and production plans.
RESOLUTION
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A BLENDING PROBLEM

Blending was the first problem that was analysed using LP. The original problem, formulated in the 1940s was concerned with optimal human diet. We know that we need a minimum amount of calories, proteins, and various types of vitamins, and that these are obtained from a variety of foods. The contribution of every type of food to the diet is known, and so is the cost of the food. For example, we know that tomatoes do not have many calories, that they have even less protein, but that they are rich in vitamins. We also know that potatoes are rich in calories, and that they also contribute vitamin C to the diet. There are various constraints that need to be satisfied. For example, one can have an overdose of certain vitamins, and some ingredients need to be balanced with others. The question is how we should feed ourselves if we want a good diet at the cheapest possible cost. If you use LP to formulate and solve this problem you will find that the optimal diet is surprisingly cheap but very boring. But optimal diet formulation has found many industrial applications, from animal feed to recipes for cheese. Here I have taken an example from the book “Linear Programming Models in Business” by Derrick Smith (1973). The problem is a simplified version of a standard industrial situation.

A company sells two types of fertiliser: Grade 1 and Grade 2.

Fertilisers need to contribute the soil with N (Nitrogen), Phosphorus (P), and Potassium (K). Obviously, the company does not know where the fertiliser is going to be used; otherwise it would make the mix appropriate to the characteristics of the soil. The chemical components do not come as raw chemical elements. For example, N is a gas, and P is toxic. The chemical elements come as part of chemical compounds: Ammonium Nitrate (AN), Diammonium Phosphate (DAP), Single Superphosphate (SS), Potash, and Mono-ammonium phosphate (MAP). The sacks of fertiliser also include filler, which just makes bulk.

Grade 1 fertiliser contains 10% N, 10% P, and 10% K. 50 tons of grade 1 fertiliser need to be made. Grade 2 fertiliser contains 22% N, 10% P, and 5% K. 100 tons of grade 2 fertiliser needs to be made. There is no problem finding AN, DAP, SS, Potash, or filler in the market, but MAP is in short supply and the company has purchased 10 tons of this product.

The table below gives the contribution (in percentage of weight) of each chemical compound to each element of the fertiliser, as well as the unit cost per ton.

Raw material N% P% K% Cost per ton

AN 34 0 0 7

DAP 18 46 0 17

SS 0 20 0 8

Potash 0 0 60 6

Filler 0 0 0 2

MAP 13 52 0 19

You are required to formulate the problem so that the company can make its purchasing and production plans.

RESOLUTION

Decision is: How much to buy of AN, DAP, SS, Pot, MAP and filler; and how much of what we buy we use for grade 1 and grade2.

Decision variables: X (^) 1AN X1DAP X1SS X1POT X1FILL X1MAP

X2AN X2DAP X2SS X2POT X2FILL X2MAP Constraints:

  • We need to use at most 10 tons of MAP: X1MAP+X (^) 2MAP≤ 10
  • Non-negativity constraints
  • (^) We need 50 tons of Grade 1: X (^) 1AN+X1DAP+X (^) 1SS+X1POT+X (^) 1FILL +X1MAP=
  • We need 100 tons of Grade 2: X (^) 2AN+X2DAP+X (^) 2SS+X2POT+X (^) 2FILL +X2MAP=
  • Nitrogen balance:
    • Grade 1: 0.34X1AN+0.18X (^) 1DAP+0.13X1MAP=50·0.
  • Grade 2: 0.34X2AN+0.18X (^) 2DAP+0.13X2MAP=100·0.
  • Phosphorous balance:
  • Grade 1: 0.46X1DAP+0.2X1SS+0.52X (^) 1MAP=0.1·
  • (^) Grade 2: 0.46X2DAP+0.2X2SS+0.52X (^) 2MAP=0.1·
  • Potassium balance:
    • Grade 1: 0.6X1POT=0.1·
    • Grade 2: 0.6X2POT=0.05·

Objective function: Min Cost= 7·X 1 AN+7·X 2 AN+17·X 1 DAP+17·X 2 DAP+8·X 1 SS+8·X 2 SS+6·X 1 POT+6·X 2 POT+ +2·X 1 FILL+2·X 2 FILL+19·X 1 MAP+19·X 2 MAP

LINGO resolution: