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Chapter 1

LINEAR PROGRAMMING

1.1. Using optimal limited resources

One practical optimization problem, which appears frequently in the managementactivity in every organization is the following: we have several limited resourcessuch as materials, labor, financial etc and using these resources we can performseveral activities. The optimization problem must comply an optimal (maximizeor minimize) criteria and consist in determination of the level of these activities,constrained by the available limited resources.

Let us denote by i, 1 i m, the resource type and with bi the level of availableresource of type i.Denote by j, 1 j n, the type of activity and by xj the level (unknown) at

which will be performed this activity. Finally, denote by aij the resource of type i,1 i m, required for the production of one unit of type j, 1 j n, (in general,activity of type j).

We suppose here that aij depends only of the resource type and process type anddoes not depend of the level at which these activity will be performed.

Using the introduced notations we can express the total quantity of resource ofi-th type which is used in the production phase:

ai1x1 + ai2x2 + . . . + ainxn.

Because is not allowed to use from i-th resource more than the available quantitybi, it result that the following constraints must be accomplished:

nj=1

aijxj bi, 1 i m. (1.1)

1

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2 LINEAR PROGRAMMING

Because xj represents the level at which is performed the j-th activity will result

that the following constraints must be accomplished:

xj 0, 1 j n. (1.2)

Inequalities 1.1 are called the constraints of the problem and 1.2 are called thenon negativity conditions of the problem.

The system of inequalities can have an infinite of solutions, a unique solution orno solution (incompatible system). In many cases when the practical problems arewell defined we are in the situation in which the system 1 .1 and 1.2 has an infinitenumber of solutions.

Thus is possible to organize the production process for the products of type j,

1 j n, in several ways which respects the constraints 1.1 of using the limitedresources.

Adoption of plan is made using some economic criteria such as: the revenue,the labor time, production costs etc. Let us assume that we can integrate all theseeconomic criteria in one criteria which allows us to adopt the decision. Moreover weshall assume that criteria is a linear one. An example of such criteria we obtain inthe following way. If we denote by cj the revenue per unit obtained by the activityof type j, 1 j n, then the total revenue is:

nj=1

cjxj. (1.3)

The problem which must be solved, is to find the solutions of the systems oflinear inequalities 1.1, 1.2 which gives the maximum value for the total revenue 1.3.In others words, from mathematical point of view we need to solve the problem:

supn

j=1

cjxj

nj=1

aijxj bi, 1 i m,

xj 0, 1 j n,

which is called linear programming problem (or linear program).The linear func-tion which must be maximized is called objective function, criteria function or effi-ciency function. The above problem can be solved with simplex algorithm or simplexdual algorithm. These algorithms are presented in every basic optimization book.

Some software products, such as MAPLE have specialized routines for solvinglinear optimizations problems.

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FORMS OF LINEAR PROGRAMMING PROBLEMS 3

1.2. Forms of linear programming problems

Standard form of a linear programming problem is:

min(max)cxAx = bx 0

Let us remark that a maximum problem can be transformed in a minimum prob-lem using the formulamax(f) = min(f).

Canonical form of a linear programming problem is:

min cxAx b

x 0

or

max cxAx bx 0

MAPLE code for transforming in canonical form a maximum problem is:

> standardize(set of constraints);

or

> convert(set of constraints,stdle);

A constraint of a linear programming problem is called concordant if it is an inequality for the minimum problem and an for the maximum problem.

The mixed form of a linear programming problem contains constraints (sometimes we use the term of restrictions) which are equations.

Because, using mathematical operations, every linear programming problem canbe transformed in canonical form, for the minimum optimization problem, we shallwork only with such of these problems, thus with problems in canonical form.

Definition 1.2.1. Let be the linear programming problem in standard form.Then, we define the set of programs by:

P = {x Rn|Ax = b, x 0}.

A minimum point of the objective function z = c

x over the set of programs Pis called optimal solution and the set of optimal solutions will be denoted with:

P = {x P| minxP

c

x = c

x}.

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Define the base solutionof the system Ax = b a solution Rn for which to nonzero

components corresponds linear independent columns.If B is a basis formed with the columns aj1, . . . , ajm of the matrix A then the

system of equations Ax = b can be written in the explicit form:

xB = B1b B1RxR

where R is the matrix obtained from A by elimination the columns j1, . . . , jm. If wedenote:

B1b =xB

, B1aj = yBj , 1 j n,

results:xB =x

BjR

yBj xj

equivalently

xBi =xB

i jR

yBijxj, i B.

where B= {j1, . . . , jm} si R = {1, . . . , n} B.

The base solution which corresponds to the basis B is xB =xB

si xR = 0. Itresults that these base solution is a program if we have:

B1b 0.

A basis B which meet the above inequality is called primal feasible basis. Inpractice these kind of basis are determined using the artificial basis method. In

many situations this basis is available directly, the feasible basis being the identity(unit) matrix.We have the following theorem called the fundamental theorem of the linear pro-

gramming.

Theorem 1.2.1. i) If the linear programming problem:

min(max)cxAx = bx 0

has an optimum, then the problem has a basis program.ii) If the above problem has a optimum program, then it has a optimum basic

program.

The fundamental algorithm for solving the linear programming programs is calledsimplex primal algorithmand it proposed by George Dantzigin 1951. This algorithmis described in every fundamental book of Operational Research and is implementedin every dedicated software package.

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SIMPLEX ALGORITHM (DANTZIG) 5

1.3. Simplex algorithm (Dantzig)

For solving linear programming problems we can use simplex algorithm developedby Dantzig (1951). This algorithm allows systematically searching, in the set of baseprograms of the linear programming problem in standard form, the transition forma base program to another program as least as good as the previous. The algorithmgives us criteria for situations in which the optimum is infinity or for the case inwhich the set of programs is the empty set.

STEP 0. Transform the optimization problem in standard form:

infcTxAx = b,x 0.

(If a problem has inequality constraints firs we transform this inequalities; sub-tracting the slack variables y, the maximum problem will be transformed in a min-imum problem etc.). A is a matrix with m rows and n columns for which we haverang(A) = m < n. Denote by z the objective function, thus z = cTx.

STEP 1. We find a primal feasible basis B (available directly or using artificialbase method) and compute:

xB

= B1b,

zB= cTB

xB

,yBj = B

1aj, 1 j n,

zBj cj , 1 j n.

These values are written in the simplex table (table 1.1) and we go to the nextstep.

Let us denote by B the set of j indexes which determines the matrix B and byR = {1, . . . , n} B. The initial simplex table has the form:

TABLE 1.1

c1 . . . cj . . . cnB.V. V.B.V x1 . . . xj . . . xn

cB xB xB

yB1 . . . yBj . . . yBnz z

BzB1 c1 . . . z

Bj cj . . . z

Bn cn

where we denoted by B.V. the set of base variables and by V.B.V. the values of thebase variables (x

B).

STEP 2. If zBj cj 0 j R, we stop (STOP): xB

is an optimum program.

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Otherwise we find the set (non empty):

R+ = {j R|zB

j cj > 0}

and go to the next step.STEP 3. If there exists j R+ for which y

Bj 0 we stop (STOP): optimum

of the problem is infinity. Otherwise, we find k R+ with entering variable choicerules:

maxj

(zBj cj) = zB

k ck

si r B+ = {i B|yB

ik > 0} with leaving variable choice rules:

miniB+

(xBi

yBik) =

xBr

yBrk

The element yBrk is called pivot. Go to the next step.

STEP 4. Let us consider the basis

B obtained by B replacing the ar columnwith the column ak, and compute the values (using the rules for changing variables)

x

B, z

B, y

Bj , z

Bj cj and go to the Step 2 replacing the basis B with the basis

B .The computations can by simplified using the rules for simplex table transforma-

tion:i) the elements from the pivot lined are divided at the pivot value;ii) the elements from the pivot column became zero, with the exception of the

pivot which became 1;iii) the others values are transformed using the rectangle rule: if we imagine the

rectangle with the diagonal determined by the element under transformation yBij

and the pivot yBrk, then the new valuesyB

ij is obtained by dividing to the pivot the

differences between the product of the elements yBijyB

rk placed on the above defined

diagonal and the product yBrjyB

ik placed on the other diagonal of the rectangle.

Remarks. i) If at the ending of the algorithm we have zBj cj < 0 j R thenthe solution of the problem is unique.

ii) During algorithm execution in is possible to have the cycling phenomena(pass-ing to another basis which was already processed). There are several techniques to

avoid this.iii) We have also, in bidimensional case, a geometrical interpretation of the solu-

tions of a linear programming problem. The feasible domain can be:-a convex polyhedron, in this case one of its vertex is the solution of the opti-

mization problem. Also we may have the solution being one side of the polyhedron,in this case we have many solutions;

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THE DUAL OF THE LINEAR PROGRAMMING PROBLEM 7

-unbounded domain, in this case the optimum is infinity;

-empty set, in this case the optimization problem has no solution.

1.4. The dual of the linear programming problem

The primal problem deals with physical quantities. With all inputs availablein limited quantities, and assuming the unit prices of all outputs is known, whatquantities of outputs to produce so as to maximize total revenue? The dual problemdeals with economic values. With floor guarantees on all output unit prices, andassuming the available quantity of all inputs is known, what input unit pricingscheme to set so as to minimize total expenditure?

To each variable in the primal space corresponds an inequality to satisfy in thedual space, both indexed by output type. To each inequality to satisfy in the primalspace corresponds a variable in the dual space, both indexed by input type.

The coefficients that bound the inequalities in the primal space are used to com-pute the objective in the dual space, input quantities in this example.

The coefficients used to compute the objective in the primal space bound theinequalities in the dual space, output unit prices in this example.

Both the primal and the dual problems make use of the same matrix. In theprimal space, this matrix expresses the consumption of physical quantities of inputsnecessary to produce set quantities of outputs. In the dual space, it expresses thecreation of the economic values associated with the outputs from set input unit

prices.Since each inequality can be replaced by an equality and a slack variable, this

means each primal variable corresponds to a dual slack variable, and each dual vari-able corresponds to a primal slack variable. This relation allows us to complementaryslackness.

Thus if we have a linear programming problem we can construct the dual problemusing the following rules:

a) the free terms from the primal problem became coefficients of the objectivefunction from the dual problem;

b) coefficients of the objective function from the dual problem became free termsin the dual problem;

c) the dual of a maximization (minimization) primal problem is a minimization(maximization) problem;

d) the matrix of the constraints from the dual problem is the transposed of thematrix of the primal problem;

e) dual (primal) variables associated with concordant primal (dual) constraintsare supposed to the non negative constraints;

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f) primal (dual) variables associated with non-concordant dual (primal) con-

straints are supposed to the non positivityg) dual (primal) variables associated with primal (dual) constraints which are

equations have no constraints regarding the sign.

Let us remark that the dual of a problem in the canonical form is also in canonicalform.

Here is the fundamental the fundamental theorem of duality:

Theorem 1.4.1. If we have the couple of dual problems:

min cxAx bx 0

and

max byAy cy 0

only one of the following problems is true:

a) both problems have programs. In this case, both problems have optimal pro-grams and the optimal values of the objective functions are the same;

b) one of the problems have programs and the other one has no programs. In thiscase, the problem which has programs has optimum infinity;

c) none of the problems have programs.

The simplex dual algorithm solve the primal problem by the dual problem. Sim-ilarly, after solving the primal algorithm we can construct the solution of the dualproblem.

MAPLE software has a procedure for the construction of the dual problem. Thesyntax is the following:

> dual(objective function, set of constraints, dual variable set);

1.5. Transportation problems

Assume that we have m warehouses and n shops. A homogenus product is storedin the quantities ai, 1 i m, in the warehouses and it is requested by the shops inthe quantities bj , 1 j n,. We assume that the following conditions are satisfied:

ai 0, 1 i m, bj 0, 1 j n,a1 + . . . + am = b1 + . . . + bn.

(1.4)

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TRANSPORTATION PROBLEMS 9

In others words the available quantities and the requested quantities are non

negative and the request it is equal with the availability. Such a problem is called abalanced transportation problem.

The problem consists in planning the transportation from the warehouses to theshops with the minimum transportation cost.

Denote by xij the quantity (unknown) which will be supplied from the warehousei to the shop j.

The total quantity transported from the warehouse i to the all shops must beequal with the availability of the warehouse i, thus:

nj=1

xij = ai, 1 i m. (1.5)

Similarly, all the total requested quantity of the shop j must be equal with thetotal quantities transported from all the warehouses to this shop, thus:

mi=1

xij = bj, 1 j n. (1.6)

The transported quantities are non negative:

xij 0, 1 i m, 1 j n. (1.7)

The system of linear inequalities 1.5-1.7 in the constraints 1.4 has an infinitenumber of solutions. For adopting a transportation plan we need a economic criteria

given by the transportation cost. Denote by cij the transportation cost per unitfrom the warehouse i to the shop j; we shall assume that the cost per unit does notdepends of the total quantity transported from i to j. The transportation cost is:

mi=1

nj=1

cijxij.

The transportation problem consist in determination the solution of the system1.5-1.7 for which the total cost is minimum, thus:

infm

i=1n

j=1cijxij,

nj=1

xij = ai, 1 i m,

mi=1

xij = bj, 1 j n,

xij 0, 1 i m, 1 j n.

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1.6. Applications

Exercise 1.6.1. Find the dual of the linear programming problem:

min(2x1 + 3x2 + x3),x1 + x2 + 3x4 3,2x2 + 5x3 + 4x4 = 5,x1 + x3 2,x1, x2 0, x3 arbitrary, x4 0.

Find the optimal solution of the primal problem.

Solution. Using the rules for construction the dual problem we get:

max(3u1 + 5u2 2u3),u1 + u3 2,u1 + 2u2 3,5u2 + u3 = 1,3u1 + 4u2 0,u1 0, u2 arbitrary, u3 0.

The solution of the primal problem can be found with simple dual or primalalgorithm. MAPLE code for solving this problem is:

> with(simplex) :> objective := 2 x1 + 3 x2 + x3;> constraints := {x1+x2+3 x4 >= 3, 2x2+5 x3+4 x4 = 5, x1+x3 minimize(objective, constraints union {x1 >= 0, x2 >= 0, x4

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APPLICATIONS 11

Exercise 1.6.3. Solve the linear programming problem:

min(2x1 + 3x2 + x3),x1 + x2 + 3x4 3 + ,2x2 + 5x3 + 4x4 = 5 ,x1 + x3 2 + 2,x1, x2 0, x3 arbitrary, x4 0.

where is a real parameter. Also solve the dual problem.

Solution. Exercises 1.6.2-1.6.3 can be solved using simplex primal algorithm orusing post optimization technique in the final simplex table, for the problem 1.6.1,

with = 0.

Exercise 1.6.4. Using simplex algorithm solve the following linear program-ming problem:

max(2x1 + x2),x1 x2 4,

3x1 x2 18,x1 + 2x2 6,

x1, x2 0.

Exercise 1.6.5. Minimize 2x1 + 3x2 under the constraints:

x1 x2 + x3 = 1,x1 + x2 x4 = 1,x1 2x2 + x5 = 1,2x1 + x3 x4 = 2,xi 0, i = 1, . . . , 5.

Solution. Applying simplex algorithm we find the solution x1 = 1, x2 = 0, x

3 =

0, x4 = 0, x5 = 0, and the minimum for the objective function z

= 2.

Exercise 1.6.6. Solve the following problem:

min(x1 + 6x2),2x1 + x2 3,x1 + 3x2 4,x1, x2 0.

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Fig. 1.1: Feasibility domain and the objective function.

Solution. We shall solve the problem graphically. We represent in bi dimensionalspace the feasibility domain. On the same graphic 1.1 we represent the function

x1 + 6x2 = 0. Now we draw parallel lines with x2 = 1

6x1 until we intersect the

feasibility domain. The first line which intersect this domain gives us the minimumof the objective function zmin = 4. Optimal solution is x1

= 4, x2

= 0.

Exercise 1.6.7. Find the minimum of the function 2x1 + 3x2 + x3 with theconstraints:

x1 + x2 + 3x4 3,2x2 + 5x3 + 4x4 = 5,x1 + x3 2,x1, x2, x3, x4 {0, 1}.

Exercise 1.6.8. Solve the following linear programming problem:

min(2x1 + 3x2),x1 x2 + x3 = 1,x1 + x2 x4 = 1,x1 2x2 + x5 = 1,2x1 + x3 x4 = 2,xi 0, 1 i 5.

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APPLICATIONS 13

Solution. For solving the linear programming problem by simplex algorithm first

we solve the following problem:

min(x6 + x7 + x8),x1 x2 + x3 + x6 = 1,x1 + x2 x4 + x7 = 1,x1 2x2 + x5 = 1,2x1 + x3 x4 + x8 = 2,xi 0, 1 i 8.

(we have introduced the slack variables x6, x7, x8 in the constraints of the initialproblem, with an exception for the third constraint where the variable x5 is asso-ciated with a unit vector of the matrix of the coefficients). The primal admissible

basis is constructed from the columns of matrix A corresponding to the variablesx6, x7, x5, x8. The simplex associated table 1.2 is:

TABLE 1.2

B.V. V.B.V. x1 x2 x3 x4 x5 x6 x7 x8x6 1 1 1 1 0 0 1 0 0

x7 1 1 1 0 1 0 0 1 0

x5 1 1 2 0 0 1 0 0 0

x8 2 2 0 1 1 0 0 0 1

4 4 0 2 2 0 0 0 0

Using entering variable choice rules the vector which entry in the basis will bea1. Leaving variable choice rules indicates that we can eliminate any of the basisvariables: we choose for example the variable that corresponds to x6, thus a

6. Weobtain the simplex table 1.3:

TABLE 1.3

B.V. V.B.V. x1 x2 x3 x4 x5 x6 x7 x8x1 1 1 1 1 0 0 1 0 0

x7 0 0 2 1 1 0 1 1 0

x5 0 0 1 1 0 1 1 0 0

x8 0 0 2 1 1 0 2 0 1

0 0 4 2 2 0 4 0 0

From table 1.3 we see that the value of the objective function is 0. All the slackvariables are 0 but x7 and x8 are still basis variables. It is easy to see that thevariable x7 can be eliminated from the basis and replaced with x2, because the pivotyB72 = 2 is different from 0. After the computations we obtain the simplex table:

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TABLE 1.4

B.V. V.B.V. x1 x2 x3 x4 x5 x6 x7 x8x1 1 1 0 1/2 1/2 0 1/2 1/2 0

x2 0 0 1 1/2 1/2 0 1/2 1/2 0

x5 0 0 0 3/2 1/2 1 3/2 1/2 0

x8 0 0 0 0 0 0 1 1 1

0 0 0 0 0 0 2 2 0

The slack variable x8 can not be eliminated from the basis variables because all thevalues y8j , 1 j 5, are equal to 0. This shows us that the four-th equation, fromthe initial linear programming problem, is a consequence of the others equations

(the four-th equation is the sum of the first two equations, situation that can beobserved from the beginning of computations). In this case, the four-th equationcan be eliminated and also the corresponding line from the simplex table. Theremainder table does not contain any slack variables in the basis, thus the firstphase is finished.

In the second phase, we solve the initial linear programming problem, using initialbasis the basis obtained in table 1.4, after the elimination the last row of the table.

The corresponding simplex table is the following:

TABLE 1.5

B.V. V.B.V. x1 x2 x3 x4 x5

x1 1 1 0 1/2 1/2 0x2 0 0 1 1/2 1/2 0

x5 0 0 0 3/2 1/2 1

z 2 0 0 1/2 5/2 0

Because zj cj 0 for all j, 1 j 5, results that we obtained the optimalsolution (degenerated): x1 = 1, x

2 = 0, x

3 = 0, x

4 = 0, x

5 = 0, and the optimal value

of the objective function is z = 2.

Exercise 1.6.9. (diet problem)In a bakery the are two kinds of cakes: brownies, which cost 50 cents each, and

minicheesecakes, which cost 80 cents. The bakery is service-oriented and can sell afraction of any item. The bakery requires three ounces of chocolate to make eachbrownie (no chocolate is needed in the cheesecakes). Two ounces of sugar are neededfor each brownie and four ounces of sugar for each cheesecake. Finally, two ouncesof cream cheese are needed for each brownie and five ounces for each cheesecake. Asnack consumer health-conscious, has decided that he needs at least six total ounces

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APPLICATIONS 15

of chocolate in his snack, along with ten ounces of sugar and eight ounces of cream

cheese. The consumer wishes to optimize his purchase by finding the least expensivecombination of brownies and cheesecakes that meet these requirements. The datais sumarized in the following table:

TABLE 1.6

Chocolate Sugar Cream Cheese Cost

Brownie 3 2 2 50

Cheesecake 0 4 5 80

Requirements 6 10 8

Find the optimal solution.

Solution. The problem that the consumer must to solve is:

min(50x + 80y),3x 6,2x + 4y 10,2x + 5y 8,x, y 0,

where x and y represent the number of brownies and cheesecakes purchased, respec-tively. By applying the simplex method of the previous selection, we find that theunique solution is (2, 3/2), thus the value of the objective function, computed forthe optimal solution is 220.

We now adopt the perspective of the wholesaler who supplies the bakerwith the chocolate, sugar, and cream cheese needed to make the goodies. The bakerinforms the supplier that he intends to purchase at least six ounces of chocolate, tenounces of sugar, and eight ounces of cream cheese, to meet the consumer minimumnutritional requirements.He also shows the supplier the other data from the table.The supplier now solves the following optimization problem: How can I set the pricesper ounce of chocolate, sugar, and cream cheese so that the baker will buy from me,and so that I will maximize my revenue? The baker will buy only if the total cost ofraw materials for brownies is below 50 cents; otherwise he runs the risk of makinga loss if the student opts to buy brownies. This restriction imposes the followingconstraint on the prices:

3u1 + 2u2 + 2u3 50.

Similarly, he requires the cost of the raw materials for each cheesecake to be below80 cents, leading to a second constraint:

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4u2 + 5u3 80.

Clearly, all the prices must be nonnegative. Moreover, the revenue from theguaranteed sales is 6u1 + 10u2 + 8u3. In summary, the problem that the suppliersolves to maximize his guaranteed revenue from the consumer snack is as follows(dual problem):

max(6u1 + 10u2 + 8u3)3u1 + 2u2 + 2u3 504u2 + 5u3 80

u1

, u2

, u3

0

.

The solution of this problem is u = (10/3, 20, 0), thus the value of the dualobjective function is 220 (which is the same with the value of the objective functioncomputed for optimal solution of the primal problem).

Exercise 1.6.10. A gardener produce two types of mixtures for planting: gar-dening mixture and potting mixture. A package of gardening mixture requires 2kg of soil, 1 kg of peat moss and 1 kg of fertilizer. A package of potting mixturerequires 1 kg of soil, 2 kg of peat moss and 3 kg of fertilizer. The gardener has atmost 16 kg of soil, 11 kg of peat moss and 15 kg of fertilizer. A package of gardenmixture sells for 3 EURO and a package of potting mixture sells for 5 EURO. Howmany packages of each type of mixture must the gardener to produce to maximizethe revenue?

Solution. The problem that the gardener must to solve is:

max(3x + 5y),2x + y 16,x + 2y 11,x + 3y 15,x, y 0,

where x and y are the packeges of gardening mixture and potting mixture.The gardener must produce 7 packages of gardening mixture and 2 packages of

potting mixture, the revenue is 31 EURO.Let see the problem from the point of view of the supplier which is informed

by the gardener that he want to sell with at least 3 EURO a package of gardeningmixture and at least 5 EURO a package of potting mixture. Also, the gardener

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APPLICATIONS 17

will inform the supplier that he intend to buy 16 kg of soil, 11 kg of peat moss and

15 kg of fertilizer. The supplier will setup the prices u,v and w for 1 kg of soil, 1kg of peat moss respectivelly 1 kg of fertilizer such that to minimize the function16u + 11v + 15w (otherwise the gardener will buy from another supplier). Also therestriction regarding the cost of gardening mixture and potting mixture must beimposed: 2u + v + w 3 respectivelly u + 2v + 3w 5. Thus, the problem whichmust be solved by the supplier is the dual problem of the gardener:

min(16u + 11v + 15w),2u + v + w 3,u + 2v + 3w 15,u,u,w 0.

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1 lecture 1

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