Handbook of mathematics for engineers and scienteists part 151 doc

If the inequality constraints of the primal problem have the≤signs, then the objective function Zx must be maximized; if the signs are≥, then the objective function Zx must be minimized.

The rules for constructing the dual problem are as follows:

1 In all constraints of the primal problem, the constant terms must be on the right-hand side, and the terms with the unknowns must be on the left-hand side

2 All inequality constraints must have the same inequality signs

3 The matrix

A=

⎛

⎜

⎜

⎝

a11 a12 · · · a1 n

a21 a22 · · · a2 n

. .

a m1 a m2 · · · a mn

⎞

⎟

⎟

⎠

of coefficients of the unknowns in the system of constraints (19.2.1.12) of the primal problem and the similar matrix

A T =

⎛

⎜

⎜

⎝

a11 a21 · · · a m1

a12 a22 · · · a m2

. .

a1n a2n · · · a mn

⎞

⎟

⎟

⎠

for the dual problem are obtained from each other by transposition

4 If the inequality constraints of the primal problem have the≤signs, then the objective

function Z(x) must be maximized; if the signs are≥, then the objective function Z(x)

must be minimized

5 To each constraint in the primal problem, there corresponds an unknown in the dual problem The unknown corresponding to an inequality constraint must be nonnegative, and the unknown corresponding to an equality constraint may have any sign

6 The number of variables in the dual problem is equal to the number of functional constraints (19.2.1.12) in the primal problem, and the number of constraints in sys-tem (19.2.1.15) of the dual problem is equal to the number of variables in the primal problem

7 The coefficients of the unknowns in the objective function (19.2.1.14) of the dual problem are equal to the constant terms in system (19.2.1.12) of constraints in the primal problem The right-hand sides of constraints (19.2.1.15) in the dual problem are equal to the coefficients of the unknowns in the objective function (19.2.1.11) of the primal problem

8 The objective function W (y) of the dual problem is to be minimized if Z(x) is to be maximized (i.e., if Z(x) → max, then W (y) → min), and vice versa.

Duality theorems establish a relationship between the optimal solutions of a pair of dual problems

FIRST DUALITY THEOREM For a pair of mutually dual linear programming problems, one of the following alternative cases holds:

1 If one of the problems has an optimal solution, then so does the other, and the objective values on the optimal solutions are the same for both problems

2 If one of the problems is unbounded, then the other is infeasible

SECOND DUALITY THEOREM Let x = (x1, , x n)be a feasible solution of the primal

problem (19.2.1.11)–(19.2.1.13), and let y = (y1, , y m) be a feasible solution of the dual problem (19.2.1.14)–(19.2.1.16) These solutions are optimal in the respective problems if and only if

x j

m i=1

a ij y i – c j



=0 (j =1,2, , n); (19.2.1.17)

y i

n j=1

a ij x j – b i



=0 (i =1,2, , m). (19.2.1.18)

In other words, if the ith constraint in the primal problem is inactive (holds with strict inequality) for an optimal solution of the primal problem, then the ith coordinate of the optimal solution of the dual problem is zero, and, conversely, if the ith coordinate of the optimal solution of the dual problem is nonzero, then the ith constraint in the primal problem

is active (holds with equality) for the optimal solution of the primal problem

Example 4 Suppose that the primal problem has the form

Z(x) = –2x1+ 4x2+ 14x3+ 2x4→ min,

– 2x1– x2+ x3 + 2x4= 6 ,

– x1+ 2x2+ 4x3– 5x4= 30 ,

x j≥ 0 (j =1 , 2 , 3 , 4 ).

Then the dual problem is

W(y) =6y1+ 30y2→ max,

– 2y1– y2 ≤ – 2 ,

– y1+ 2y2≤ 4 ,

y1 + 4y2 ≤ 14 ,

2y1– 5y2≤ 2

The optimal solution of the dual problem is yopt = ( 2 , 3 ) After the substitution of the optimal solution into the system of constraints in the dual problem, we see that the first and last constraints are satisfied with strict inequality,

– 2 × 2 – 3 < – 2 ⇒ x ∗1 = 0 , – 2 + 2 × 3 = 4 ⇒ x ∗

2 > 0 ,

2 + 4 × 3 = 14 ⇒ x ∗

3 > 0 ,

2 × 2 – 5 × 3 < 2 ⇒ x ∗

4 = 0

By the second duality theorem, the corresponding coordinates of the optimal solution of the primal problem

are zero, x ∗1 = x ∗4 = 0 The system of constraints of the primal problem acquires the form

– x2+ x3 = 6 ,

2x2 + 4x3 = 30 ,

which implies that x ∗2 = 1, x ∗3 = 7

Thus xopt= ( 0 , 1 , 7 , 0 ) is the optimal solution of the primal problem.

19.2.1-5 Transportation problem General statement of problem

Suppose that m supply sources A1, , A m have amounts a1, , a m of identical goods

that must be shipped to n consumers B1, , B n with respective demands b1, , b n for

these goods Let c ij (i =1,2, , m; j =1,2, , n) be the unit transportation cost from supply source i to consumer destination j The problem is to find a flow of least cost that

ships from supply sources to consumer destinations

The input data of the transportation problem are usually presented in the form of the table shown in Fig 19.1

Let x ij (i =1,2, , m; j = 1, , n) be the flow from supply source i to consumer destination j The mathematical model of the transportation problem in the general case

has the form

Z(x) =

m



i=1

n



j=1

Figure 19.1 The input data of the transportation problem.

n



j=1

m



i=1

x ij >0, a i≥ 0, b j ≥ 0 (i =1,2, , m; j =1,2, , n) (19.2.1.22) The objective function (19.2.1.19) is the total transportation cost to be minimized Equa-tions (19.2.1.20) mean that each supply source must ship all goods present at that source Equations (19.2.1.21) mean that the demands of all consumers must be completely satisfied Inequalities (19.2.1.22) are the conditions that all variables in the problem are nonnegative

A necessary and sufficient condition for the solvability of (19.2.1.19)–(19.2.1.22) is the

balance condition

m



i=1

a i=

n



j=1

A transportation problem satisfying relation (19.2.1.23) is said to be balanced, and the corresponding model is said to be closed If this relation does not hold, then the problem is said to be unbalanced and the corresponding model is said to be open.

Under the balance condition (19.2.1.23), the rank of the system of equations (19.2.1.20),

(19.2.1.20) is equal to n + m –1; therefore, n + m –1out of the mn unknown variables

must be basic variables It follows that for any feasible basic flow the number of occupied

cells in the transportation tableau shown in Fig 19 1 is equal to n + m –1; these cells are

usually said to be basic and the other cells are said to be nonbasic.

There are various methods for obtaining a feasible initial solution (a feasible shipment)

in the transportation problem, including the cross-out method, the northwest corner rule, the minimal cost method, Vogel’s approximation method, etc Of these methods, the minimal cost method is simplest and most convenient

Minimal cost method The method consists of several steps of the same type At each

step, one fills only one cell in the tableau corresponding to the minimal cost mini,j{cij}and excludes only one row (supply source) or column (consumer destination) from subsequent considerations A supply source is excluded if its supply of goods is completely exhausted

A consumer is eliminated if his demand is completely satisfied At each step, either a supply source or a consumer is excluded If a supply source is yet to be excluded but its supply of

goods is already zero, then, at the step where it should (but cannot) supply goods, a basic zero is written in the corresponding cell and only after this the supply source is excluded Consumers are treated similarly

Reduction of an unbalanced transportation problem to a balanced transportation prob-lem.

1 If the total supply exceeds the total demand, i.e.,

m



i=1

a i >

n



j=1

b j, then one introduces fictitious consumer n +1with demand

b n+1=

m



i=1

a i–

n



j=1

b j

equal to the difference between the total supply and total demand and with zero unit

transportation costs c i(n+1) =0(i =1,2, , m).

2 If the total demand exceeds the total supply, i.e.,

m



i=1

a i <

n



j=1

b j, then one introduces fictitious supply source m +1with supply

a m+1 =

n



j=1

b j–

m



i=1

a i

equal to the difference between the total demand and the total supply and with zero unit

transportation costs c(m+1)j =0(j =1,2, , n).

Remark When constructing the initial basic solution, the supply of the fictitious source and the demands

of the fictitious consumer are the last to be assigned, even though they correspond to minimum (zero) unit transportation costs.

19.2.1-6 Method of potentials

A balanced transportation problem can be solved as a linear programming problem But there are less cumbersome methods for solving the transportation problem The most widely used method is the method of potentials

If a feasible solution X ≡[x ij ] (i =1,2, , m; j = 1,2, , n) of the transportation problem is optimal, then there exist supplier potentials u i (i = 1,2, , m) and consumer potentials v j (j =1,2, , n) satisfying the following conditions:

u i + v j = c ij for basic cells, (19.2.1.24)

u i + v j ≤c ij for nonbasic cells. (19.2.1.25) Relations (19.2.1.24) are used as a system of equations for the potentials This system has

n + m unknowns and n + m –1equations Since the number of unknowns exceeds the number of equations by one, one of the variables can be taken arbitrarily, and the others are then found from the system

Inequalities (19.2.1.25) are used to verify the optimality of a basic solution To this end,

the reduced costs

Δij = u i + v j – c ij (19.2.1.26)

of nonbasic cells are used

Remark 1 For basic cells, the quantities Δijare zero, Δij= 0

Remark 2. The variables u i (i =1 , 2, , m) are dual to the respective constraints in (19.2.1.20), and the variables v j (j =1 , 2, , n) are dual to the respective constraints in (19.2.1.21) The dual problem has the

form

m



i=1

a i u i+

n



j=1

b j v j → min,

u i + v j≤c ij (i =1 , 2, , m, j =1 , 2, , n).

THEOREM(AN OPTIMALITY CRITERION) An admissible solution is optimal if the reduced costs are nonpositive for all cells in the tableau

A cycle is a sequence (i1, j1), (i1, j2), (i2, j2), , (i k , j1) of cells in the transportation problem tableau in which exactly two neighboring cells lie in the same row or column and, moreover, the first and last cells also lie in the same row or column

If the current basic solution is not optimal, then one needs to pass to a new solution with smaller value of the objective function To this end, in the tableau one takes the cell with the largest positive reduced cost

max{Δij}=Δlk Next, one constructs a cycle including this cell and some basic cells The cells in the cycle are alternately marked by “+” and “–” signs starting from the cell with the largest positive

reduced cost For the “–” cells, one finds the quantity θ = min{xij} Next, one performs a

shift (redistribution of goods) over the cycle by θ The “–” cell at which min{xij}is attained becomes empty If the minimum is attained at several cells, then one of them becomes empty and the other cells are filled with basic zeros, so that the number of occupied cells remains

equal to n + m –1

Example 5 Let us solve the transportation problem with the following input data (see Fig 19.2):

Figure 19.2 Input data for Example 5.

The total demand 4

j=1b j= 200 + 200 + 300 + 400 = 1100 exceeds the total supply 3

i=1a i= 200 + 300 + 500 = 1000

by 100 units; hence the transportation problem is unbalanced One should introduce the fictitious fourth supply

source with supply a4= 100 and with zero unit transportation costs (see Fig 19.3).

The initial basic solution x1 is found by the minimal cost method (Fig 19.3) On this solution, the objective function takes the value

Z(x1) = 1 × 200 + 2 × 200 + 3 × 100 + 7 × 100 + 9 × 300 + 12 × 100 + 0 × 100 = 5300

Figure 19.3 Introduction of the fictitious fourth supply source The initial basis solution.

To verify whether the basic solution is optimal, we find the potentials To this end, we use system (19.2.1.24), which acquires the form

u1+ v4= 1 ,

u2+ v1 = 2 ,

u2+ v2 = 3 ,

u3+ v2= 7 ,

u3+ v3 = 9 ,

u3+ v4= 12 ,

u4+ v4 = 0

Let u3= 0 Then the other potentials are determined uniquely: v2 = 7, v3 = 9, v4= 12, u1= – 11, u4= – 12 ,

u2 = – 4, and v1 = 6 The values of the potentials are written down in the tableau for the solution of the transportation problem (see Fig 19.4).

Figure 19.4 Determination of potentials Construction of the cycle.

Next, we verify whether the solution x1 is optimal To this end, using (19.2.1.26), we find the reduced costs Δijfor all empty (nonbasic) cells of the tableau The solution x1is not optimal since there is a positive reduced cost Δ 24= u2+ v4– c24= – 4 + 12 – 6 = 2

For cell ( 2 , 4 ) with positive reduced cost, we construct the cycle ( 2 , 4 ), ( 3 , 4 ), ( 3 , 2 ), ( 2 , 2 ), ( 2 , 4 ) The cycle

is shown in Fig 19.4 At the corners of the cycle, we alternately write the “+” and “–” signs, starting from cell ( 2 , 4) For the cells with the “–” sign, we have θ = min{100 , 100} Then we perform a shift (redistribution of

Figure 19.5 The second basis solution.

goods) over the cycle by θ =100and obtain the second basic solution x2 (see Fig 19.5) Once the system of

potentials has been found (Fig 19.5), we see that the solution x2is optimal The objective function takes the value

Z(x2) = 1 × 200 + 2 × 200 + 6 × 100 + 7 × 200 + 9 × 300 + 0 × 100 = 5200

on this solution Thus, Zmin(X) =5200 for

X ∗=

200 0 0 100

0 200 300 0

)

19.2.1-7 Game theory

Mathematical models of conflict situations are called games, and their participants are called players Mathematical models of conflict situations and various methods for solving problems that arise in these situations are constructed in game theory.

According to the number of players, the games are divided into two-person and n-person games In n-n-person games, the players’ interests may coincide In this case, they can cooperate and form coalitions Such games are called coalition games.

A player’s strategy is a set of rules uniquely determining the player’s behavior in each

specific situation arising in the game A strategy ensuring the maximum possible mean

payoff for a player in repeated games is said to be optimal The number of possible strategies

of each player can be either finite or infinite Depending on this, the games are divided into

finite and infinite games.

Games in which one of the players is indifferent to the results are usually called “games with nature.”

A two-person game in which the payoff of one player is equal to the loss of the other

player is called an antagonistic two-person zero-sum game Suppose that two players A and B have finitely many pure strategies: player A can choose any of m strategies A1, ,

A m , and player B can choose any of n strategies B1, , B n These strategies determine

⎜

⎜

⎝

a11 a12 · · · a1 n

a21 a22 · · · a2 n

. .

a m1 a m2 · · · a mn

⎞

⎟

⎟