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On solving of bilinear programming problems 197When the algorithm stop (i.e. inequality (5) hold), we have approximate critical point of(BLP ).Further notice features of the algorithm. First, algorithm is needed only x 0 from (x 0 , y 0 ) forstarting. Second, this point may be not feasible. In spite of the last fact, the convergence ofthis method was proven.Below we presented another "symmetric" method for local search in (BLP ). This algorithmis needed only y 0 from (x 0 , y 0 ) for starting.Y-procedure.Step 0. Let s := 0, y s := y 0 .Step 1. Find ρ s /2-solution x s+1 of Linear ProgramHere〈c + Qy s , x〉 ↑ maxx , x ∈ X, (LP x).〈c + Qy s , x s+1 〉 + ρ s /2 ≥ sup{〈c + Qy s , x〉 | x ∈ X}. (8)xStep 2. Find ρ s /2-solution x s+1 of Linear ProgramStep 3. IfHere〈d + x s+1 Q, y〉 ↑ maxy , y ∈ Y, (LP y).〈d + x s+1 Q, y s+1 〉 + ρ s /2 ≥ sup{〈d + x s+1 Q, y〉 | y ∈ Y }. (9)yF (x s+1 , y s+1 ) − F (x s+1 , y s ) ≤ τ, (10)where τ is accuracy of solution, Then Stop, else let s := s + 1 and go to step 1.Theorem 2. In conditions of theorem 1 sequence (x s , y s ) from Y-procedure converge to (̂x, ŷ), such as(6) and (7) takes place.Notice, that both above methods converges to the point with identical properties.4. Global Search AlgorithmSince we not obtain the global solution by local search, further the global search algorithm ispresented.Suppose (x 0 , y 0 ) ∈ D △ = X × Y is a starting point, {τ k }, {δ k } are sequences of numbers,τ k , δ k > 0, k = 0, 1, 2, ..., τ k ↓ 0, δ k ↓ 0, (k → ∞), Dir = {(u 1 , v 1 ), ..., (u N , v N ) ∈IR m+n |(u s , v s ) ≠ 0,scalars.s = 1, ..., N} is set of vectors, γ −△= inf(g, D), γ+△= sup(g, D), ν and q areGlobal search algorithmStep 0. Let k := 1, (¯x k , ȳ k ) := (x 0 , y 0 ), s := 1, p := 1, γ := γ − , △γ = (γ + − γ − )/q.Step 1. Beginning from (¯x k , ȳ k ) ∈ D by X-procedure or Y-procedure obtain τ k -critical point(x k , y k ) ∈ D in (P). Here F (x k , y k ) ≥ F (¯x k , ȳ k ). Let ζ k := F (x k , y k ).Step 2. Compute point (ū s , ¯v s ) of level surface approximation of f(·) by (u s , v s ) ∈ Dir. Heref(ū s , ¯v s ) = γ + ζ k .Step 3. If g(ū s , ¯v s ) > γ + νγ, then let s := s + 1 and go to step 2, else go to step 4.