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P. 34

S. Ben Hanachi, B. Sellami, M. Belloufi / IJOCTA, Vol.15, No.1, pp.25-34 (2025)
            Next, considering the above relation again and    Finally: 30  if 0 < θ k < 1, then there exist two real
            adding and subtracting the value θ k g k+1 , we get:  numbers µ 1 , µ 2 such that 0 < µ 1 ≤ θ k ≤ µ 2 < 1,
                                                              and we have:
                                                  ∥y k ∥ 2
               hyb
              d    = −(1 − θ k + θ k )g k+1 + (1 − θ k )  d k
               k+1                                 T
                                                  d y k
                                                   k              T   hyb            T   BA       T   HZ
                                                                                        d
                                                                     d
                                                                                                     d
                         1    T        ∥y k ∥ 2  T              g k+1 k+1  = (1 − θ k )g k+1 k+1  + θ k g k+1 k+1 ,
                   +θ k     (y g k+1 − 2     d g k+1 )d k ,
                        T     k          T    k
                       d y k            d y k
                        k                k
                                       ∥y k ∥ 2                           ≤ (1 − µ 2 )g T  d BA  + µ 1 g T  d HZ  ,
                   = (1 − θ k )(−g k+1 +    d k )                                    k+1 k+1      k+1 k+1
                                        T
                                       d y k
                                        k
                                   1
                   +θ k (−g k+1 +                                29
                                  T                           In,  they proved that:
                                 d y k
                                  k
                               ∥y k ∥ 2
                                     T
                                                                                                  2
                     T
                   (y g k+1 − 2  T  d g k+1 )d k ).                    g T  d BA  ≤ −(1 − ϵ 1 )∥g k+1 ∥ ,  (14)
                     k
                                                                        k+1 k+1
                                     k
                               d y k
                                k
            Finally, from the above relation, we obtain:
                                                              where, 0 < ϵ 1 << 1, from the above relations (13)
                                                              and (14), it follows that:
                       hyb           BA       HZ
                     d    = (1 − θ k )d  + θ k d  .
                       k+1           k+1      k+1
                                                         □
                                                                                                       7
                                                                                                              2
                                                                   hyb
                                                                                                 2
            2.1. The sufficient descent condition             g T  d   ≤ −(1 − µ 2 )(1 − ϵ 1 )∥g k+1 ∥ − µ 1 ∥g k+1 ∥ .
                                                               k+1 k+1
                                                                                                       8
            In the following we prove that the search direction  Therefore,
            d k obtained by the new hybrid conjugate gradi-
                                                                                               2
            ent satisfies the sufficient descent condition, which         g T  d hyb  ≤ −K∥g k+1 ∥ ,     (15)
                                                                           k+1 k+1
            plays a crucial role in ensuring global convergence.
            Based on the relation (12), we distinguish three  where K = (1 − µ 2 )(1 − ϵ 1 ) + µ 1  7 8   .
            cases:
            Firstly: If θ k = 0, then                         Algorithm 1. (CCH, NDH)
                                                                                                           n
                               hyb    BA                      Initialization. Choose a starting point x 0 ∈ R ,
                              d    = d   ,
                               k+1    k+1                     ϵ > 0, and 0 < δ ≤ σ < 1. Compute f(x 0 ) and
                                                              g 0 = ∇f(x 0 ). Consider d 0 = −g 0 , the initial
            Therefore, the sufficient descent condition holds  guess α 0 = 0, and k = 0.
            for the hybrid method if it holds for the BA      Step 1: If ∥g k ∥ < ϵ, then stop; otherwise, con-
            method. Delladji, S., Belloufi, and Sellami prove  tinue with Step 2.
            in 29  that d BA  satisfies the sufficient descent con-  Step 2: Compute α k by the strong Wolfe line
                       k+1
            dition for all k, under the strong Wolfe line search  search (3), (4).
            conditions.                                       Step 3: Generate x k+1 = x k + α k d k .
                                                              Compute f(x k+1 ), g k+1 = ∇f(x k+1 ), and y k =
                                                              g k+1 − g k .
            Secondly: If θ k = 1, then
                                                                                                       T
                                                              Step 4: Compute the θ k parameter. If y g k −
                                                                                                       k
                               hyb    HZ                       ∥y k ∥ 2  T
                              d    = d   ,                    2  T  d g k+1 = 0, then set θ k = 0. Otherwise,
                                                                     k
                               k+1    k+1
                                                                d y k
                                                                k
                                                              compute θ k as follows:
            Therefore, the sufficient descent condition holds
                                                              CCH algorithm (using the Conjugacy Condition):
            for our hybrid method if it holds for the HZ
            method, the following theorem (2), established by                         T
                                                                                      k
            William W. Hager and Hong Zhang in,   26  proves            θ k =        g y k 2      .
                                                                                      ∥y k ∥
                                                                               T
                                                                                            T
            the sufficient descent for the HZ method:                         y g k − 2  T  d g k+1
                                                                               k
                                                                                            k
                                                                                      d y k
                                                                                       k
                             T
            Theorem 2. If d y k ̸= 0, and                     NDH algorithm (using the Newton Direction):
                             k
                 d k+1 = −g k+1 + τd k ,
                                                                          T            2    T      T
                                                                        (y g k+1 − ∥y k ∥ − d g k+1 )d y k
                                                                                                   k
                                                                          k
                                                                                            k
                   d 0 = −g 0 , ∀τ ∈ [β k HZ , max{0, β k HZ }].  θ k =  T    T          2    T        T   .
                                                                    (y g k+1 )(d y k ) − ∥y k ∥ (2d g k+1 + d y k )
                                                                      k       k               k        k
            Then,
                                      7       2
                              HZ
                         T
                             d
                        g k+1 k+1  ≤ − ∥g k+1 ∥ .      (13)   Step 5. If 0 < θ k < 1, compute β     hyb  as in
                                      8                                                             k
                                                            28
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