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Page 10 of 15 Chen et al. Complex Eng Syst 2023;3:8 I http://dx.doi.org/10.20517/ces.2022.50

The objective function of this MPC strategy has the following quadratic form:

Õ Õ
( ) = k ( + ) − ( + )k 2 + kΔ ( + − 1)k 2 + Θ (25)

=1 =1

where and are the weighting matrices of the first and second items, respectively. Θ represents a positive
relaxation factor. The objective of this function is to follow the ideal model smoothly and accurately. The first
term of this function describes the ability of the actual vehicle model to track the reference model. The second
term indicates the change in the input vector, which can restrict changes to the input variables. Meanwhile,
the input, input increment, and output variables are constrained in a domain that can be expressed as follows:

 min ( + ) ≤ ( + ) ≤ max ( + ) ( = 0, 1, · · · , − 1)

 
Δ min ( + ) ≤ Δ ( + ) ≤ Δ max ( + ) ( = 0, 1, · · · , − 1) (26)

 min ( + ) ≤ ( + ) ≤ max ( + ) = 0, 1, · · · , − 1

Because of constraints, it is generally impossible to obtain the analytical solution to this problem. For this
reason, it is necessary to transform it into a quadratic programming (QP) problem to obtain a numerical
solution. Therefore, we convert the above constraint equations into the form ≥ , as follows.

 − Δ −Δ max ( )


 Δ ( ) ≥

 Δ Δ min ( )



  0
− ( − 1) − max ( )
Δ ( ) ≥ (27)
0
 min ( ) − ( − 1)


 0
 − ( − 1) − max ( )

 Δ ( ) ≥

0
 min ( ) − ( − 1)

where Δ , Δ max ( ), Δ min ( ), , ( − 1), max ( ), min ( ), ( − 1), max ( ), and min ( ) can be
0
0
calculated according as described in [21] . Then, this question can be described as a standard QP problem. In
this manner, the solution of this problem without the constraint equation can be set as the initial solution,
which can be expressed as follows:

Δ ( , 0) = ( − 1) ( − 1) + ( − 1) ( ) (28)

where ( ) = ( )− ( −1)Δ ( )− ( ) heoptimalsolutionoftheinputvector Δ ( ) canbecalculated
∗
using the algorithm of the QP problem with constraints. Then, the closed-loop control input can be obtained
as follows:

∗
Δ ( ) = 2×2 0 · · · 0 Δ ( ) (29)
1×