_{1}

This paper provides an extension to an optimal control problem using the negative logarithm of deterioration and spoilage function as total cost. This function must be minimized at the end of planning period depending on the alternative quadratic exponential form. The co-state variable has negative values along the optimal trajectory according to the Pontryagin Minimum Principle (PMP). The different values of this co-state variable are investigated using initial values for the optimal control rates, separately. The controlled system according to each value is presented. Studying the behavior of optimal inventory levels, the optimal production rates, and the optimal spoilage function, it is our optimal solution along the optimal trajectory. The effectiveness of increasing and decreasing the co-state values on the optimal trajectory especially at the end of planning period is investigated. Also, the sensitivity analysis that reflects the effect of changes of different parameters (the deterioration and spoilage parameters values, and the initial values of inventory levels and production rates) on the optimal solution is explained with many different cases. Finally, we compared, numerically, the results for using these different co-state values with the results obtained when this value is negative.

An optimal control problem of multi-item inventory model has a wide importance in practice. El-Sayed [

So, we concentrated on the sensitivity analysis of changing the spoilage parameters on the optimal solution especially at the end of planning period. Finally, to complete this study, we must compare the obtained results for these different co-state values with the results which are obtained when the co-state variable equals −1.

Zhao and Prentice [

f ( x 1 , x 2 ) = 1 ∑ x 1 , x 2 exp { θ 1 x 1 + θ 2 x 2 + θ 12 x 1 x 2 } exp { θ 1 x 1 + θ 2 x 2 + θ 12 x 1 x 2 } . (1)

El-Sayed [

Since θ 1 , θ 2 and θ 12 are the deterioration parameters, θ ′ s > 0 .

So, we can use the normalizing term, ∑ x 1 , x 2 exp { θ 1 x 1 + θ 2 x 2 + θ 12 x 1 x 2 } , in the function (1) to be rewritten in the exponential form, El-Sayed et al. [

f ( x 1 , x 2 ) = exp { θ 1 x 1 + θ 2 x 2 + θ 12 x 1 x 2 − log ( 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 ) } . (2)

The minimizing of an integral of negative logarithm of this function can be used as the total cost. This cost reflects the levels of deterioration and spoilage of items at the end of planning period.

This paper can be organized as follows: Section 2 presents the mathematical model for the optimal control problem and the coresponding controlled systems. Section 3 presents the numerical solution for the controlled systems for different rates and different co-state values. Section 4 presents the sensitivity analysis of the model parameters and co-state variable λ 0 ( t ) . Finally Section 5 gives some conclusions.

Let us define the following parameters, which are used in the mathematical optimal control model:

X i ( t ) : Inventory levels at time t.

U i ( t ) : Production rates at time t.

T: Length of planning period.

x i 0 : Initial inventory levels.

a i i : Deteriration coefficient due to self-contact of item x i .

a i j : Deteriration coefficient of x i due to presence a unit of x j , i ≠ j = 1 , 2 .

D i ( x 1 , x 2 , t ) : Demand rates of ( x 1 , x 2 ) .

ψ 1 : Spoilage rate of of x 1 , ψ 1 > 0 .

ψ 2 : Spoilage rate of of x 2 , ψ 2 > 0 .

ψ 12 : Spoilage rate of ( x 1 , x 2 ) , jointly, ψ 12 > 0 .

θ 1 : Natural deterioration rate of x 1 ^{, θ 1 > 0 }.

θ 2 : Natural deterioration rate of x 2 , θ 2 > 0 .

θ 12 : Natural deterioration rate of ( x 1 , x 2 ) , jointly, θ 12 > 0 .

As we mentioned before, the integral of negative logarithm of the function (2), which represents the deterioration and spoilage function, is used as a cost function:

x 0 ( T ) = ∫ 0 T − ln f ( x 1 , x 2 ) d t = ∫ 0 T [ − θ 1 x 1 − θ 2 x 2 − θ 12 x 1 x 2 + log ( 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 ) ] d t . (3)

So, the problem can be formulated as:

Minimize { x 0 ( T ) = ∫ 0 T [ − θ 1 x 1 − θ 2 x 2 − θ 12 x 1 x 2 + log ( 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 ) ] d t } , (4)

subject to:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − D 1 + u 1 , (5)

x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − D 2 + u 2 , (6)

and

x 1 ( t ) ≥ 0, x 2 ( t ) ≥ 0, u 1 ( t ) ≥ 0, u 2 ( t ) ≥ 0 , (7)

where,

t ∈ T , D ( x 1 , x 2 , t ) ≥ 0, θ 1 , θ 2 , θ 12 > 0, ψ 1 , ψ 2 , ψ 12 > 0.

Using the Pontryagin Minimum Principle (PMP), let us define x ˙ 0 = ∂ x 0 ( T ) ∂ t ,

and introduce the co-state variables λ 0 , λ 1 and λ 2 corresponding to the state variables X 0 , X 1 and X 2 respectively. From (4), (5) and (6), we can write the Hamiltonian function as follows:

H = λ 0 x ˙ 0 + λ 1 x ˙ 1 + λ 2 x ˙ 2 , (8)

Moreover, to obtain the co-state equations and the Lagrange multipliers associated with the constraints (5) and (6), we formulate the Lagrangian function as follows:

L = H + μ 1 x 1 + μ 2 x 2 + μ 3 u 1 + μ 4 u 2 , (9)

where, μ 1 ( t ) , μ 2 ( t ) , μ 3 ( t ) , μ 4 ( t ) are known as Lagrange multipliers. These Lagrange multipliers satisfy the conditions:

μ 1 ( t ) ≥ 0 , μ 2 ( t ) ≥ 0 , μ 3 ( t ) ≥ 0 , μ 4 ( t ) ≥ 0 , μ 1 x 1 = 0 , μ 2 x 2 = 0 , μ 3 u 1 = 0 , μ 4 u 2 = 0. (10)

From (9), we can easily obtain the co-state equations

λ ˙ i ( t ) = − ∂ L ∂ x i , i = 0 , 1 , 2 , (11)

then,

λ ˙ 0 ( t ) = − ∂ L ∂ x 0 = 0 , λ ˙ 1 ( t ) = − ∂ L ∂ x 1 , λ ˙ 2 ( t ) = − ∂ L ∂ x 2 , (12)

The first equation of (12) shows that the co-state variable λ 0 ( t ) remains constant along the optimal trajectory, and the Pontryagin principle requires that this constant should be a negative value, Sethi and Thompson [

In this paper, we will use different values for this co-state variable λ 0 ( t ) .

λ 0 = − 10 , λ 0 = − 2 or λ 0 = − 0.1 , (13)

Substituting from (4), (5), (6), (8) and (13) in (9), we can write the Hamiltonian function, L, in the following form, first: when λ 0 ( t ) = − 10 :

L = 10 [ θ 1 x 1 + θ 2 x 2 + θ 12 x 1 x 2 − log ( 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 ) ] + λ 1 [ − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − D 1 + u 1 ] + λ 2 [ − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − D 2 + u 2 ] + μ 1 x 1 + μ 2 x 2 + μ 3 u 1 + μ 4 u 2 . (14)

From conditions (7) and (10), we get

μ 1 ( t ) = μ 2 ( t ) = μ 3 ( t ) = μ 4 ( t ) = 0 . (15)

Substituting from (13) and (14) into (12) we get

λ ˙ 1 = λ 1 ( ∂ D 1 ∂ x 1 + 10 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + λ 2 ( ∂ D 2 ∂ x 1 + a 21 x 2 ) − θ 1 − 10 θ 12 x 2 , (16)

λ ˙ 2 = λ 2 ( ∂ D 2 ∂ x 2 + 10 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + λ 1 ( ∂ D 1 ∂ x 2 + a 12 x 1 ) − θ 2 − 10 θ 12 x 1 , (17)

with boundary conditions

λ i ( T ) ≠ 0 , i = 1 , 2 (18)

where T is the length of planning period which can be suggested.

To obtain the optimal production rates (control variables) U i , i = 1 , 2 , we differentiate the Lagrange function (14) with respect to u 1 , u 2 respectively and putting it equal to zero, we get

∂ L ∂ u 1 = − 10 ( ψ 1 + ψ 12 u 2 ) 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 + λ 1 = 0.

∂ L ∂ u 2 = − 10 ( ψ 2 + ψ 1 u 1 ) 1 + ψ 1 u 1 + ψ 2 u 2 + ψ 12 u 1 u 2 + λ 2 = 0.

Then,

U 1 * ( t ) = 1 λ 1 − 1 + ψ 2 u 2 10 ( ψ 1 + ψ 12 u 2 ) , λ 1 ≠ 0 (19)

U 2 * ( t ) = 1 λ 2 − 1 + ψ 1 u 1 10 ( ψ 2 + ψ 12 u 1 ) , λ 2 ≠ 0 (20)

Since, u 1 and u 2 are start values of the production rates. Then using the Equations (5), (6), (16) and (17), we get the controlled system of non-linear ordinary differential equations:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − D 1 + u 1 x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − D 2 + u 2 λ ˙ 1 = λ 1 ( ∂ D 1 ∂ x 1 + 10 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + λ 2 ( ∂ D 2 ∂ x 1 + a 21 x 2 ) − θ 1 − 10 θ 12 x 2 λ ˙ 2 = λ 2 ( ∂ D 2 ∂ x 2 + 10 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + λ 1 ( ∂ D 1 ∂ x 2 + a 12 x 1 ) − θ 2 − 10 θ 12 x 1 } (21)

We can construct this system when λ 0 = − 2 :

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − D 1 + u 1 x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − D 2 + u 2 λ ˙ 1 = λ 1 ( ∂ D 1 ∂ x 1 + 2 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + λ 2 ( ∂ D 2 ∂ x 1 + a 21 x 2 ) − θ 1 − 2 θ 12 x 2 λ ˙ 2 = λ 2 ( ∂ D 2 ∂ x 2 + 2 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + λ 1 ( ∂ D 1 ∂ x 2 + a 12 x 1 ) − θ 2 − 2 θ 12 x 1 } (22)

and when λ 0 = − 0.1

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − D 1 + u 1 x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − D 2 + u 2 λ ˙ 1 = λ 1 ( ∂ D 1 ∂ x 1 + 0.1 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + λ 2 ( ∂ D 2 ∂ x 1 + a 21 x 2 ) − θ 1 − 0.1 θ 12 x 2 λ ˙ 2 = λ 2 ( ∂ D 2 ∂ x 2 + 0.1 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + λ 1 ( ∂ D 1 ∂ x 2 + a 12 x 1 ) − θ 2 − 0.1 θ 12 x 1 } (23)

The optimal control variables can be constructed when λ 0 = − 2 :

U 1 * ( t ) = 1 λ 1 − 1 + ψ 2 u 2 2 ( ψ 1 + ψ 12 u 2 ) , λ 1 ≠ 0 , (24)

U 2 * ( t ) = 1 λ 2 − 1 + ψ 1 u 1 2 ( ψ 2 + ψ 12 u 1 ) , λ 2 ≠ 0 , (25)

and when λ 0 ( t ) = − 0.1 :

U 1 * ( t ) = 1 λ 1 − 1 + ψ 2 u 2 0.1 ( ψ 1 + ψ 12 u 2 ) , λ 1 ≠ 0 , (26)

U 2 * ( t ) = 1 λ 2 − 1 + ψ 1 u 1 0.1 ( ψ 2 + ψ 12 u 1 ) , λ 2 ≠ 0. (27)

This system can be used to describe the time evolution of inventory levels and production rates. The analytical solution of this system is very difficult and then we can solve it numerically.

The solution of optimal control problem of this model will be carried out using Pontryagin Minimum Principle (PMP). The numerical solution is to be necessary when the analytical solution is absence for the non-linear systems (21, 22 and 23). In this solution we solve the non-linear ordinary differential equations using Runge-Kutta method, using the initial and boundary values for x 1 ( t ) , x 2 ( t ) , λ 1 ( t ) and λ 2 ( t ) . For simplicity we supposed that the initial values u 10 , u 20 can be used alternative to u 1 , u 2 in the Equations (19), (20), (24), (25), (26) and (27) to obtain the optimal production rates U 1 * ( T ) and U 2 * ( T ) respectively. Also, we will use these initial values to obtain the optimal total cost x 0 * ( T ) as it is in the Equation (4).

The numerical solution can be explained by different types of demand as:

1) The demand rates are constant:

D ( x 1 , x 2 , t ) = γ i .

2) The demand rates are linear functions of inventory levels and time:

D ( x 1 , x 2 , t ) = γ i + w i x i .

3) The demand rates are logistic functions of inventory levels and time:

D ( x 1 , x 2 , t ) = 2 x i ( κ i − x i ) .

4) The demand rates are periodic functions of time:

D ( x 1 , x 2 , t ) = 1 − b i cos ( t ) .

where γ i , w i , κ i and b i ( i = 1 , 2 ) are positive constants.

The next subsections explain the controlled system for each case of the demand rates functions with different co-state value λ 0 ( t ) as shown below.

In this subsection we will use different demand rates with λ 0 = − 10 .

We will present the model with demand function as constant rates, D ( x 1 , x 2 , t ) = γ i .

Substituting in the controlled system (21) by the constant demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 2 u 20 10 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 10 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 10 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 10 θ 12 x 2 λ ˙ 2 = λ 2 ( 10 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 10 θ 12 x 1 } (28)

Solving the controlled system (28) numerically, we get some results as displayed in

Also, we will present the model with demand function as linear rates, D ( x 1 , x 2 , t ) = γ i + w i x i . Substituting in the controlled system (21) by the linear demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( w 1 + θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 2 u 20 10 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( w 2 + θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 10 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( w 1 + 10 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 10 θ 12 x 2 λ ˙ 2 = λ 2 ( w 2 + 10 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 10 θ 12 x 1 } (29)

Solving the controlled system (29) numerically, we get some results as displayed in

We present the model with demand function as logistic rates, D ( x 1 , x 2 , t ) = 2 x i ( κ i − x i ) .

u 10 | u 20 | θ 1 | θ 2 | θ 12 | a 12 | a 21 | a 11 | a 22 | γ 1 |
---|---|---|---|---|---|---|---|---|---|

20 | 18 | 0.02 | 0.01 | 0.05 | 0.8 | 0.7 | 0.02 | 0.01 | 0.8 |

x 10 | x 20 | x 20 | w 2 | κ 1 | κ 2 | b 1 | b 2 | T | γ 2 |

1 | 1 | 0.3 | 0.2 | 0.9 | 0.8 | 0.9 | 0.8 | 1 | 0.7 |

ψ 1 | ψ 2 | ψ 12 | λ 1 ( T ) | λ 2 ( T ) | |||||

0.02 | 0.01 | 0.05 | 1 | 1 |

Demand Rates | x 1 ∗ ( T ) | x 2 ∗ ( T ) | u 1 ∗ ( T ) | u 2 ∗ ( T ) | x 0 ∗ ( T ) |
---|---|---|---|---|---|

Constant | 0.88 | 0.95 | 0.87 | 0.86 | 0.07 |

Linear | 0.81 | 0.87 | 0.87 | 0.86 | 0.10 |

Logistic | 0.97 | 1.10 | 0.87 | 0.86 | 0.00 |

Periodic | 1.16 | 1.14 | 0.87 | 0.86 | 0.06 |

Substituting in the controlled system (21) by the logistic demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( 2 ( κ 1 − x 1 ) + θ 1 + a 12 x 2 + a 11 x 1 ) + 1 λ 1 − 1 + ψ 2 u 20 10 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( 2 ( κ 2 − x 2 ) + θ 2 + a 21 x 2 + a 22 x 2 ) + 1 λ 2 − 1 + ψ 1 u 10 10 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 2 ( κ 1 − 2 x 1 + a 11 x 1 ) + 10 θ 1 + a 12 x 2 ) + a 21 λ 2 x 2 − θ 1 − 10 θ 12 x 2 λ ˙ 2 = λ 2 ( 2 ( κ 2 − 2 x 2 + a 22 x 2 ) + 10 θ 2 + a 21 x 1 ) + a 12 λ 1 x 1 − θ 2 − 10 θ 12 x 1 } (30)

Solving the controlled system (30) numerically, we get some results as displayed in

Finally, we will present the model with demand function as periodic rates, D ( x 1 , x 2 , t ) = 1 − b i cos ( t ) . Substituting in the controlled system (21) by the periodic demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − 1 + b 1 cos ( t ) + 1 λ 1 − 1 + ψ 2 u 20 10 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − 1 + b 2 cos ( t ) + 1 λ 2 − 1 + ψ 2 u 10 10 ( ψ 1 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 10 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 10 θ 12 x 2 λ ˙ 2 = λ 2 ( 10 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 10 θ 12 x 1 } (31)

Solving the controlled system (31) numerically, we get some results as displayed in

As we see from

Also, in this subsection we use the previous demand rates with λ 0 = − 2 .

We will present the model with demand function as constant rates, D ( x 1 , x 2 , t ) = γ i . Substituting in the controlled system (22) by the constant demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 2 u 20 2 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 2 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 2 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 2 θ 12 x 2 λ ˙ 2 = λ 2 ( 2 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 2 θ 12 x 1 } (32)

Solving the controlled system (32) numerically, we get some results which are displayed in

Also, we will present the model with demand function as linear rates, D ( x 1 , x 2 , t ) = γ i + w i x i . Substituting in the controlled system (22) by the linear demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( w 1 + θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 2 u 20 2 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( w 2 + θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 2 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( w 1 + 2 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 2 θ 12 x 2 λ ˙ 2 = λ 2 ( w 2 + 2 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 2 θ 12 x 1 } (33)

Solving the controlled system (33) numerically, we get the results displayed in

We present the model with demand function as logistic rates, D ( x 1 , x 2 , t ) = 2 x i ( κ i − x i ) . Substituting in the controlled system (22) by the logistic demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( 2 ( κ 1 − x 1 ) + θ 1 + a 12 x 2 + a 11 x 1 ) + 1 λ 1 − 1 + ψ 2 u 20 2 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( 2 ( κ 2 − x 2 ) + θ 2 + a 21 x 2 + a 22 x 2 ) + 1 λ 2 − 1 + ψ 1 u 10 2 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 2 ( κ 1 − 2 x 1 + a 11 x 1 ) + 2 θ 1 + a 12 x 2 ) + a 21 λ 2 x 2 − θ 1 − 2 θ 12 x 2 λ ˙ 2 = λ 2 ( 2 ( κ 2 − 2 x 2 + a 22 x 2 ) + 2 θ 2 + a 21 x 1 ) + a 12 λ 1 x 1 − θ 2 − 2 θ 12 x 1 } (34)

Solving the controlled system (30) numerically, we get the results are displayed in

Finally, we will present the model with demand function as periodic rates, D ( x 1 , x 2 , t ) = 1 − b i cos ( t ) . Substituting in the controlled system (22) by the periodic demand rates, we have the controlled system:

Demand Rates | x 1 ∗ ( T ) | x 2 ∗ ( T ) | u 1 ∗ ( T ) | u 2 ∗ ( T ) | x 0 ∗ ( T ) |
---|---|---|---|---|---|

Constant | 1.42 | 1.57 | 0.36 | 0.31 | 0.00 |

Linear | 1.30 | 1.42 | 0.36 | 0.31 | 0.00 |

Logistic | 0.65 | 0.68 | 0.36 | 0.31 | 0.00 |

Periodic | 1.57 | 1.69 | 0.36 | 0.31 | 0.00 |

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − 1 + b 1 cos ( t ) + 1 λ 1 − 1 + ψ 2 u 20 2 ( ψ 1 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − 1 + b 2 cos ( t ) + 1 λ 2 − 1 + ψ 1 u 10 2 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 2 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 2 θ 12 x 2 λ ˙ 2 = λ 2 ( 2 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 2 θ 12 x 1 } (35)

Solving the controlled system (35) numerically, we get the results can be displayed in

We will use the parameters values as they are in

As we see from

Finally, we use the previous demand rates with λ 0 = − 0.1 .

We will present the model with demand function as constant rates, D ( x 1 , x 2 , t ) = γ i .

Substituting in the controlled system (23) by the constant demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 1 u 20 0.1 ( ψ 2 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 0.1 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( 0.1 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 0.1 θ 12 x 2 λ ˙ 2 = λ 2 ( 0.1 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 0.1 θ 12 x 1 } (36)

Solving the controlled system (36) numerically, we get the results displayed in

Also, we will present the model with demand function as linear rates, D ( x 1 , x 2 , t ) = γ i + w i x i . Substituting in the controlled system (23) by the linear demand rates, we have the controlled system:

x ˙ 1 = − x 1 ( w 1 + θ 1 + a 12 x 2 + a 11 x 1 ) − γ 1 + 1 λ 1 − 1 + ψ 1 u 20 0.1 ( ψ 2 + ψ 12 u 20 ) x ˙ 2 = − x 2 ( w 2 + θ 2 + a 21 x 2 + a 22 x 2 ) − γ 2 + 1 λ 2 − 1 + ψ 1 u 10 0.1 ( ψ 2 + ψ 12 u 10 ) λ ˙ 1 = λ 1 ( w 1 + 0.1 θ 1 + a 12 x 2 + 2 a 11 x 1 ) + a 21 λ 2 x 2 − θ 1 − 0.1 θ 12 x 2 λ ˙ 2 = λ 2 ( w 2 + 0.1 θ 2 + a 21 x 1 + 2 a 22 x 2 ) + a 12 λ 1 x 1 − θ 2 − 0.1 θ 12 x 1 } (37)

Demand Rates | x 1 ∗ ( T ) | x 2 ∗ ( T ) | |||
---|---|---|---|---|---|

Constant | 0.56 | 0.61 | 0.00 | 0.00 | 0.67 |

Linear | 0.33 | 0.40 | 0.00 | 0.00 | 0.60 |

Logistic | 0.37 | 0.13 | 0.00 | 0.00 | 0.06 |

Periodic | 0.21 | 0.24 | 0.00 | 0.00 | 0.16 |

Solving the controlled system (37) numerically, we get the results displayed in

We present the model with demand function as logistic rates,

Substituting in the controlled system (23) by the logistic demand rates, we have the controlled system:

Solving the controlled system (38) numerically, we get the results displayed in

Finally, we will present the model with demand function as periodic rates,

Solving the controlled system (39) numerically, we get the results displayed in

We will use the parameters values as they are in

Constant case

Linear case

Logistic case

Periodic case

As we see from

In this section will compare between the results when the co-state value

The optimal production rates, when

Also, we can use the initial values

The controlled systems are become in each case as follow.

The controlled system is become:

The controlled system is become:

The controlled system is become:

The controlled system is become:

Constant case

Linear case

Logistic case

Periodic case

The optimal cost

As we see from

So, we can conclude that the co-state value

In this study, we discussed the optimal control problem using the deterioration and spoilage function, as the total cost must be minimized at the end of planning period, depending on the alternative quadratic exponential form (AQEF). We have used different values for the co-state variable

Demand Rates | |||||
---|---|---|---|---|---|

Constant | 1.77 | 1.90 | 0.85 | 0.76 | 1.83 |

Linear | 1.5 | 1.56 | 0.48 | 0.26 | 1.10 |

Logistic | 0.89 | 0.90 | 0.76 | 0.61 | 0.02 |

Periodic | 1.86 | 1.95 | 0.75 | 0.60 | 1.41 |

co-state variable

El-Sayed, A.M.M. (2018) Studying the Changes of an Optimal Trajectory. Open Access Library Journal, 5: e4716. https://doi.org/10.4236/oalib.1104716