Fig. 1 Recover amount statistics and analytics. Source: Environmental Protection Police (EPA (2023))

Reusing/remanufacturing recovers the value of end-of-life products by reusing their durable components to manufacture a product with the original functionality. Green investments are important investments such as remanufacturing policies and mutual funds that target sustainable strategies. Here is an example from a real manufacturing company in Taiwan. At the end of a bike’s life, it can either end up in a landfill or its components can be recycled, saving large amounts of energy. The company has an advanced recycling system to reduce the waste of components and raw materials. Figure 2 summarizes the flow chart of the recycling system. The system comprises one manufacturer and one retailer. Rebate agreements play a crucial role in the supply chain as they provide companies with an effective tool to incentivize their customers. The program encourages retailers to engage in manufacturer rebates within a green supply chain and examines its impact on the green product sales.

Fig. 3 Graph illustrating inventory levels for components, end products, and returned items

Deriving the maximum inventory level of the component i based on Eqs. (1) and (2) can be formulated as:

Hi=(pi−χe)ti=χetid. (3)

At the period t_id, inventory depletion of the component i is:

tid=Heχe=(pi−χe)tiχe, (4)

where i=1,2,…,n.

The maximum inventory level of the end product can be expressed as follows:

He=(χe−D)te=(χe−D)(tn+tnd)=(χe−D)pnχetn.(fromEq.(4)) (5)

At the period t_rd, inventory depletion of the return product is:

Hr=pi(1−θr)trd=rtr. (6)

This model is composed of the following nine elements:

SR=(sp−s)DT=(sp−s)pntn.

H C e = h e H e T 2 = h e ( χ e − D ) 2 χ e D p n 2 t n 2 .

H C s = ∑ i = 1 n h i H i ( t i + t id ) 2 = p n 2 t n 2 2 [ ∑ i = 1 n h i ( 1 χ e − 1 p i ) ] .

H C r = h r r t r 2 2 .

R C = r [ c r θ r + θ e D T + ∑ i = 1 n θ i DT ] .

R C r = c w r ( 1 − θ r ) .

P C = ( β 0 + β 1 χ e + β 2 χ e ) p n t n .

I C = γ ( F r + V r H r t rd ) = γ [ F r + V r ( r t r ) t rd ] .

To summarize the above results, total profit per unit time (denoted byAP(s,tn,r)) can be obtained as follows:

AP(s,tn,r)=1T(SR−SC−HCe−HCs−HCr−RC−RCr−PC−IC) =1T{(sp−s)pntn−k−he(χe−D)2χeDpn2tn2−pn2tn22[∑i=1nhi(1χe−1pi)]−r[crθr+θeDT+∑i−1nθiDT]−(β0+β1χe+β2χe)pntn −hrrtr22−cwr(1−θr)−γ[Fr+Vr(rtr)trd]}. (7)

Now, the first-order partial derivatives of the total profit should be calculated per unit time with respect to s, t_n, and r. Then the following is obtained:

∂AP(s,tn,r)∂s=1T{−pntn+[he2D2δ+he2χe]pn2tn2−r[θeδT+δ∑i−1nθiT]}, (8)

∂AP(s,tn,r)∂tn=1T{(sp−s)pn−he(χe−D)χeDpn2tn−pn2tn[∑i=1nhi(1χe−1pi)]−(β0+β1χe+β2χe)pn}, (9)

and

∂AP(s,tn,r)∂s=1T{−pntn+[he2D2δ+he2χe]pn2tn2−r[δθeT+δ∑i−1nθiT]},∂AP(s,tn,r)∂r=1T{[he2D2δ+he2χe]pn2tn2+[he2D2(1−δ−θr)−he2χe]pn2tn2−[crθr+DθeT+D∑i−1nθiT]−r[θeδT+δT∑i−1nθi]−r(1−δ−θr)[θeT+T∑i−1nθi]−hrtr22−cw(1−θr)−γVrtrtrd}. (10)

The optimal solutions for (s,tn,r) satisfy the equations ∂AP(s,tn,r)/∂s=0, ∂AP(s,tn,r)/∂tn=0 and ∂AP(s,tn,r)/∂r=0, simultaneously. The necessary condition for AP(s, tn, r) to be maximum is ∂AP(s,tn,r)/∂tn=0, which gives:

(sp−s)pn−he(χe−D)χeDpn2tn−pn2tn[∑i=1nhi(1χe−1pi)]−(β0+β1χe+β2χe)pn=0 (11)

It is not easy to find the closed-form solution of t_n from Eq. (11). But the value of t_n that satisfies Eq. (11) not only exists, but is also unique, as stated by the following lemma:

Lemma 1.

The solution of tn (say tn*) profit per profit per unit time AP(s,tn,r) has the global maximum value at the point tn=tn*, where tn*∈(0,∞) and satisfies Eq. (11).

Proof.

Let:

F(tn)=(sp−s)pn−he(χe−D)χeDpn2tn−pn2tn[∑i=1nhi(1χe−1pi)]−(β0+β1χe+β2χe)pn (12)

for tn∈(0,∞). Taking the first derivative of F(tn) with respect to t_n, it gives:

∂ F ( t n ) ∂ t n = − p n 2 { h e ( χ e − D ) χ e D + [ ∑ i = 1 n h i ( 1 χ e − 1 p i ) ] } < 0 ,

i.e.,F(tn) is a strictly decreasing function of t_n. Then, it is obtained limtn→∞F(tn)=−∞ and limtn→0F(tn)=[(sp−s)−(β0+β1χe+β2χe)]pn.

Following two cases are possible:

CASE 1 If 0]]> ( s p − s ) − ( β 0 + β 1 χ e + β 2 χ e ) > 0 , then the solution t n * which maximizes AP ( s , t n , r ) not only exists, but is also unique when applying the Intermediate Value Theorem, where t n * ∈ ( 0 , ∞ ) .

In this case, t_n is also greater than or equal to zero. The most important finding in this case is that the return profits for defective products are greater than the production costs. If the return is made within a limited time frame of 30 days, the manufacturer will provide a courtesy credit to the account for a cash refund for all defective products. Since the sufficient condition is met, taking the second derivative of AP(s,tn,r)with respect to t_n, and then substituting tn=tn*into it, the following is obtained:

∂ 2 A P ( s , t n , r ) ∂ t n 2 | t n = t n * = − { [ ∑ i = 1 n h i ( 1 χ e − 1 p i ) + h e ( χ e − D ) χ e D ] + ( β 0 + β 1 χ e + β 2 χ e ) p n } < 0 .

Therefore, t n * is the global maximum point of AP ( s , t n , r ) . The proof is now complete.

CASE 2If(sp−s)−(β0+β1χe+β2χe)<0, then the optimal solution istn*=0.

In this case, t_n is negative and against tn≥0 because return profits for defective products are less than production costs. At this point, the manufacturer stops offering a refund policy. Therefore, the optimal length of the production run time of the component n should approach zero. Then substituting tn=tn*into Eq. (7), the total profit per unit time AP(s,tn,r)can bemodified to a new function of (s, r), given by:

AP(s,r|tn=tn*)=1T{(sp−s)pntn*−k−he(χe−D)2χeDpn2tn2*−pn2tn2*2[∑i=1nhi(1χe−1pi)]−r[crθr+θeDT+∑i−1nθiDT]−(β0+β1χe+β2χe)pntn* −hrrtr22−cwr(1−θr)−γ[Fr+Vr(rtr)trd]}. (13)

Due to the complexity of the model, it is difficult to find the close form of (s, r) and check the concavity of the manufacturer’s profit function directly. Alternatively, in the following section, the concavity by numerical analysis will be verified and then a simple algorithm developed to obtain the solutions for the manufacturer.