## Abstract

Commercial cheating, that is, counterfeit products and lower quality products sold as genuine products, exists extensively in many countries of the world, especially in the developing countries. In this paper, we investigate the phenomena of commercial cheating, study the optimal cheating actions of inventory managers under a monitoring and limiting regime from the industrial administration office (IAO for short) and demonstrate the efficiency of monitoring and limiting such cheating activities. A newsvendor model has been considered for the inventory manager to order different quality products with different set-up costs. The model, a kind of extension of the general newsvendor problem, is viewed as a shocked inventory model. We analyse some properties of the optimal cheating policies from the point of view of an inventory manager, and investigate the effectiveness of both the punishment level and the checking rate.

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## Acknowledgements

We thank the anonymous referees for their valuable comments that helped improve this paper. This paper was supported partially by the National Natural Science Foundation of China under Grants 60274050, 70221001 and 70328001.

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## Appendix A

### Appendix A

In this Appendix, we present, the proof of Proposition 2 and Theorems 1 and 2.

### A.1. Proof of Proposition 2

**Proof** On the curve *M*_{2}, . Taking the derivative of with respect to *y*_{1}, we can obtain . That is

On the curve *M*_{1}, , also we have

Then we can derive that

On the curve Γ_{2}, *G*(*y*_{1}, Ψ_{2}(*y*_{1}))=*K*_{2}+*G*(*y*_{1}, *Y*_{2}(*y*_{1})), and then,

Now we obtain

Similarly, on the curve Γ_{1}, we have *G*(Ψ_{1}(*y*_{2}), *y*_{2})= *K*_{1}+*G*(*Y*_{1}(*y*_{2}), *y*_{2}) and then

Simplifying the above equation, we have

The conclusion holds if and only if

The last inequality holds because

Now the proof of the proposition is complete.

### A.2. Proof of Theorem 1

**Proof** Now we prove this theorem in four cases, *X*∈*O*_{1}, *X*∈*O*_{2}, *X*∈*O*_{12} and *X*∈*Ō*.

(1) For the case *X*∈*O*_{1}, we increase *X* to *Y*′. Then, we measure three possible subcases:

*Subcase (1.1)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. When *x*_{2}⩾*S*_{2}, recall the definition of *M*_{1} and with the help of Proposition 1, we have *K*+*G*(*Y*′)⩾*K*+*G*(*Y*_{1}(*y*_{2}′), *y*_{2}′)⩾ *K*+*G*(*Y*_{1}(*x*_{2}), *x*_{2})>*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2}). Otherwise, *x*_{2}<*s*_{2}, and we also have *K*+*G*(*Y*′)⩾*K*+*G*(*S*)=*K*_{1}+*G*(*S*′)⩾ *K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2}).

*Subcase (1.2)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. If *x*_{2}⩾Ψ_{2}(*x*_{1}), recalling the definition of Ψ_{2}, we have *K*_{2}+*G*(*Y*′)⩾*G*(*X*)>*K*_{1}+*G*(*Y*_{1}(*x*_{2}),*x*_{2}). Otherwise, *x*_{2}<Ψ_{2}(*x*_{1}). Then, let *Ȳ*=(*x̄*_{1}, *x*_{2})∈Γ_{2}, and we have

*Subcase (1.3)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. Obviously, we have *K*_{1}+*G*(*Y*′)⩾*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2}).

Combining the three subcases, we always have *G*(*X*)⩾*K*+*G*(*Y*_{1}(*x*_{2}), *x*_{2}), which means that to increase inventory *X* to (*Y*_{1}(*x*_{2}), *x*_{2}) is optimal in this case.

(2) For the case *X*∈*O*_{2}, the proof is similar to that of part (1).

(3) For the case *X*∈*O*_{12}, we increase *X* to *Y*′. Then three subcases will be considered.

*Subcase (3.1)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. Obviously, we have *K*+*G*(*Y*′)⩾*K*+*G*(*S*).

*Subcase (3.2)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. Recalling the definition of *Y*_{1}(·), we have *K*_{1}+*G*(*Y*′)⩾*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2})⩾*K*_{1}+*G*(*S*′)=*K*+*G*(*S*).

*Subcase (3.3)*: Assume that *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}. By the definition of *Y*_{2}(·), we have *K*_{2}+*G*(*Y*)⩾*K*_{2}+*G*(*x*_{1}, *Y*_{2}(*x*_{1}))⩾ *K*_{2}+*G*(*S*″)=*K*+*G*(*S*).

Combining the three subcases, and we always have *G*(*X*)>*K*+*G*(*S*), which means that increasing inventory *X* to *S* is an optimal selection in this case.

(4) For the case *X*∈*Ō*, we have four subcases.

*Subcase (4.1)*: Assume that *X*⩾Γ_{1}, *X*⩾Γ_{2}, *X*⩾Γ and *X*<*S*. We increase inventory *X* to *Y*′. If *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, then we have *K*+*G*(*Y*′)⩾*K*+*G*(*S*)⩾*G*(*X*). If *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, then we have *K*_{1}+*G*(*Y*′)⩾*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2})⩾*G*(*X*). If *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, then we also have *K*_{2}+*G*(*Y*′)⩾*K*_{2}+*G*(*x*_{1}, *Y*_{2}(*x*_{1}))⩾*G*(*X*). It means that doing nothing is an optimal selection.

*Subcase (4.2)*: Assume that Γ_{1}⩽*X*<*M*_{1} and *x*_{2}⩾*s*_{2}. We increase inventory *X* to *Y*′. When *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, we have *K*+*G*(*Y*′)⩾*K*+*G*(*Y*_{1}(*y*_{2}′) *y*_{2}′)>*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2})⩾ *G*(*X*). When *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, we have *K*_{1}+*G*(*Y*′)⩾ *K*_{1}+*G*(*Y*_{1}(*y*_{2}′) *y*_{2}′)⩾*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2})⩾*G*(*X*). When *y*_{1}′> *x*_{1} and *y*_{2}′>*x*_{2}, and if *X*<*M*_{2}, we have *K*_{2}+*G*(*Y*′)⩾ *K*_{2}+ *G*(*x*_{1}, *Y*_{2}(*x*_{1}))⩾*G*(*X*); otherwise, *X*⩾*M*_{2}, and we also have *K*_{2}+*G*(*Y*′)⩾*K*_{2}+*G*(*X*)>*G*(*X*). Now we show that doing nothing is an optimal selection.

*Subcase (4.3)*: Assume that Γ_{2}⩽*X*<*M*_{2} and *x*_{1}⩾*s*_{1}. We increase inventory *X* to *Y*′. Using a similar analysis, we can derive that doing nothing is an optimal selection.

*Subcase (4.4)*: In the subcase of *X*⩾*M*_{1} and *X*⩾*M*_{2}, we also can derive that doing nothing is optimal similarly.

So for the case 4, we know that doing nothing is optimal.

### A.3. Proof of Theorem 2

**Proof** Now we prove this theorem in five cases, that is *X*∈*O*_{1}, *X*∈*O*_{2}, *X*∈*O*_{12}, *X*∈*O* and *X*∈*Ō*.

(1) For the case *X*∈*O*_{1}, when *x*_{2}⩾*s̄*_{2}, the proof is similar to that of part (1) in Theorem 1.

When *x*_{2}<*s̄*_{2}, we increase *X* to *Y*′ and consider three subcases.

*Subcase (1.1)*: If *y*_{1}′>*x*_{1} and *y*_{2}′>*x*_{2}, we have

*Subcase (1.2)*: If *y*_{1}′=*x*_{1} and *y*_{2}′>*x*_{2}, we obtain

*Subcase (1.3)*: If *y*_{1}′>*x*_{1} and *y*_{2}′=*x*_{2}, obviously, we have *K*_{1}+*G*(*Y*′)⩾*K*_{1}+*G*(*Y*_{1}(*x*_{2}), *x*_{2}).

The three subcases mean that to increase inventory *X* to (*Y*_{1}(*x*_{2}), *x*_{2}) is optimal in this case.

(2) For the case *X*∈*O*_{2}, the proof is similar to that of the part (1).

(3) For the case *X*∈*O*_{12}, the proof is similar to that of Theorem 1.

(4) For the case *X*∈*O*, clearly we have

and we also have

So, the optimal action is to choose one *Y* from (*x*_{1}, *Y*_{2}(*x*_{1})) and (*Y*_{1}(*x*_{2}), *x*_{2}) such that *V*(*Y*) achieves the minimal value.

(5) For the case *X*∈*Ō*, the proof is similar to that of Theorem 1.

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### Cite this article

Liu, K., Li, JA., Wu, Y. *et al.* Analysis of monitoring and limiting of commercial cheating: a newsvendor model.
*J Oper Res Soc* **56, **844–854 (2005). https://doi.org/10.1057/palgrave.jors.2601913

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### Keywords

- inventory
- counterfeit
- optimal policy
- checking rate