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Analysis of monitoring and limiting of commercial cheating: a newsvendor model

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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Correspondence to K K Lai.

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 M2, . Taking the derivative of with respect to y1, we can obtain . That is

On the curve M1, , also we have

Then we can derive that

On the curve Γ2, G(y1, Ψ2(y1))=K2+G(y1, Y2(y1)), and then,

Now we obtain

Similarly, on the curve Γ1, we have G1(y2), y2)= K1+G(Y1(y2), y2) 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, XO1, XO2, XO12 and XŌ.

(1) For the case XO1, we increase X to Y′. Then, we measure three possible subcases:

Subcase (1.1): Assume that y1′>x1 and y2′>x2. When x2S2, recall the definition of M1 and with the help of Proposition 1, we have K+G(Y′)⩾K+G(Y1(y2′), y2′)⩾ K+G(Y1(x2), x2)>K1+G(Y1(x2), x2). Otherwise, x2<s2, and we also have K+G(Y′)⩾K+G(S)=K1+G(S′)⩾ K1+G(Y1(x2), x2).

Subcase (1.2): Assume that y1′>x1 and y2′>x2. If x2⩾Ψ2(x1), recalling the definition of Ψ2, we have K2+G(Y′)⩾G(X)>K1+G(Y1(x2),x2). Otherwise, x22(x1). Then, let Ȳ=(1, x2)∈Γ2, and we have

Subcase (1.3): Assume that y1′>x1 and y2′>x2. Obviously, we have K1+G(Y′)⩾K1+G(Y1(x2), x2).

Combining the three subcases, we always have G(X)⩾K+G(Y1(x2), x2), which means that to increase inventory X to (Y1(x2), x2) is optimal in this case.

(2) For the case XO2, the proof is similar to that of part (1).

(3) For the case XO12, we increase X to Y′. Then three subcases will be considered.

Subcase (3.1): Assume that y1′>x1 and y2′>x2. Obviously, we have K+G(Y′)⩾K+G(S).

Subcase (3.2): Assume that y1′>x1 and y2′>x2. Recalling the definition of Y1(·), we have K1+G(Y′)⩾K1+G(Y1(x2), x2)⩾K1+G(S′)=K+G(S).

Subcase (3.3): Assume that y1′>x1 and y2′>x2. By the definition of Y2(·), we have K2+G(Y)⩾K2+G(x1, Y2(x1))⩾ K2+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 XŌ, we have four subcases.

Subcase (4.1): Assume that X⩾Γ1, X⩾Γ2, X⩾Γ and X<S. We increase inventory X to Y′. If y1′>x1 and y2′>x2, then we have K+G(Y′)⩾K+G(S)⩾G(X). If y1′>x1 and y2′>x2, then we have K1+G(Y′)⩾K1+G(Y1(x2), x2)⩾G(X). If y1′>x1 and y2′>x2, then we also have K2+G(Y′)⩾K2+G(x1, Y2(x1))⩾G(X). It means that doing nothing is an optimal selection.

Subcase (4.2): Assume that Γ1X<M1 and x2s2. We increase inventory X to Y′. When y1′>x1 and y2′>x2, we have K+G(Y′)⩾K+G(Y1(y2′) y2′)>K1+G(Y1(x2), x2)⩾ G(X). When y1′>x1 and y2′>x2, we have K1+G(Y′)⩾ K1+G(Y1(y2′) y2′)⩾K1+G(Y1(x2), x2)⩾G(X). When y1′> x1 and y2′>x2, and if X<M2, we have K2+G(Y′)⩾ K2+ G(x1, Y2(x1))⩾G(X); otherwise, XM2, and we also have K2+G(Y′)⩾K2+G(X)>G(X). Now we show that doing nothing is an optimal selection.

Subcase (4.3): Assume that Γ2X<M2 and x1s1. 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 XM1 and XM2, 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 XO1, XO2, XO12, XO and XŌ.

(1) For the case XO1, when x22, the proof is similar to that of part (1) in Theorem 1.

When x2<2, we increase X to Y′ and consider three subcases.

Subcase (1.1): If y1′>x1 and y2′>x2, we have

Subcase (1.2): If y1′=x1 and y2′>x2, we obtain

Subcase (1.3): If y1′>x1 and y2′=x2, obviously, we have K1+G(Y′)⩾K1+G(Y1(x2), x2).

The three subcases mean that to increase inventory X to (Y1(x2), x2) is optimal in this case.

(2) For the case XO2, the proof is similar to that of the part (1).

(3) For the case XO12, the proof is similar to that of Theorem 1.

(4) For the case XO, clearly we have

and we also have

So, the optimal action is to choose one Y from (x1, Y2(x1)) and (Y1(x2), x2) such that V(Y) achieves the minimal value.

(5) For the case XŌ, the proof is similar to that of Theorem 1.  

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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