**Appendix A**

**Proof of Lemma 2.** In proving Lemma 2, we consider the assertions of Lemma 2 item-wise.

(A) Taking the first-order derivative of *G*(*T*) with respect to *T*, we obtain

$$\begin{array}{l} G'(T) &= \frac{D(1+m-t\_d)t\_dT}{\left(1+m-T\right)^2} + \frac{2pD^2}{x\left(1-p\right)^2} \left\{ \left[t\_d + \left(1+m-t\_d\right)\right]t\_d \right\} \\ & \quad \ln\left(\frac{1+m-t\_d}{1+m-T}\right) \cdot \frac{\left(1+m-t\_d\right)T}{\left(1+m-T\right)^2} + \frac{\left(1+m-t\_d\right)^2T}{\left(1+m-T\right)^2} \right\} \\ & \quad + \frac{D}{2}\left(1+m-t\_d\right)^2 \cdot \frac{T}{\left(1+m-T\right)^2} + \frac{DT}{2} \\ & \quad > 0 \end{array}$$

and

(B)

$$G(t\_d) = \frac{pD^2}{x(1-p)^2} > 0$$

Furthermore, we see that *G*(*T*) > 0 if *T* ≥ *td*. We define *g*(*T*) as follows:

$$\log(T) = D(1+m-t\_d)\frac{T}{1+m-T} - Dt\_d - D(1+m-t\_d)\cdot \ln(\frac{1+m-t\_d}{1+m-T})$$

Taking the first-order derivative of *g*(*T*) with respect to *T*, we derive

$$g'(T) = \frac{D(1+m-t\_d)T}{\left(1+m-T\right)^2} > 0$$

and

$$\lg(t\_d) = 0$$

Furthermore, we have

$$D(1+m-t\_d)\frac{T}{1+m-T} - Dt\_d - D(1+m-t\_d)\ln(\frac{1+m-t\_d}{1+m-T}) > 0 \text{ if } T \ge t\_d$$

(C) Taking the first-order derivative of *H*(*T*) with respect to *T*, we obtain

$$\begin{array}{rcl}H'(T) &=& \frac{4pD^2}{x(1-p)^2}(1+m-t\_d)[t\_d+(1+m-t\_d)\cdot \ln(\frac{1+m-t\_d}{1+m-T})]\frac{T^2}{\left(1+m-T\right)^3} \\ &+\frac{6pD^2}{x(1-p)^2}(1+m-t\_d)^2\frac{T^2}{\left(1+m-T\right)^3} \end{array}$$

and

$$H(t\_d) = \frac{Dt\_d^3}{(1+m-t\_d)} + \frac{2pD^2}{\varkappa(1-p)^2} \cdot \frac{t\_d^3}{(1+m-t\_d)} > 0$$

Furthermore, we have *H*(*T*) > 0 if *T* ≥ *td*.

(D) We define *h*(*T*) by

$$\begin{array}{l} h(T) &= D(1+m-t\_d) \frac{(1+m)T}{(1+m-T)^2} - 3D(1+m-t\_d) \frac{T}{1+m-T} + 2Dt\_d \\ &+ 2D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T}) \end{array}$$

Taking the first-order derivative of *h*(*T*) with respect to *T*, we find that

$$h'(T) = D(1 + m - t\_d) \frac{2T^2}{\left(1 + m - T\right)^3} > 0$$

and

$$h(t\_d) = \frac{D t\_d^2}{(1 + m - t\_d)} > 0$$

So, we finally have

$$\begin{cases} D(1+m-t\_d) \frac{(1+m)T}{(1+m-T)^2} - 3D(1+m-t\_d) \frac{T}{1+m-T} \\ + 2Dt\_d + 2D(1+m-t\_d) \cdot \ln(\frac{1+m-t\_d}{1+m-T}) > 0 \end{cases} \tag{A1}$$

if *T* ≥ *td*.

> We thus have completed the proof of Lemma 2. -
