Dynamics of sensor-based information in supply chains with perishables substitutable by non-perishables

Abstract

Supply chains with either perishables or non-perishables have been well-studied as evidenced through extant published literature. Among these studies, very few consider supply chains with both perishable and non-perishable products. Since the early 2000s, RFID (Radio-Frequency IDentification) tags have been increasingly used in supply chains that deal with perishables as well as non-perishables. While there is a reasonably large amount of published literature on RFID use in supply chains, we are unaware of any that considers the dynamics of RFID-generated information in supply chains that simultaneously involve perishables substitutable by non-perishables in retail environments. We attempt to address this void. We consider the relative benefits of sensor-enabled RFID tag use in supply chains that simultaneously contain perishables substitutable by non-perishables. We also derive expressions for conditions on their dynamics through specific consideration of their pre-determined and actual expiry dates. We operationalize our analysis from the perspective of retailers and customers.

Appendix

Proof for negative second derivative of (3)

This is seen from the six terms in this expression in which the first, fourth, fifth and sixth are effectively always negative and the third is positive. The second term is positive or negative depending on the value of n.

$$\begin{aligned}&2\bigg [\frac{I^p_r(1-n)^2((1 + \lambda t_e)e^{-\lambda t_e} - 1)}{\lambda ^2 } + {Ts^nn}(1.5n-1)\nonumber \\&\quad + \frac{s^p}{2\lambda }(1 - n)^2(1 - e^{-\lambda t_e})- \frac{I_r^n T^2n^2}{2} - {P_r^nnT} - \frac{(1-n)^2}{2\lambda }A \bigg ] \end{aligned}$$

(20)

We show that the second derivative is negative in two steps. First we show that the magnitude of the third term is less than that of the last term. We then show that the magnitude of the second term is less than the fifth term.

$$\begin{aligned} \frac{s^p}{2\lambda }(1 - n)^2(1 - e^{-\lambda t_e}) \lessgtr \frac{(1-n)^2}{2\lambda }A \end{aligned}$$

which is the same as $s^p(1 - e^{-\lambda t_e}) \lessgtr P^p_r(1 - e^{-\lambda t_e})$ substituting $P_r^{{\underline{p}}}$ for $P_r^p$ as this is more restrictive which reduces to $s^p < P^p_r$.

Since the second term could be positive sometimes, we consider the magnitude of the second term and the fifth term for a fixed s to show that the latter dominates the former thereby rendering the entire second derivative negative.

$$\begin{aligned} {Ts^nn}(1.5n-1) \lessgtr {P^n_rnT} \end{aligned}$$

which is the same as $s^n(1.5n-1) \lessgtr {P^n_r}$.

The highest value of LHS occurs when $n=1$. When $n=1$, we have $0.5s^n < P^n_r$. Similarly, the lowest value of LHS occurs when $n=0$. When $n=0$, we have $-s^n < P^n_r$. $\square$

Proof for negative second derivative of (5)

The second derivative of (5), which is the denominator of (6), is negative as all except for the second term are always effectively negative.

$$\begin{aligned} {\bigg [{(s^p- P^p_r)(1 - n)^2 t_{q^*}} + {2s^nTn(1.5n-1)} - {I^p_r(1-n)^2t_{q^{*}}^2} - {I^n_rT^2n^2} - {P^n_rn^2T}\bigg ]} \end{aligned}$$

Since the second term could be positive sometimes, we consider the magnitude of the second term and the last term for a fixed s to show that the latter dominates the former thereby rendering the entire second derivative negative.

$$\begin{aligned} {2s^nTn(1.5n-1)} \lessgtr {P^n_rn^2T} \end{aligned}$$

which is the same as $2s^n(1.5n-1)\lessgtr {P^n_rn}$.

The highest value of LHS occurs when $n=1$. When $n=1$, we have $s^n < P^n_r$. Similarly, the lowest value of LHS occurs when $n=0$. When $n=0$, we have $-2s^n < 0$. $\square$

Proof for negative second derivative of (7)

The second term could be positive sometimes depending on the value of n.

$$\begin{aligned}&\bigg [\frac{(s^p - P^p_r)}{\lambda }(1 - n)^2(1 - e^{-\lambda t_e}) + \frac{s^nTn}{2}(1.5n -1) \\&\quad - {P^n_rn^2T} - {I^n_rn^2T^2} + \frac{I_r^p(1-n)^2}{2\lambda ^2}(e^{-\lambda t_e}(1 + \lambda t_e) - 1) \bigg ] \end{aligned}$$

Since the second term could be positive sometimes, we consider the magnitude of the second term and the third term for a fixed s to show that the latter dominates the former thereby rendering the entire second derivative negative.

$$\begin{aligned} \frac{s^nTn}{2}(1.5n -1) \lessgtr {P^n_rn^2T} \end{aligned}$$

which is the same as $\frac{s^n}{2}(1.5n -1) \lessgtr {nP^n_r}$.

The highest value of LHS occurs when $n=1$. When $n=1$, we have $s^n < 4P^n_r$. Similarly, the lowest value of LHS occurs when $n=0$. When $n=0$, we have $-0.5s^n < 0$. $\square$

Proof for negative second derivative of (9)

$$\begin{aligned} {\bigg [{(s^p - P^p_r)(1 - n)^2 t_{q^{*}}} - I_r^p{t_{q^{*}}^2(1-n)^2} - P^n_rn^2T - I_r^n{T^2n^2} + {2s^nTn}(1.5n-1) \bigg ]} \end{aligned}$$

Since the last term could be positive sometimes, we consider the magnitude of the last term and the third term for a fixed s to show that the latter dominates the former thereby rendering the entire second derivative negative.

$$\begin{aligned} {2s^nTn}(1.5n-1) ~ \lessgtr ~ P^n_rn^2T \end{aligned}$$

which is the same as ${2s^n}(1.5n-1) \lessgtr P^n_rn$.

The highest value of RHS occurs when $n=1$. When $n=1$, we have ${2s^n}(1.5-1) \lessgtr P^n_r \Rightarrow s^n < P^n_r$. Similarly, the lowest value of RHS occurs when $n=0$. When $n=0$, we have $-2s^n < 0$. $\square$

Proof of Lemma 3.3

We show that $Q_r^{*} > Q_r^{[RFID]*}$ based on (4) and (6) for the $t_e > t_{q^*}$ case. We then show (8) > (10) for the $t_{q^*} \ge t_e$ case.

We first consider expressions (4) and (6). To operationalize this, we first compare the numerator of these expressions. We then compare their denominators.

The numerators:

$$\begin{aligned}&D^{+}\bigg [(1 - n)C_r^p + nC_r^n - nTP^n_r - \frac{(1-n)}{\lambda }\big ( P_r^{{\underline{p}}} (e ^{-\lambda t_{q^*}} - e ^{-\lambda t_e})\\&\quad - P_r^p(1 - e ^{-\lambda t_{q^*}}) \big )\bigg ] \gtrless \\&\quad {D^{+}} \bigg [(1 - n)(C_r^p + R_r) + nC_r^n - nP^n_rT - (1-n)P^p_rt_{q^{*}} \bigg ] \end{aligned}$$

which reduces to

$$\begin{aligned}&-(1-n) \big [ P_r^{{\underline{p}}} (e ^{-\lambda t_{q^*}} - e ^{-\lambda t_e}) - P_r^p(1 - e ^{-\lambda t_{q^*}}) \big ] + \lambda P^p_rt_{q^{*}} ~~ \gtrless ~~ \lambda R_r \end{aligned}$$

Now, the numerator from (4) on the left hand side of the above-expression is larger than the numerator from (6) on the right hand side since the expression in the square-brackets is negative and $P^p_rt_{q^{*}} > R_r$.

We now compare the denominators.

$$\begin{aligned}&2\bigg [\frac{I^p_r(1-n)^2((1 + \lambda t_e)e^{-\lambda t_e} - 1)}{\lambda ^2 } + {Ts^nn}(1.5n-1)\\&\quad + \frac{s^p}{2\lambda }(1 - n)^2(1 - e^{-\lambda t_e})- \frac{I_r^n T^2n^2}{2} - {P_r^nn^2T} + \frac{(1-n)^2}{2\lambda }A \bigg ] \\&\quad \gtrless \bigg [{(s^p- P^p_r)(1 - n)^2 t_{q^*}} \\&\quad + {2s^nTn(1.5n-1)} - {I^p_r(1-n)^2t_{q^{*}}^2} - {I^n_rT^2n^2} - {P^n_rn^2T}\bigg ] \end{aligned}$$

which reduces to

$$\begin{aligned}&2\bigg [\frac{s^p}{2\lambda }(1 - e^{-\lambda t_e}) + \frac{A}{2\lambda }\bigg ] + {I^p_rt_{q^{*}}^2}\\&\quad + (P^p_r - s^p) t_{q^*} ~~ \gtrless \frac{P^n_rn^2T}{(1-n)^2} + 2\frac{I^p_r(1 - (1 + \lambda t_e)e^{-\lambda t_e} )}{\lambda ^2 } \end{aligned}$$

Here, the denominator from (4) on the left hand side of the above-expression is less than the denominator from (6) on the right hand side.

Based on the comparisons above, the numerator and denominator of (4) are more than and less than respectively vs. the numerator and denominator of (6). The result follows.

We next show $Q_r^{*} > Q_r^{[RFID]*}$ based on (8) and (10) for the $t_{q^*} \ge t_e$ case. Again, to operationalize this, we first compare the numerator of these expressions. We then compare their denominators.

The numerators:

$$\begin{aligned}&D^{+}\bigg [(1 - n)C_r^p + nC_r^n - P^n_rnT - \frac{P^p_r(1-n)}{\lambda }(1 - e^{-\lambda t_e})\bigg ] \\&\quad \gtrless ~~ {D^{+}}\bigg [(1 - n)(C_r^p + R_r) - P^n_rnT + nC_r^n + P_r^pt_{q^{*}}(1-n) \bigg ] \end{aligned}$$

which reduces to

$$\begin{aligned} 0 ~~ < ~~ (1 - n)R_r + P_r^pt_{q^{*}}(1-n) + \frac{P^p_r(1-n)}{\lambda }(1 - e^{-\lambda t_e}) \end{aligned}$$

As all the terms on the right hand side are positive, the above inequality holds. I.e, the numerator of $Q_r^{*}$ is much less than that of $Q_r^{[RFID]*}$.

The denominators:

$$\begin{aligned}&\bigg [\frac{(s^p - P^p_r)}{\lambda }(1 - n)^2(1 - e^{-\lambda t_e}) + \frac{s^nTn}{2}(1.5n -1)\\&\quad - {P^n_rn^2T} - {I^n_rn^2T^2} + \frac{I_r^p(1-n)^2}{2\lambda ^2}(e^{-\lambda t_e}(1 + \lambda t_e) - 1) \bigg ] \\&\quad \gtrless ~~ \bigg [{(s^p - P^p_r)(1 - n)^2 t_{q^{*}}} - I_r^p{t_{q^{*}}^2(1-n)^2} - P^n_rn^2T - I_r^n{T^2n^2} + {2s^nTn}(1.5n-1) \bigg ] \end{aligned}$$

which reduces to

$$\begin{aligned}&\bigg [\frac{(s^p - P^p_r)}{\lambda }(1 - n)^2(1 - e^{-\lambda t_e}) + \frac{s^nTn}{2}(1.5n -1) + \frac{I_r^p(1-n)^2}{2\lambda ^2}(e^{-\lambda t_e}(1 + \lambda t_e) - 1) \bigg ] \\&\quad \gtrless ~~ \bigg [{(s^p - P^p_r)(1 - n)^2 t_{q^{*}}} + {2s^nTn}(1.5n-1) - I_r^p{t_{q^{*}}^2(1-n)^2} \bigg ] \end{aligned}$$

In the above, there are three terms on both sides. The first and third terms on the left hand side are more in magnitude than the first and third terms on the left hand side. The second term on the right hand side is four times that of the second term on the left hand side. However all the terms on both sides are effectively negative and the last term on the left hand side dominates all the individual terms. Therefore, the left hand side is less than the right hand side with a magnitude that dwarfs the numerator comparison. Therefore the result follows.