Two-day appointment scheduling with patient preferences and geometric arrivals

Abstract

We consider an appointment system where the patients have preferences about the appointment days. A patient may be scheduled on one of the days that is acceptable to her, or be denied appointment. The patient may or may not show up at the appointed time. The net cost is a convex function of the actual number of patients served on a given day. We study the optimal scheduling policy that minimizes the long-run average cost and study its structural properties. We advocate an index policy, which is easy to implement, performs well in comparison with other heuristic policies, and is close to the optimal policy.

Appendix

1.1 Proof of Lemma 3

(a) The result follows since monotonicity is closed under the minimum operator.

(b) We need to show $T_{12}f(i+1,j)\le T_{12}f(i,j)$ for $0\le i\le m-1, j\ge 0$. We consider three cases:

Case (b)(1) $0\le i\le m-2, j\ge 0$. The result follows since monotonicity is closed under the minimum operator.

Case (b)(2) $i=m-1,0\le j\le m-1$. We need to show $T_{12}f(m,j)\le T_{12}f(m-1,j)$. We have

$$\begin{aligned} \textit{Dec}(2) \text { in } S_0\cup S_2&\Rightarrow f(m,j+1)\le f(m,j);\\ \textit{Inc}(1) \text { in } S_2\cup S_3&\Rightarrow f(m,j)\le f(m+1,j). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(m,j)=\min \{f(m+1,j),f(m,j+1),f(m,j)\}=f(m,j+1). \end{aligned}$$

Similarly, we have

$$\begin{aligned} \textit{Dec}(1) \text { in } S_0\cup S_1&\Rightarrow f(m,j)\le f(m-1,j);\\ \textit{Dec}(2) \text { in } S_0\cup S_2&\Rightarrow f(m-1,j+1)\le f(m-1,j). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(m-1,j)&=\min \{f(m,j),f(m-1,j+1),f(m-1,j)\}\\&=\min \{f(m,j),f(m-1,j+1)\}. \end{aligned}$$

We know $f(m,j+1)\le f(m,j)$ by $\textit{Dec}(2) \text { in } S_0\cup S_2$ and $f(m,j+1)\le f(m-1,j+1)$ by $\textit{Dec}(1) \text { in } S_0\cup S_1$. Therefore,

$$\begin{aligned} T_{12}f(m,j)=f(m,j+1)\le \min \{f(m,j),f(m-1,j+1)\}=T_{12}f(m-1,j). \end{aligned}$$

Case (b)(3) $i=m-1,j\ge m$. We need to show $T_{12}f(m,j)\le T_{12}f(m-1,j)$. We have

$$\begin{aligned} \textit{Inc}(1) \text { in } S_2\cup S_3&\Rightarrow f(m,j)\le f(m+1,j);\\ \textit{Inc}(2) \text { in } S_1\cup S_3&\Rightarrow f(m,j)\le f(m,j+1). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(m,j)=\min \{f(m+1,j),f(m,j+1),f(m,j)\}=f(m,j). \end{aligned}$$

We have

$$\begin{aligned} \textit{Dec}(1) \text { in } S_0\cup S_1&\Rightarrow f(m,j)\le f(m-1,j);\\ \textit{Inc}(2) \text { in } S_1\cup S_3&\Rightarrow f(m-1,j)\le f(m-1,j+1). \end{aligned}$$

Therefore,

$$\begin{aligned} T_{12}f(m-1,j)=\min \{f(m,j),f(m-1,j+1),f(m-1,j)\}=f(m,j). \end{aligned}$$

The result follows.

(c) The result follows since monotonicity is closed under the minimum operator.

(d) We need to show $T_{12}f(i,j+1)\le T_{12}f(i,j)$ for $i\ge 0,0\le j\le m-1$. We consider three cases:

Case (d)(1) $i\ge 0, 0\le j\le m-2$. The result follows since monotonicity is closed under the minimum operator.

Case (d)(2) $0\le i\le m-1,j=m-1$. We need to show $T_{12}f(i,m)\le T_{12}f(i,m-1)$. We have

$$\begin{aligned} \textit{Dec}(1) \text { in } S_0\cup S_1&\Rightarrow f(i+1,m)\le f(i,m);\\ \textit{Inc}(2) \text { in } S_1\cup S_3&\Rightarrow f(i,m)\le f(i,m+1). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(i,m)=\min \{f(i+1,m),f(i,m+1),f(i,m)\}=f(i+1,m). \end{aligned}$$

Similarly, we have

$$\begin{aligned} \textit{Dec}(1) \text { in } S_0\cup S_1&\Rightarrow f(i+1,m-1)\le f(i,m-1);\\ \textit{Dec}(2) \text { in } S_0\cup S_2&\Rightarrow f(i,m)\le f(i,m-1). \end{aligned}$$

Therefore,

$$\begin{aligned} T_{12}f(i,m-1)&=\min \{f(i+1,m-1),f(i,m),f(i,m-1)\}\\&=\min \{f(i+1,m-1),f(i,m)\}. \end{aligned}$$

We know $f(i+1,m)\le f(i+1,m-1)$ by $\textit{Dec}(2) \text { in } S_0\cup S_2$ and $f(i+1,m)\le f(i,m)$ by $\textit{Dec}(1) \text { in } S_0\cup S_1$.

$$\begin{aligned} T_{12}f(i,m)=f(i+1,m)\le \min \{f(i+1,m-1),f(i,m)\}=T_{12}f(i,m-1). \end{aligned}$$

Case (d)(3) $i\ge m,j=m-1$. We need to show $T_{12}f(i,m)\le T_{12}f(i,m-1)$. We have

$$\begin{aligned} \textit{Inc}(1) \text { in } S_2\cup S_3&\Rightarrow f(i,m)\le f(i+1,m);\\ \textit{Inc}(2) \text { in } S_1\cup S_3&\Rightarrow f(i,m)\le f(i,m+1). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(i,m)=\min \{f(i+1,m),f(i,m+1),f(i,m)\}=f(i,m). \end{aligned}$$

Similarly, we have

$$\begin{aligned} \textit{Dec}(2) \text { in } S_0\cup S_2&\Rightarrow f(i,m)\le f(i,m-1);\\ \textit{Inc}(1) \text { in } S_2\cup S_3&\Rightarrow f(i,m-1)\le f(i+1,m-1). \end{aligned}$$

Therefore,

$$\begin{aligned} T_{12}f(i,m-1)=\min \{f(i+1,m-1),f(i,m),f(i,m-1)\}=f(i,m). \end{aligned}$$

The result follows.

(e) We show that $T_{12}f(i,j)+T_{12}f(i+2,j)\ge 2T_{12}f(i+1,j)$ for $0\le i\le m-2, 0\le j\le m-1$ and skip the proof that $T_{12}f(i,j)+T_{12}f(i,j+2)\ge 2T_{12}f(i,j+1)$ for $0\le i\le m-1, 0\le j\le m-2$ since it is similar. We consider four cases:

Case (e)(1) $i\le m-3,j\le m-1$. We know

$$\begin{aligned} T_{12}f(i,j)&=\min \{f(i+1,j),f(i,j+1),f(i,j)\}\\&=\min \{f(i+1,j),f(i,j+1)\}. \end{aligned}$$

The $T_{12}$ here coincides with the $T_{R(I)}$ in Koole [14]. The result follows.

Case (e)(2) $i\le m-3,j=m$. We show

$$\begin{aligned} T_{12}f(i,m)+T_{12}f(i+2,m)\ge 2T_{12}f(i+1,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(i,m)&=f(i+1,m);\\ T_{12}f(i+2,m)&=f(i+3,m);\\ T_{12}f(i+1,m)&=f(i+2,m). \end{aligned}$$

The result follows from the $\textit{Convexity}$ of f.

Case (e)(3) $i=m-2,j\le m-1$. We show

$$\begin{aligned} T_{12}f(m-2,j)+T_{12}f(m,j)\ge 2T_{12}f(m-1,j), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m-2,j)&=\min \{f(m-1,j),f(m-2,j+1)\};\\ T_{12}f(m-1,j)&=\min \{f(m,j),f(m-1,j+1)\};\\ T_{12}f(m,j)&=f(m,j+1). \end{aligned}$$

We consider two cases:

Case (e)(3)(i) $f(m-1,j)\ge f(m-2,j+1)$. Then $T_{12}f(m-2,j)=f(m-2,j+1)$. By the property $\textit{SuperC}$ of f, we know

$$\begin{aligned} f(m,j)-f(m-1,j+1)\ge f(m-1,j)-f(m-2,j+1)\ge 0. \end{aligned}$$

Hence $T_{12}f(m-1,j)=f(m-1,j+1)$. By the $\textit{Convexity}$ of f, we have

$$\begin{aligned} T_{12}f(m-2,j)+T_{12}f(m,j)&=f(m-2,j+1)+f(m,j+1)\\&\ge 2f(m-1,j+1)=2T_{12}f(m-1,j). \end{aligned}$$

Case (e)(3)(ii) $f(m-1,j)< f(m-2,j+1)$. Then $T_{12}f(m-2,j)=f(m-1,j)$. By the $\textit{Super}$ of f, we have

$$\begin{aligned} T_{12}f(m-2,j)+T_{12}f(m,j)&=f(m-1,j)+f(m,j+1)\\&\ge f(m,j)+f(m-1,j+1)\\&\ge 2\min \{f(m,j),f(m-1,j+1)\}\\&=2T_{12}f(m-1,j). \end{aligned}$$

Case (e)(4) $i=m-2,j=m$. We show

$$\begin{aligned} T_{12}f(m-2,m)+T_{12}f(m,m)\ge 2T_{12}f(m-1,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m-2,m)&=f(m-1,m);\\ T_{12}f(m-1,m)&=f(m,m);\\ T_{12}f(m,m)&=f(m,m). \end{aligned}$$

The result follows from $f(m-1,m)\ge f(m,m)$ by $\textit{Dec}(1) \text { in } S_0\cup S_1$.

(f) We show that $T_{12}f(i,j)+T_{12}f(i+1,j+1)\ge T_{12}f(i+1,j)+T_{12}f(i,j+1)$ for $0\le i,j\le m-1$. We consider four cases:

Case (f)(1) $i\le m-2,j\le m-2$. We know $T_{12}f(i,j)=\min \{f(i+1,j),f(i,j+1)\}$ coincides with $T_{R(I)}$ in Koole [14]. The result follows.

Case (f)(2) $i=m-1,j\le m-2$. We show

$$\begin{aligned} T_{12}f(m-1,j)+T_{12}f(m,j+1)\ge T_{12}f(m,j)+T_{12}f(m-1,j+1), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m-1,j)&=\min \{f(m,j),f(m-1,j+1)\};\\ T_{12}f(m,j+1)&=f(m,j+2);\\ T_{12}f(m,j)&=f(m,j+1);\\ T_{12}f(m-1,j+1)&=\min \{f(m,j+1),f(m-1,j+2)\}. \end{aligned}$$

We consider two cases:

Case (f)(2)(i) $f(m,j)\ge f(m-1,j+1)$. Then $T_{12}f(m-1,j)=f(m-1,j+1)$. By the $\textit{Super}$ of f, we know

$$\begin{aligned} T_{12}f(m-1,j)+T_{12}f(m,j+1)&=f(m-1,j+1)+f(m,j+2)\\&\ge f(m,j+1)+f(m-1,j+2) \text { (by } \textit{Super})\\&\ge f(m,j+1)+\min \{f(m,j+1),\\&\quad \ f(m-1,j+2)\}\\&=T_{12}f(m,j)+T_{12}f(m-1,j+1). \end{aligned}$$

Case (f)(2)(ii) $f(m,j)< f(m-1,j+1)$. Then $T_{12}f(m-1,j)=f(m,j)$. By the $\textit{SuperC}$ of f, we know

$$\begin{aligned} f(m,j+1)-f(m-1,j+2)\le f(m,j)-f(m-1,j+1)<0. \end{aligned}$$

Hence $T_{12}f(m-1,j+1)=f(m,j+1)$. By the $\textit{Convexity}$ of f, we have

$$\begin{aligned}&T_{12}f(m-1,j)+T_{12}f(m,j+1)\\&\quad =f(m,j)+f(m,j+2)\ge f(m,j+1)+f(m,j+1)\\&\quad = T_{12}f(m,j)+T_{12}f(m-1,j+1). \end{aligned}$$

Case (f)(3) $i\le m-2,j=m-1$. We show

$$\begin{aligned} T_{12}f(i,m-1)+T_{12}f(i+1,m)\ge T_{12}f(i+1,m-1)+T_{12}f(i,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(i,m-1)&=\min \{f(i+1,m-1),f(i,m)\};\\ T_{12}f(i+1,m)&=f(i+2,m);\\ T_{12}f(i+1,m-1)&=\min \{f(i+2,m-1),f(i+1,m)\};\\ T_{12}f(i,m)&=f(i+1,m). \end{aligned}$$

We know

$$\begin{aligned}&T_{12}f(i,m-1)+T_{12}f(i+1,m)\\&\quad =\min \{f(i+1,m-1),f(i,m)\}+f(i+2,m)\\&\quad =\min \{f(i+1,m-1)+f(i+2,m),f(i,m)+f(i+2,m)\}\\&\quad \ge \min \{f(i+1,m-1)+f(i+2,m),2f(i+1,m)\} \text { (by } \textit{Convexity})\\&\quad \ge \min \{f(i+2,m-1)+f(i+1,m),2f(i+1,m)\} \text { (by } \textit{Super})\\&\quad = \min \{f(i+2,m-1),f(i+1,m)\}+f(i+1,m)\\&\quad =T_{12}f(i+1,m-1)+T_{12}f(i,m). \end{aligned}$$

Case (f)(4) $i=m-1,j=m-1$. We show

$$\begin{aligned} T_{12}f(m-1,m-1)+T_{12}f(m,m)\ge T_{12}f(m,m-1)+T_{12}f(m-1,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m-1,m-1)&=\min \{f(m,m-1),f(m-1,m)\};\\ T_{12}f(m,m)&=f(m,m);\\ T_{12}f(m,m-1)&=f(m,m);\\ T_{12}f(m-1,m)&=f(m,m). \end{aligned}$$

We know

$$\begin{aligned} T_{12}f(m-1,m-1)+T_{12}f(m,m)&=\min \{f(m,m-1),f(m-1,m)\}+f(m,m)\\&\ge f(m,m)+f(m,m) \text { (by } \textit{Dec}(1) \text { and } \textit{Dec}(2) )\\&=T_{12}f(m,m-1)+T_{12}f(m-1,m). \end{aligned}$$

(g) We show that $T_{12}f(i+2,j)+T_{12}f(i,j+1)\ge T_{12}f(i+1,j)+T_{12}f(i+1,j+1)$ for $0\le i\le m-2, 0\le j\le m-1$. We skip the proof of $T_{12}f(i+1,j)+T_{12}f(i,j+2)\ge T_{12}f(i,j+1)+T_{12}f(i+1,j+1)$ for $0\le i\le m-1, 0\le j\le m-2$. We consider four cases:

Case (g)(1) $i\le m-3,j\le m-2$. We know $T_{12}f(i,j)=\min \{f(i+1,j),f(i,j+1)\}$ coincides with $T_{R(I)}$ in Koole [14]. The result follows.

Case (g)(2) $i=m-2,j\le m-2$. We show

$$\begin{aligned} T_{12}f(m,j)+T_{12}f(m-2,j+1)\ge T_{12}f(m-1,j)+T_{12}f(m-1,j+1), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m,j)&=f(m,j+1);\\ T_{12}f(m-2,j+1)&=\min \{f(m-1,j+1),f(m-2,j+2)\};\\ T_{12}f(m-1,j)&=\min \{f(m,j),f(m-1,j+1)\};\\ T_{12}f(m-1,j+1)&=\min \{f(m,j+1),f(m-1,j+2)\}. \end{aligned}$$

We consider two cases:

Case (g)(2)(i) $f(m-1,j+1)<f(m-2,j+2)$. Then $T_{12}f(m-2,j+1)=f(m-1,j+1)$. We have

$$\begin{aligned}&T_{12}f(m,j)+T_{12}f(m-2,j+1)\\&\quad =f(m,j+1)+f(m-1,j+1)\\&\quad \ge \min \{f(m,j+1),f(m-1,j+2)\}+ \min \{f(m,j),f(m-1,j+1)\}\\&\quad =T_{12}f(m-1,j+1)+T_{12}f(m-1,j). \end{aligned}$$

Case (g)(2)(ii) $f(m-1,j+1)\ge f(m-2,j+2)$. Then $T_{12}f(m-2,j+1)=f(m-2,j+2)$. We have

$$\begin{aligned}&\textit{SuperC} \Rightarrow f(m,j+1)-f(m-1,j+2)\ge f(m-1,j+1)\\&\quad -f(m-2,j+2)\ge 0\Rightarrow T_{12}f(m-1,j+1)=f(m-1,j+2)\\&\quad \textit{SuperC} \Rightarrow f(m,j)-f(m-1,j+1)\ge f(m,j+1)\\&\quad -f(m-1,j+2)\ge 0\Rightarrow T_{12}f(m-1,j)=f(m-1,j+1). \end{aligned}$$

We have

$$\begin{aligned}&T_{12}f(m,j)+T_{12}f(m-2,j+1)\\&\quad =f(m,j+1)+f(m-2,j+2)\\&\quad \ge f(m-1,j+2)+ f(m-1,j+1) \text { (by } \textit{SuperC} )\\&\quad =T_{12}f(m-1,j+1)+T_{12}f(m-1,j). \end{aligned}$$

Case (g)(3) $i\le m-3,j=m-1$. We show

$$\begin{aligned} T_{12}f(i+2,m-1)+T_{12}f(i,m)\ge T_{12}f(i+1,m-1)+T_{12}f(i+1,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(i+2,m-1)&=\min \{f(i+3,m-1),f(i+2,m)\};\\ T_{12}f(i,m)&=f(i+1,m);\\ T_{12}f(i+1,m-1)&=\min \{f(i+2,m-1),f(i+1,m)\};\\ T_{12}f(i+1,m)&=f(i+2,m). \end{aligned}$$

We consider two cases:

Case (g)(3)(i) $f(i+3,m-1)<f(i+2,m)$. Then $T_{12}f(i+2,m-1)=f(i+3,m-1)$. We have

$$\begin{aligned}&\textit{SuperC} \Rightarrow f(i+2,m-1)-f(i+1,m)\le f(i+3,m-1)-f(i+2,m)<0\\&\qquad \quad \ \ \Rightarrow T_{12}f(i+1,m-1)=f(i+2,m-1). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(i+2,m-1)+T_{12}f(i,m)&=f(i+3,m-1)+f(i+1,m)\\&\ge f(i+2,m-1)+ f(i+2,m) \text { (by } \textit{SuperC} )\\&=T_{12}f(i+1,m-1)+T_{12}f(i+1,m). \end{aligned}$$

Case (g)(3)(ii) $f(i+3,m-1)\ge f(i+2,m)$. Then $T_{12}f(i+2,m-1)=f(i+2,m)$. We have

$$\begin{aligned} T_{12}f(i+2,m-1)+T_{12}f(i,m)&=f(i+2,m)+f(i+1,m)\\&\ge f(i+2,m)+ \min \{f(i+2,m-1),\\&\quad f(i+1,m)\}\\&=T_{12}f(i+1,m)+T_{12}f(i+1,m-1). \end{aligned}$$

Case (g)(4) $i=m-2,j=m-1$. We show

$$\begin{aligned} T_{12}f(m,m-1)+T_{12}f(m-2,m)\ge T_{12}f(m-1,m-1)+T_{12}f(m-1,m), \end{aligned}$$

where we know

$$\begin{aligned} T_{12}f(m,m-1)&=f(m,m);\\ T_{12}f(m-2,m)&=f(m-1,m);\\ T_{12}f(m-1,m-1)&=\min \{f(m,m-1),f(m-1,m)\};\\ T_{12}f(m-1,m)&=f(m,m). \end{aligned}$$

Hence

$$\begin{aligned} T_{12}f(m,m-1)+T_{12}f(m-2,m)&=f(m,m)+f(m-1,m)\\&\ge f(m,m)+ \min \{f(m,m-1),f(m-1,m)\}\\&=T_{12}f(m-1,m)+T_{12}f(m-1,m-1). \end{aligned}$$

1.2 Proof of Lemma 4

Due to the symmetry, we show the proof for operator $T_1$ and skip the proof for operator $T_2$. Given the properties f has, we know

$$\begin{aligned} T_1f(i,j)=\min \{f(i+1,j),f(i,j)\}= {\left\{ \begin{array}{ll} f(i+1,j), &{}\quad \forall (i,j)\in S_0\cup S_1;\\ f(i,j), &{}\quad \forall (i,j)\in S_2\cup S_3. \end{array}\right. } \end{aligned}$$

We prove (f) and skip the rest due to the similarity. We show that $T_1f(i,j)+T_1f(i+1,j+1)\ge T_1f(i+1,j)+T_1f(i,j+1)$ for any $0\le i,j\le m-1$. We consider four cases:

Case (f)(1) $0\le i\le m-2,0\le j\le m-2$. We have

$$\begin{aligned} T_1f(i,j)+T_1f(i+1,j+1)&=f(i+1,j)+f(i+2,j+1)\\&\ge f(i+2,j)+f(i+1,j+1)\text { (by } \textit{Super} )\\&=T_1f(i+1,j)+T_1f(i,j+1). \end{aligned}$$

Case (f)(2) $i=m-1,0\le j\le m-2$. We need to show $T_1f(m-1,j)+T_1f(m,j+1)\ge T_1f(m,j)+T_1f(m-1,j+1)$. We have

$$\begin{aligned} T_1f(m-1,j)+T_1f(m,j+1)&=f(m,j)+f(m,j+1)\\&=T_1f(m,j)+T_1f(m-1,j+1) \end{aligned}$$

Case (f)(3) $0\le i\le m-2,j=m-1$. We need to show $T_1f(i,m-1)+T_1f(i+1,m)\ge T_1f(i+1,m-1)+T_1f(i,m)$. We have

$$\begin{aligned} T_1f(i,m-1)+T_1f(i+1,m)&=f(i+1,m-1)+f(i+2,m)\\&\ge f(i+2,m-1)+f(i+1,m)\text { (by } \textit{Super})\\&=T_1f(i+1,m-1)+T_1f(i,m). \end{aligned}$$

Case (f)(4) $i=m-1,j=m-1$. We need to show $T_1f(m-1,m-1)+T_1f(m,m)\ge T_1f(m,m-1)+T_1f(m-1,m)$. This follows because

$$\begin{aligned} T_1f(m-1,m-1)+T_1f(m,m)&=f(m,m-1)+f(m,m)\\&=T_1f(m,m-1)+T_1f(m-1,m). \end{aligned}$$

1.3 Proof of Lemma 6

We know $T_cf(i,j)=c(i)+f(j,0)$ and c(x) is a convex function which achieves its minimum at $x=m$, hence (a) and (b) follow. For (c), if f is $\textit{Inc}(1) \text { in } S_2\cup S_3$, then $f(i+1,j)\ge f(i,j), \forall i\ge m, j\ge 0$. For any $j\ge m$, we have $T_cf(i,j+1)=c(i)+f(j+1,0)\ge c(i)+f(j,0)=T_cf(i,j)$. Hence $T_cf$ is $\textit{Inc}(2) \text { in } S_1\cup S_3$. For (d), if f is $\textit{Dec}(1) \text { in } S_0\cup S_1$, then $f(i+1,j)\le f(i,j), \forall i\le m-1, j\ge 0$. For any $j\le m-1$, we have $T_cf(i,j+1)=c(i)+f(j+1,0)\le c(i)+f(j,0)=T_cf(i,j)$. Hence $T_cf$ is $\textit{Dec}(2) \text { in } S_0\cup S_2$. For (e), the Convexity of $T_cf$ follows directly from the Convexity of f. (f) is trivially shown given the definition of the operator $T_c$. For (g), we have

$$\begin{aligned} T_cf(i+2,j)+T_cf(i,j+1)&=c(i+2)+f(j,0)+c(i)+f(j+1,0)\\&\ge c(i+1)+f(j,0)+c(i+1)+f(j+1,0)\\&=T_cf(i+1,j)+T_cf(i+1,j+1), \end{aligned}$$

and

$$\begin{aligned} T_cf(i+1,j)+T_cf(i,j+2)&=c(i+1)+f(j,0)+c(i)+f(j+2,0)\\&\ge c(i)+f(j+1,0)+c(i+1)+f(j+1,0)\\&=T_cf(i,j+1)+T_cf(i+1,j+1). \end{aligned}$$