Cognitive Radio Networks: Analysis of a Paid-Sharing Approach Based on Admission Control Decisions

Abstract

Cognitive radio networks have emerged in the last decade as a solution for two problems: spectrum underutilization and spectrum scarcity. The main idea is to manage the radio spectrum more efficiently, where secondary users (SUs) are allowed to exploit the spectrum holes in primary user’s (PUs) frequency bands. We consider a paid-sharing approach where SUs pay for spectrum utilization. A challenging aspect in these mechanisms is how to proceed when a PU needs certain amount of bandwidth and the free capacity is insufficient. We assume a preemptive system where PUs have strict priority over SUs; when a PU arrives to the system and there are not enough free channels to accommodate the new user, one or more SUs will be deallocated. The affected SUs will then be reimbursed, implying some cost for the PUs service provider (SP). This paper bears on the design and analysis of an optimal SU admission control policy; i.e. that maximizes the long-run profit of the SP. We model the optimal revenue problem as a Markov Decision Process and we use dynamic programming and further techniques such as sample-path analysis to characterize properties of the optimal admission control policy. We introduce different changes to one of the best known dynamic programming algorithms incorporating the knowledge of the characterization. In particular, those proposals accelerate the rate of convergence of the algorithm when is applied in the considered context. Our results are validated through numerical examples.

Appendices

Appendix A: Proof of Proposition 1

Inequality holds for $n=0$ by definition. Assuming the inequality holds for $n-1$, we show that it holds for n. We divide the problem in two cases: preempted and non-preempted situations.

Proof

• Case 1 $b_1x+b_2(y+1) \le C-b_1$;

  $$\begin{aligned} V_n(x,y+1)= & \lambda '_1 V_{n-1}(x+1,y+1) \end{aligned}$$

  (6)
  $$\begin{aligned}&+\, \lambda '_2 \max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \end{aligned}$$

  (7)
  $$\begin{aligned}&+\,\mu '_1 x V_{n-1}(x-1,y+1) \end{aligned}$$

  (8)
  $$\begin{aligned}&+\, \mu '_2 y V_{n-1}(x,y) \end{aligned}$$

  (9)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) \end{aligned}$$

  (10)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) V_{n-1}(x,y+1) \end{aligned}$$

  (11)
  $$\begin{aligned} V_n(x,y)= & {} \lambda '_1 V_{n-1}(x+1,y) \end{aligned}$$

  (12)
  $$\begin{aligned}&+\, \lambda '_2 \max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \end{aligned}$$

  (13)
  $$\begin{aligned}&+\, \mu '_1 x V_{n-1}(x-1,y) \end{aligned}$$

  (14)
  $$\begin{aligned}&+\, \mu '_2 y V_{n-1}(x,y-1) \end{aligned}$$

  (15)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) \end{aligned}$$

  (16)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) V_{n-1}(x,y) \end{aligned}$$

  (17)

  Please note that (6) $\le$ (12), (8) $\le$ (14), (9) $\le$ (15), (11) $\le$ (17) are proved directly by induction assumption. On the other hand (10) $=$ (16).

  Following we show the justification of (7) $\le$ (13):

  • 1. If ($\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}=V_{n-1}(x,y+1)$ and

    $\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}=V_{n-1}(x,y)$) or

    ($\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}=V_{n-1}(x,y+2)+b_2R$ and

    $\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}=V_{n-1}(x,y+1)+b_2R$), the inequality is a direct consequence of induction hypothesis.

  • 2. If ($\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}=V_{n-1}(x,y+1)$ and

    $\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}=V_{n-1}(x,y+1)+b_2R$), it is simple to note the validity of the demonstration because $R>0$.

  • 3. If ($\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}=V_{n-1}(x,y+2)+b_2R$ and

    $\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}=V_{n-1}(x,y)$), then we have that $V_{n-1}(x,y)>V_{n-1}(x,y+1)+b_2R\ge V_{n-1}(x,y+2)+b_2R$.

• Case 2 $C-b_1 < b_1x+b_2(y+1) \le C$;

  According to Eq. (1), we define the number of preempted SUs (z and $z'$) as

  $z=\left[ \frac{b_1x+b_2(y+1)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2(y+1)-C+b_1)\%b_2 \ne 0\right\} }\right]$ and

  $z'=\max \left\{ \left[ \frac{b_1x+b_2y-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\}$; we have:

  $$\begin{aligned} V_n(x,y+1)= & {} \lambda '_1 (V_{n-1}(x+1,y+1-z)-zb_2K) \end{aligned}$$

  (18)
  $$\begin{aligned}&+\, \lambda '_2 \max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \end{aligned}$$

  (19)
  $$\begin{aligned}&+\, \mu '_1 x V_{n-1}(x-1,y+1) \end{aligned}$$

  (20)
  $$\begin{aligned}&+\, \mu '_2 y V_{n-1}(x,y) \end{aligned}$$

  (21)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) \end{aligned}$$

  (22)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) V_{n-1}(x,y+1) \end{aligned}$$

  (23)
  $$\begin{aligned} V_n(x,y)= & {} \lambda '_1 (V_{n-1}(x+1,y-z')-z'b_2K) \end{aligned}$$

  (24)
  $$\begin{aligned}&+\, \lambda '_2 \max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \end{aligned}$$

  (25)
  $$\begin{aligned}&+ \,\mu '_1 x V_{n-1}(x-1,y) \end{aligned}$$

  (26)
  $$\begin{aligned}&+\, \mu '_2 y V_{n-1}(x,y-1) \end{aligned}$$

  (27)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) \end{aligned}$$

  (28)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) V_{n-1}(x,y) \end{aligned}$$

  (29)

  On one hand, we have (22) $=$ (28); and (20) $\le$ (26), (21) $\le$ (27), (23) $\le$ (29) are proved directly with the hypothesis of induction. On the other hand, the proof of (19) $\le$ (25) is analogous to the previous case.² Finally, please note that $z'=z-1$ and $K>0$, then the inequality (18) $\le$ (24) is proved.

$\square$

Appendix B: Proof of Proposition 2

Proof

Inequality holds for $n=0$ by definition. Assuming the inequality holds for $n-1$, we show that it holds for n. We divide the problem in two cases: preempted and non-preempted situations.

• Case 1 $b_1x+b_2(y+1) \le C-b_1$;

  $$\begin{aligned} V_n(x,y+1)-V_n(x,y)&= \lambda '_1 (V_{n-1}(x+1,y+1)-V_{n-1}(x+1,y)) \\&\quad+\, \lambda '_2 \max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \\&\quad- \,\lambda '_2 \max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \\&\quad+\, \mu '_1 x (V_{n-1}(x-1,y+1)-V_{n-1}(x-1,y)) \\&\quad+\, \mu '_2 y (V_{n-1}(x,y)-V_{n-1}(x,y-1)) \\&\quad+\, \mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y) \\&\quad+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) (V_{n-1}(x,y+1)\\&\quad-\,V_{n-1}(x,y)) \ge -b_2K \end{aligned}$$

  Using induction hypothesis we have $V_n(x,y+1)-V_n(x,y) \ge (1-\lambda '_2-\alpha ')(-b_2K)-\mu '_2 (V_{n-1}(x,y+1)-V_{n-1}(x,y)) +\lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}-\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\})$. We can observe that the second term is always positive or zero (being zero the worst case). Then, if $K\ge R$ we can prove that $\lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}-\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\})\ge -\lambda '_2b_2K$. This proves the result $\forall \alpha$.

• Case 2 $C-b_1 < b_1x+b_2(y+1) \le C$;

  In the same way as in the previous case, we define z and $z'$ as

  $z=\left[ \frac{b_1x+b_2(y+1)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2(y+1)-C+b_1)\%b_2 \ne 0\right\} }\right]$ and

  $z'=\max \left\{ \left[ \frac{b_1x+b_2y-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\}$; we have:

  $$\begin{aligned} V_n(x,y+1)-V_n(x,y)= & {}\, \lambda '_1 (V_{n-1}(x+1,y+1-z)-zb_2K-V_{n-1}(x+1,y-z')+z'b_2K) \\&+\, \lambda '_2 \max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \\&- \,\lambda '_2 \max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \\&+\, \mu '_1 x (V_{n-1}(x-1,y+1)-V_{n-1}(x-1,y)) \\&+\, \mu '_2 y (V_{n-1}(x,y)-V_{n-1}(x,y-1)) \\&+\, \mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y) \\&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+1)\right) (V_{n-1}(x,y+1) \\&-\, V_{n-1}(x,y)) \ge -b_2K \end{aligned}$$

  We can observe that $z'=z-1$ and using the same arguments as in case 1, the proof is completed. When we have to evaluate the terms that implies the decision (the maximization), we have used that $K\ge R$.

$\square$

Appendix C: Proof of Proposition 3

Proof

Inequality holds for $n=0$ by definition. Assuming the inequality holds for $n-1$, we show that it holds for n. We divide the problem in two cases: preempted and non-preempted situations.

• Case 1 $b_1(x+1)+b_2(y+1) \le C-b_1$;

  $$\begin{aligned} V_n(x+1,y+1) -V_n(x+1,y)= & {} \lambda '_1 (V_{n-1}(x+2,y+1) -V_{n-1}(x+2,y)) \end{aligned}$$

  (30)
  $$\begin{aligned}&+\, \lambda '_2 (\max \{V_{n-1}(x+1,y+1),V_{n-1}(x+1,y+2)+b_2R\} \nonumber \\&- \,\max \{V_{n-1}(x+1,y),V_{n-1}(x+1,y+1)+b_2R\}) \end{aligned}$$

  (31)
  $$\begin{aligned}&+\, \mu '_1 x (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (32)
  $$\begin{aligned}&+ \,\mu '_2 y (V_{n-1}(x+1,y) -V_{n-1}(x+1,y-1)) \end{aligned}$$

  (33)
  $$\begin{aligned}&+ \,\mu '_1 (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (34)
  $$\begin{aligned}&+ \,\left( C\left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 (x+1)-\mu '_2 (y+1)\right) \nonumber \\&(V_{n-1}(x+1,y+1) - V_{n-1}(x+1,y)) \end{aligned}$$

  (35)
  $$\begin{aligned} V_n(x,y+1)-V_n(x,y)= & {} \lambda '_1 (V_{n-1}(x+1,y+1) -V_{n-1}(x+1,y)) \end{aligned}$$

  (36)
  $$\begin{aligned}&+\, \lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \nonumber \\&-\, \max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}) \end{aligned}$$

  (37)
  $$\begin{aligned}&+\, \mu '_1 x (V_{n-1}(x-1,y+1) -V_{n-1}(x-1,y)) \end{aligned}$$

  (38)
  $$\begin{aligned}&+ \,\mu '_2 y (V_{n-1}(x,y) -V_{n-1}(x,y-1)) \end{aligned}$$

  (39)
  $$\begin{aligned}&+ \,\mu '_1 (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (40)
  $$\begin{aligned}&+ \,\left( C\left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 (x+1)-\mu '_2 (y+1)\right) \nonumber \\&(V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (41)

  (30) $\le$ (36), (32) $\le$ (38), (33) $\le$ (39), (35) $\le$ (41) are proved directly by induction hypothesis. In addition, we have (34) $=$ (40).

  The last inequality requires a degree of algebraic manipulation:

  $$\begin{aligned}&\lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}-\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y+2)-V_{n-1}(x,y+1)+b_2R,0\}\\&\qquad -\, \max \{V_{n-1}(x,y)-V_{n-1}(x,y+1),b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y+2)-V_{n-1}(x,y+1)+b_2R,0\}\\&\qquad +\,\min \{-V_{n-1}(x,y)+V_{n-1}(x,y+1),-b_2R\}) \\&\quad \ge \lambda '_2 (\max \{V_{n-1}(x+1,y+2)-V_{n-1}(x+1,y+1)+b_2R,0\}\\&\qquad +\,\min \{-V_{n-1}(x+1,y)+V_{n-1}(x+1,y+1),-b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x+1,y+1),V_{n-1}(x+1,y+2)+b_2R\}\\&\qquad -\,\max \{V_{n-1}(x+1,y),V_{n-1}(x+1,y+1)+b_2R\}) \end{aligned}$$

• Case 2 $C-b_1 < b_1(x+1)+b_2(y+1) \le C$;

  Defining z, $z'$, $z^+$ and $z^*$ according to Eq. (1):

  $$\begin{aligned}&z=\left[ \frac{b_1(x+1)+b_2(y+1)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1(x+1)+b_2(y+1)-C+b_1)\%b_2 \ne 0\right\} }\right] , \\&\quad z'=\max \left\{ \left[ \frac{b_1(x+1)+b_2y-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1(x+1)+b_2y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\} , \\&\quad z^+=\max \left\{ \left[ \frac{b_1x+b_2(y+1)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2(y+1)-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\} \hbox { and }\\&\quad z^*=\max \left\{ \left[ \frac{b_1x+b_2y-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\} , \end{aligned}$$
  $$\begin{aligned} V_n(x+1,y+1) -V_n(x+1,y)= & {} \,\lambda '_1 (V_{n-1}(x+2,y+1-z)-zb_2K) \nonumber \\&- \,\lambda '_1 (V_{n-1}(x+2,y-z')-z'b_2K) \end{aligned}$$

  (42)
  $$\begin{aligned}&+ \,\lambda '_2 (\max \{V_{n-1}(x+1,y+1),V_{n-1}(x+1,y+2)+b_2R\} \nonumber \\&-\, \max \{V_{n-1}(x+1,y),V_{n-1}(x+1,y+1)+b_2R\}) \end{aligned}$$

  (43)
  $$\begin{aligned}&+\, \mu '_1 x (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (44)
  $$\begin{aligned}&+\, \mu '_2 y (V_{n-1}(x+1,y) -V_{n-1}(x+1,y-1)) \end{aligned}$$

  (45)
  $$\begin{aligned}&+\, \mu '_1 (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (46)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 (x+1)-\mu '_2 (y+1)\right) \nonumber \\&(V_{n-1}(x+1,y+1)-V_{n-1}(x+1,y)) \end{aligned}$$

  (47)
  $$\begin{aligned} V_n(x,y+1)-V_n(x,y)= & {} \,\lambda '_1 (V_{n-1}(x+1,y+1-z^+)-z^+b_2K) \nonumber \\&- \,\lambda _1(V_{n-1}(x+1,y-z^*)-z^*b_2K) \end{aligned}$$

  (48)
  $$\begin{aligned}&+\, \lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \end{aligned}$$

  (49)
  $$\begin{aligned}&-\,\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}) \nonumber \\&+\, \mu '_1 x (V_{n-1}(x-1,y+1) -V_{n-1}(x-1,y)) \end{aligned}$$

  (50)
  $$\begin{aligned}&+\, \mu '_2 y (V_{n-1}(x,y) -V_{n-1}(x,y-1)) \end{aligned}$$

  (51)
  $$\begin{aligned}&+\, \mu '_1 (V_{n-1}(x,y+1) -V_{n-1}(x,y)) \end{aligned}$$

  (52)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 (x+1)-\mu '_2 (y+1)\right) \nonumber \\&(V_{n-1}(x,y+1)-V_{n-1}(x,y)) \end{aligned}$$

  (53)

  (42) $\le$ (48) is valid according to Proposition 2 and observing that $z'=z-1$ and $z^*=z^+-1$. The other inequalities can be demonstrated in the same way as in the previous case. In particular, we have to be careful in the proof of (43) $\le$ (49) in the scenarios where the number of allocated channels is greater $C-b_2$.

$\square$

Appendix D: Proof of Proposition 4

Proof

Inequality holds for $n=0$ by definition. Assuming the inequality holds for $n-1$, we show that it holds for n. We divide the problem in two cases: preempted and non-preempted situations.

• Case 1 $b_1x+b_2(y+2) \le C-b_1$;

  $$\begin{aligned} V_n(x,y) -V_n(x,y+1)= & {} \lambda '_1 (V_{n-1}(x+1,y) -V_{n-1}(x+1,y+1)) \end{aligned}$$

  (54)
  $$\begin{aligned}&+\, \lambda '_2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \end{aligned}$$

  (55)
  $$\begin{aligned}&-\,\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}) \nonumber \\&+\, \mu '_1 x (V_{n-1}(x-1,y) -V_{n-1}(x-1,y+1)) \end{aligned}$$

  (56)
  $$\begin{aligned}&+\, \mu '_2 y (V_{n-1}(x,y-1) -V_{n-1}(x,y)) \end{aligned}$$

  (57)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y+1) \end{aligned}$$

  (58)
  $$\begin{aligned}&+\, \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+2)\right) \nonumber \\&(V_{n-1}(x,y)-V_{n-1}(x,y+1)) \end{aligned}$$

  (59)
  $$\begin{aligned} V_n(x,y+1)-V_n(x,y+2)= & {} \lambda '_1 (V_{n-1}(x+1,y+1) -V_{n-1}(x+1,y+2)) \end{aligned}$$

  (60)
  $$\begin{aligned}&+\, \lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \end{aligned}$$

  (61)
  $$\begin{aligned}&-\,\max \{V_{n-1}(x,y+2),V_{n-1}(x,y+3)+b_2R\}) \nonumber \\&+\, \mu '_1 x (V_{n-1}(x-1,y+1) -V_{n-1}(x-1,y+2)) \end{aligned}$$

  (62)
  $$\begin{aligned}&+\, \mu '_2 y (V_{n-1}(x,y) -V_{n-1}(x,y+1)) \end{aligned}$$

  (63)
  $$\begin{aligned}&+\, \mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y+1) \end{aligned}$$

  (64)
  $$\begin{aligned}&+\, \left( C\left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+2)\right) \nonumber \\&(V_{n-1}(x,y+1)-V_{n-1}(x,y+2)) \end{aligned}$$

  (65)

  The relations (54) $\le$ (60), (56) $\le$ (62), (57) $\le$ (63) and (59) $\le$ (65) are direct consequences of the induction hypothesis. The (58) and (64) terms cancel out each other. Next we show that (61) $\ge$ (55):

  $$\begin{aligned}&\lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}-\max \{V_{n-1}(x,y+2),V_{n-1}(x,y+3)+b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y+1)-V_{n-1}(x,y+2),b_2R\}-\max \{0,V_{n-1}(x,y+3)-V_{n-1}(x,y+2)+b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y+1)-V_{n-1}(x,y+2),b_2R\}+\min \{0,-V_{n-1}(x,y+3)+V_{n-1}(x,y+2)-b_2R\}) \\&\quad \ge \lambda '_2 (\max \{V_{n-1}(x,y)-V_{n-1}(x,y+1),b_2R\}+\min \{0,-V_{n-1}(x,y+2)+V_{n-1}(x,y+1)-b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y)-V_{n-1}(x,y+1),b_2R\}-\max \{0,V_{n-1}(x,y+2)-V_{n-1}(x,y+1)+b_2R\}) \\&\quad = \lambda '_2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}-\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}) \end{aligned}$$

• Case 2 $C-b_1 <b_1x+b_2(y+2) \le C$;

  Defining z, $z'$ and $z^*$ according to Eq. (1):

  $$\begin{aligned} z= & {} \left[ \frac{b_1x+b_2(y+2)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2(y+2)-C+b_1)\%b_2 \ne 0\right\} }\right] , \\ z'= & \max \left\{ \left[ \frac{b_1x+b_2(y+1)-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2(y+1)y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\} \hbox { and}\\ z^*= & \max \left\{ \left[ \frac{b_1x+b_2y-C+b_1}{b_2}+{\mathbb {1}}_{\left\{ (b_1x+b_2y-C+b_1)\%b_2 \ne 0\right\} }\right] ,0 \right\} , \end{aligned}$$
  $$\begin{aligned} V_n(x,y) -V_n(x,y+1)= & {} \lambda '_1 (V_{n-1}(x+1,y-z^*) -z^*b_2K) \end{aligned}$$

  (66)
  $$\begin{aligned}&- \lambda '_1 (V_{n-1}(x+1,y+1-z')-z'b_2K) \nonumber \\&\lambda '_2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\} \end{aligned}$$

  (67)
  $$\begin{aligned}&-\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}) \nonumber \\&+ \mu '_1 x (V_{n-1}(x-1,y) -V_{n-1}(x-1,y+1)) \end{aligned}$$

  (68)
  $$\begin{aligned}&+ \mu '_2 y (V_{n-1}(x,y-1) -V_{n-1}(x,y)) \end{aligned}$$

  (69)
  $$\begin{aligned}&+ \mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y+1) \end{aligned}$$

  (70)
  $$\begin{aligned}&+ \left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+2)\right) \nonumber \\&(V_{n-1}(x,y)-V_{n-1}(x,y+1)) \end{aligned}$$

  (71)
  $$\begin{aligned} V_n(x,y+1)-V_n(x,y+2)= & {} \lambda '_1 (V_{n-1}(x+1,y+1-z')-z'b_2K) \end{aligned}$$

  (72)
  $$\begin{aligned}&- \,\lambda '_1 (V_{n-1}(x+1,y+2-z)-zb_2K) \nonumber \\&+\, \lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\} \end{aligned}$$

  (73)
  $$\begin{aligned}&-\, \max \{V_{n-1}(x,y+2), V_{n-1}(x,y+3)+b_2R\}) \nonumber \\&+\, \mu '_1 x (V_{n-1}(x-1,y+1) -V_{n-1}(x-1,y+2)) \end{aligned}$$

  (74)
  $$\begin{aligned}&+ \,\mu '_2 y (V_{n-1}(x,y) -V_{n-1}(x,y+1)) \end{aligned}$$

  (75)
  $$\begin{aligned}&+ \,\mu '_2 V_{n-1}(x,y) - \mu '_2 V_{n-1}(x,y+1) \end{aligned}$$

  (76)
  $$\begin{aligned}&+ \,\left( C \left( \frac{\mu '_1}{b_1}+\frac{\mu '_2}{b_2}\right) -\mu '_1 x-\mu '_2 (y+2)\right) \nonumber \\&(V_{n-1}(x,y+1) -V_{n-1}(x,y+2)) \end{aligned}$$

  (77)

  (66) $\le$ (72) is true according to Proposition 2 (in particular, the result of Proposition 2 is used in the limit case: $z'=z^*=0$ ). The relations (68) $\le$ (74), (69) $\le$ (75) and (71) $\le$ (77) are direct consequences of the induction hypothesis, and (70) $=$ (76) is trivial. In order to analyze the inequality (67) $\le$ (73) the same arguments as case 1 can be applied. It is important to highlight that if the number of busy resources in the system is greater than $C-b_2$, the admission control decision has no sense (no SU will be accepted). So, if for example $b_1x+b_2(y+2)>C-b_2$ (and $b_1x+b_2(y+1)\le C-b_2$), we can demonstrate the inequality in the following way: $\lambda '_2 (\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R\}-V_{n-1}(x,y+2))=\lambda '_2 (\max \{V_{n-1}(x,y+1)-V_{n-1}(x,y+2),b_2R\}) \ge \lambda '_2 (\max \{V_{n-1}(x,y)-V_{n-1}(x,y+1),b_2R\})=\lambda '_2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}-V_{n-1}(x,y+1) )$, then

  • if $\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R \}=V_{n-1}(x,y+1)$ the proof is completed.

  • if $\max \{V_{n-1}(x,y+1),V_{n-1}(x,y+2)+b_2R \}=V_{n-1}(x,y+2)+b_2R$, then

    $$\begin{aligned}&\lambda _2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}-V_{n-1}(x,y+1)) \\&\quad \ge \lambda _2 (\max \{V_{n-1}(x,y),V_{n-1}(x,y+1)+b_2R\}-V_{n-1}(x,y+2)-b_2R) \end{aligned}$$

    and the proof is over.

$\square$