Pricing and Budget Allocation for IoT Blockchain with Edge Computing

Fig. 1 depicts the architecture of the system model of this paper. Consider that there is an IoT blockchain network that has been running for a period which adopts the proof-of-work consensus mechanism. The IoT blockchain network consists of lots of IoT systems such as smart factories or smart homes. Each IoT system includes a set of IoT devices and can be seen as a group. Due to the security and privacy protection brought by the blockchain, the IoT blockchain network will continuously attract other IoT systems to join.

In each IoT system, there are two edge servers which provide hash computing service (hash-server) and task processing service (task-server), respectively. Motivated by the reward from the blockchain network platform, devices in the IoT system would like to be miners of the blockchain network, that is, they will compete with other miners to scramble the right of generating a new block by solving a hash puzzle. Due to the limited computational power of these devices, they will purchase computational resources from the hash-server and then offload their hash puzzle to the hash-server during the mining process. Moreover, each IoT device has its own tasks, such as data collecting, data analysis, and data processing. IoT devices could benefit from performing these tasks. However, when the amount of data is relatively large, it is difficult for these IoT devices to perform the tasks. Then IoT devices will purchase task processing resources from the task-server to perform their tasks. Generally, the more resource they purchase, the faster and better they perform their tasks, and then the more benefit they get from these tasks.

Different from traditional IoT systems, merging the IoT system into a blockchain network has many advantages, especially in terms of security and reliability. Specifically, the IoT blockchain network with edge computing inherits the security and reliability performance of the blockchain, as shown in follows.

1. 1.

   Get rid of a third party: IoT devices carry out transactions in a P2P manner, in which each device has the same rights, and the trust between devices is built with the help of the smart contract of the blockchain. Therefore, the IoT blockchain network guarantees the robustness and scalability without involving a trusted third party.

2. 2.

   Privacy protection: Blockchain can help deal with the increasing risk of sensitive data being exposed to malwares. In the blockchain network, the communication between devices uses the asymmetric encryption technology to protect the sensitive data. The message sent to a device will be encrypted with the receiver’s public key, and it can only be decrypted by the receiver’s private key. In this way, even if malicious devices intercept the message, they cannot know the content of the message.

3. 3.

   Integrity: The blocks are duplicated recorded in different devices in a distributed way, so it’s very hard for attackers to tamper with the blockchain. Besides, the blocks are linked together through cryptography. If an attacker attempts to tamper with the transactions in a block, the hash value of each subsequent block will be changed, and thus the attacker needs to redo the PoW puzzle of each subsequent block, which is nearly impossible for the attackers.

4. 4.

   Authentication: In this IoT blockchain network, each new block needs to pass a two-round validation before it is added into the blockchain. It’s very hard for an attacker to control a whole group (IoT system), so the new block that contains illegal transactions cannot pass the first round validation. Even if some attackers forge the signature of a group, the illegal block cannot pass the second round validation.

We assume that each IoT device has a unique budget to purchase resources from edge servers. The amount of resources they purchase from the two edge servers depends on how many profits they can get from the trading and are limited by their budgets. Each IoT device will allocate their budget to purchase different services from the hash-server and the task-server to maximize their profits. For the hash-server and the task-server, they will set the unit price of their resources to maximize their utilities. Moreover, there is a competition between the two servers. For example, if the unit price of resources from hash-server is too high, IoT devices would purchase more resources from the task-server, and vice versa. Naturally, we model the interaction between the two servers and IoT devices as a multi-leader multi-follower Stackelberg game, where the hash-server and the task-server act as leaders who first set the unit price of their resources, and IoT devices act as followers who determine their strategies according to the leaders’ bids.

We use $\mathcal{S}=\{s_{1},s_{2},\dots,s_{n}\}$ to denote the set of IoT devices, the budget for each device $s_{i}\in\mathcal{S}$ to purchase resources is $b_{i}$ where $b_{i}>0$. The unit price of resources from the hash-server and the task-server are denoted by $p_{h}$ and $p_{t}$ per day, respectively. Let $x_{i}^{h}$ and $x_{i}^{t}$ be the amount of resources purchased by $s_{i}$ from the hash-server and the task-server, respectively. The amount of resources purchased buy each device is limited by its budget, that is, $x_{i}^{h}*p_{h}+x_{i}^{t}*p_{t}\leq b_{i},\forall s_{i}\in\mathcal{S}$.

The profits of each IoT device $s_{i}\in\mathcal{S}$ includes two parts. The first part comes from mining new blocks for the blockchain network, which is related to the amount of resources $x_{i}^{h}$ purchased by $s_{i}$ from the hash-server. The second part comes from performing tasks, which is related to the amount of resources $x_{i}^{t}$ purchased by $s_{i}$ from the task-server. We use $P_{i}^{h}$ and $P_{i}^{t}$ to denote the two parts of profits, respectively. In the following, we will describe how to calculate the two parts of profits.

As the blockchain network has been running for a period, in this paper, we assume that the total hash computational power of the blockchain network in a period time in the future can be estimated and is denoted by $H$. In the PoW consensus, the first miner who solves the hash puzzle has the right to generate a new block and will get the reward from the blockchain network. The probability of a miner winning the mining competition is directly related to the hash computational power. We use $pro_{i}$ to denote the probability that device $s_{i}\in\mathcal{S}$ is the first one to solve the hash puzzle, and $pro_{i}$ can be estimated by

Generally, the blockchain network will adjust the difficulty of the hash puzzle periodically according to the total hash computation power in the network to stabilize the block generation speed. We assume that an average of $N$ new blocks are generated per day, and the miners will get a reward $R$ for generating a new block. The expected reward obtained by device $s_{i}$ in per day is $pro_{i}RN$, and the cost is $x_{i}^{h}p_{h}$. Then the expected profits that $s_{i}$ gets from the mining process in a day is calculated as

The profits $P_{i}^{t}$ got by $s_{i}$ comes from performing tasks is related to the amount of resources purchased by $s_{i}$ from the task server, we use a logarithmic function to estimate the benefits of device $s_{i}$ for performing tasks, and then $P_{i}^{t}$ is calculated as

where $\alpha>0$ and $\beta\geq 1$ are two constant parameters.

Therefore, the total profits got by device $s_{i}$ is calculated as

We use $U_{h}$ and $U_{t}$ to denote the utility of the hash-server and task-server, respectively. Assume that the unit hash resource cost of the hash-server is $c_{h}$, and the unit task resource cost of the task-server is $c_{t}$. Then the utilities of the two servers can be calculated as

The profits $P_{i}$ gots by device $s_{i}$ involves two parts, i.e., $P_{i}^{h}$ and $P_{i}^{t}$, which comes from trading with the two servers, respectively. The first-order derivatives of $P_{i}^{h}$ and $P_{i}^{t}$ are

To make the problem reasonable, the conditions $\frac{\partial P_{i}^{h}}{\partial x_{i}^{h}}(0)\geq 0$ and $\frac{\partial P_{i}^{t}}{\partial x_{i}^{t}}(0)\geq 0$ should be hold, otherwise, IoT devices will never purchase resources from the hash-server or the task-server. Hence, we have $p_{h}\leq\frac{RN}{H}$ and $p_{t}\leq\alpha\beta$. As servers will never sell their resources at a price below the cost, that is, $p_{h}\geq c_{h}$ and $p_{t}\geq c_{t}$. Therefore, in this paper, we assume that $c_{h}\leq p_{h}\leq\frac{RN}{H}$ and $c_{t}\leq p_{t}\leq\alpha\beta$.

The interaction between the two servers and IoT devices has two stages. In the upper stage, the hash-server and the task-server offer a unit price of their resources. In the lower stage, IoT devices determine their strategies to maximize their profits according to the price of different services. In the following, we give a detailed definition of the problem in each stage.

Note that in the lower stage, each IoT device make their decision independently, and in the upper stage, the two servers are also non-cooperative. Therefore, the problems in the two stages form a non-cooperative multi-leader multi-follower Stackelberg game. Our objective is to find the Stackelberg equilibrium (SE) point of the game, where none of the players of the game wants to change its strategy unilaterally. The SE point in our model is defined as follows.

The second-order derivatives of the profits function $P_{i}$ of device $s_{i}$ are

Therefore, the profits function $P_{i}$ is strictly concave, and the problem for each device $s_{i}$ in the lower stage is actually a convex optimization problem. Sequentially, we use the Karush–Kuhn–Tucker (KKT) conditions to solve the problem.

Let $\lambda_{1},\lambda_{2}$ and $\lambda_{3}$ be the Lagrange’s multipliers that associated with conditions in Eqs. (10) and (11). Then we define the Lagrangian function as follows.

The KKT conditions (including four group of conditions) of Problem 1 are listed as follows.

Stationarity conditions:

Primal feasibility conditions:

Dual feasibility conditions:

Complementary slackness conditions:

The optimal solution of the problem is taken in one of the following four cases.

(1) Case 1: $x_{i}^{h}=x_{i}^{t}=0$. According to the KKT condition (28), we have $\lambda_{1}=0$ as $b_{i}>0$. Substitute it into KKT conditions (23) and (24), we have $\lambda_{2}=p_{h}-\frac{RN}{H}$ and $\lambda_{3}=p_{t}-\alpha\beta$. As $p_{h}\leq\frac{RN}{H}$ and $p_{t}\leq\alpha\beta$ hold, we have $\lambda_{2}\leq 0$ and $\lambda_{3}\leq 0$. The KKT conditions can be satisfied only when $\lambda_{2}=0$ and $\lambda_{3}=0$. Thus we need to check whether $p_{h}=\frac{RN}{H}$ and $p_{t}=\alpha\beta$ hold, if yes, the optimal solution is $x_{i}^{h}=x_{i}^{t}=0$, otherwise, the optimal solution is not in this case.

(2) Case 2: $x_{i}^{h}=0$ and $x_{i}^{t}>0$. In this case, we have $\lambda_{3}=0$ according to the KKT condition (30).

Consider $\lambda_{1}=0$, substitute $x_{i}^{h}=0$ and $\lambda_{1}=0$ into (23), we have $\lambda_{2}=p_{h}-\frac{RN}{H}\leq 0$. If $p_{h}<\frac{RN}{H}$, the KKT condition (27) cannot be satisfied as $\lambda_{2}<0$, which means $\lambda_{1}=0$ is not feasible in this case. Otherwise, if $p_{h}=\frac{RN}{H}$, we have $\lambda_{2}=0$. Substitute $\lambda_{1}=\lambda_{3}=0$ into (24), we have $x_{i}^{t}=\frac{\alpha}{p_{t}}-\frac{1}{\beta}$. Then we need to check whether the primal feasibility condition (25) is satisfied, if yes, the optimal solution is $(0,\frac{\alpha}{p_{t}}-\frac{1}{\beta})$, if no, $\lambda_{1}=0$ is not feasible in this case.

Consider $\lambda_{1}>0$, according to (28), we have $b_{i}-x_{i}^{h}p_{h}-x_{i}^{t}p_{t}=0$. Combing $x_{i}^{h}=0$, we have

Substitute (31) and $\lambda_{3}=0$ into KKT condition (24), we have

Substitute (32) and $x_{i}^{h}=0$ into (23), we have

It obvious that $x_{i}^{t}=\frac{b_{i}}{p_{t}}>0$ is satisfied. If $\lambda_{1}$ got by (32) satisfies $\lambda_{1}>0$ and $\lambda_{2}$ got by (33) satisfies $\lambda_{2}\geq 0$, it means all of the KKT conditions can be satisfied, and thus the optimal solution can be determined as $(0,\frac{b_{i}}{p_{t}})$. Otherwise, the optimal solution is not in Case 2.

(3) Case 3: $x_{i}^{h}>0$ and $x_{i}^{t}=0$. In this case, we have $\lambda_{2}=0$ according to the KKT condition (29).

Consider $\lambda_{1}=0$, substitute $x_{i}^{t}=0$ and $\lambda_{1}=0$ into (24), we have $\lambda_{3}=p_{t}-\alpha\beta\leq 0$. If $p_{t}<\alpha\beta$, the KKT condition (27) cannot be satisfied as $\lambda_{3}<0$, which implies $\lambda_{1}=0$ in not feasible in this case. Otherwise, if $p_{t}=\alpha\beta$, we have $\lambda_{3}=0$. Substitute $\lambda_{1}=\lambda_{2}=0$ into (23), we have $x_{i}^{h}=\sqrt{\frac{RNH}{p_{h}}}-H$. Then we need to check whether the primal feasibility condition (25) is satisfied, if yes, the optimal solution is $(\sqrt{\frac{RNH}{p_{h}}}-H,0)$, if no, $\lambda_{1}=0$ is not feasible in this case.

Consider $\lambda_{1}>0$, we have $b_{i}-x_{i}^{h}p_{h}-x_{i}^{t}p_{t}=0$ according to (28). Combing $x_{i}^{t}=0$, we have

Substitute (34) and $\lambda_{2}=0$ into KKT condition (23), we have

Substitute (35) and $x_{i}^{t}=0$ into (24), we have

It’s obvious that $x_{i}^{h}=\frac{b_{i}}{p_{h}}>0$ is satisfied. If $\lambda_{1}$ got by (35) satisfies $\lambda_{1}>0$ and $\lambda_{3}$ got by (36) satisfies $\lambda_{3}\geq 0$, it means that all of the KKT conditions can be satisfied, and thus the optimal solution is $(\frac{b_{i}}{p_{h}},0)$. Otherwise, the optimal solution is not in Case 3.

(4) Case 4: $x_{i}^{h}>0$ and $x_{i}^{t}>0$. In this case, we have $\lambda_{2}=\lambda_{3}=0$ according to the KKT conditions (29) and (30).

Suppose $\lambda_{1}=0$, substitute it into the stationarity conditions (23) and (24), we have

As $p_{h}\leq\frac{RN}{H}$ and $p_{t}\leq\alpha\beta$, it’s easy to know that the condition (26) is satisfied. Then we check whether the condition (25) is satisfied. If yes, then the solution shown in Eq. (37) is the optimal solution. Otherwise, $\lambda_{1}=0$ is not feasible in this case.

Now we consider the case that $\lambda_{1}>0$. According to the condition (28), we have $b_{i}-x_{i}^{h}p_{h}-x_{i}^{t}p_{t}=0$. Solving conditions (23) and (24), we have

Substitute (38) into $b_{i}-x_{i}^{h}p_{h}-x_{i}^{t}p_{t}=0$, we have

where $A=b_{i}+Hp_{h}+\frac{1}{\beta}p_{t}>0$, and $B=\sqrt{RNHp_{h}}>0$. Let $t=\sqrt{1+\lambda_{1}}>0$, substitute $t$ into (39) and solve the function, we have $t=\frac{B\pm\sqrt{B^{2}+4A\alpha}}{2A}$. As $t>0$, we have

By solving (40), we have

Then we check whether $\lambda_{1}$ got by (41) satisfies $\lambda_{1}>0$. If $\lambda_{1}\leq 0$, it means that the solutions found in (38) cannot satisfy all of the KKT conditions, and thus the optimal solution is not in Case 4. Otherwise, we substitute (41) into (38), and we will get a solution $(x_{i}^{h},x_{i}^{t})$. Then we check whether the solution $(x_{i}^{h},x_{i}^{t})$ satisfies $x_{i}^{h}>0$ and $x_{i}^{t}>0$. If yes, the solution $(x_{i}^{h},x_{i}^{t})$ is the optimal solution. If no, the optimal solution is not in Case 4.

The optimal solution will be found in one of the four cases from Case 1 to Case 4. Based on above analysis, we present the algorithm $\sf{FOSD}$ to solve Problem 1 in the lower stage. The pseudo-code is shown in Algorithm 1.

On the leaders’ side, hash-server and task-server set their unit prices $p_{h}$ and $p_{t}$ of resources to maximize their utilities $U_{h}$ and $U_{t}$ which are calculated by Eqs. (5) and (6), respectively. Note that the pricing strategies of leaders will directly affect the strategies of followers, which has been analyzed in Section V-A. Therefore, the strategies of the hash-server and task-server will affect each other’s utility. The game between the two servers is non-cooperative and competitive. To maximize their utilities, each server (leader) should give a suitable unit price of its resource. For example, for the hash-server, if the unit price of resource $p_{h}$ is too high, the followers will prefer to purchase more resources from the task-server, and thus the hash-server will get a very low utility. On the contrary, if $p_{h}$ is set too low, although the followers tend to purchase resources from the hash-server, the total purchased resources are limited due to the budget limitation of these followers, which also results in a low utility. In the following, we will show that the game between the two servers will reach a Nash equilibrium, and we design an algorithm to find the Nash equilibrium point.

The concept of the Nash Equilibrium (NE) of the game between the two servers is defined as follows.

According to the definition, at the NE point, none of the servers can improve its utility by unilaterally changing its strategy. Therefore, when the game reaches a NE point, the interaction between the two servers is suspended, and the pricing strategies of the two servers will never change again. Next, we will prove the existence and uniqueness of the NE point of the game between the two servers.

After the leaders (the two servers) issue their strategies, the followers (IoT devices) will determine their strategies for purchasing different resources from the two servers, as is discussed in Section V-A. The interaction between servers and IoT devices formulates a multi-leader multi-follower Stackelberg game, and the objective is to find the Stackelberg equilibrium (SE) point of the game, which is defined as Definition 1. Next, we will prove that the Stackelberg equilibrium of the game between servers and IoT devices exists and is unique.

To find the NE point of the game between the two servers, based on the optimal strategies of IoT devices, we proposed an algorithm $\sf{FNES}$ to find the final pricing strategies of the two servers. The $\sf{FNES}$ is based on sub-gradient technique [31, 32], and it invokes $\sf{FOSD}$ as its subroutine. The pseudo-code of algorithm $\sf{FNES}$ is shown in Algorithm 2.