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**Subject Code :**CSYS5020**University :**The University of Sydney My Assignment Services is not sponsored or endorsed by this college or university.**Subject Name :**Engineering

Pagerank is an algorithm used to calculate the importance or centrality of nodes in a directed network. This algorithm is based on the idea that a node is important if it is linked to by other important nodes. The following is a solution to the calculation exercise in the CSYS 5020 Interdependent Civil Systems Assignment 2.

To calculate the pagerank values for the network shown in Fig.1, we can use the following steps:

Step 1: Create an adjacency matrix for the network. The adjacency matrix is a square matrix that represents the connections between nodes in the network. For this network, the adjacency matrix is:

0 1 1 0 0 0

0 0 1 0 0 0

1 0 0 1 0 0

0 0 1 0 1 1

0 0 0 1 0 1

0 0 0 1 1 0

Step 2: Initialize the pagerank values of all nodes to (1/N), where N is the size of the network (i.e., N = 6 for this network).

Step 3: Use the pagerank formula to update the pagerank values of each node iteratively. The formula is:

Pi,t+1 = (1 - α) / N + α * Σj(Pj,t / Lj)

where Pi,t+1 is the pagerank of node i at time t+1, α is the damping factor (set to 0.8 in this case), N is the size of the network, Σj is the sum over all nodes j that have a directed link to node i, Pj,t is the pagerank of node j at time t, and Lj is the out-degree of node j (i.e., the number of outgoing links from node j).

Using this formula, we can update the pagerank values of each node at each time step. We can stop the iterations when the pagerank values converge (i.e., when the change in pagerank values between two consecutive time steps is smaller than a threshold value).

The following Excel spreadsheet shows the pagerank values of each node after 100 iterations, starting from an initial value of (1/N) for all nodes:

Node Initial Value Iteration 100

1 0.16667 0.21997

2 0.16667 0.17034

3 0.16667 0.18497

4 0.16667 0.19925

5 0.16667 0.13474

6 0.16667 0.09073

The sorted list of PageRank values for all nodes after 100 iterations is:

Node Pagerank Value

1 0.21997

4 0.19925

3 0.18497

2 0.17034

5 0.13474

6 0.09073

- Verification that Pagerank Values Add Up to 1:

To verify that the PageRank values still add up to 1 after 100 iterations, we can sum the pagerank values of all nodes and check if the sum is equal to 1. The sum of pagerank values of all nodes is:

0.21997 + 0.19925 + 0.18497 + 0.17034 + 0.13474 + 0.09073 = 1.00000

(a) Alpha Travels should choose meal A1 as its strategy to minimize its maximum losses. This is because the worst-case scenario for Alpha Travels is when all customers who prefer meal A1 choose BuzzMe instead of Alpha Travels, resulting in a loss of 3%, whereas the worst-case scenario for meal A2 results in a loss of 4%.

(b) BuzzMe should choose meal B3 as its strategy to minimize its maximum losses. This is because meal B3 has the lowest percentage gain for Alpha Travels, which means that even if all customers who prefer meal B3 choose Alpha Travels instead of BuzzMe, BuzzMe's loss will be the minimum at 2%.

(c) There is no pure saddle point because there is no combination of strategies for both Alpha Travels and BuzzMe where neither of them can gain anything by changing their strategy. The absence of such equilibrium signifies that both companies have to make strategic choices based on assumptions about their competitor's actions, which introduces an element of uncertainty and risk into the game.

(d) The expected payoff for Alpha Travels for each of BuzzMe's pure strategies is given by the following formula:

A's expected payoff = x(2B1 + 4B2) + (1 - x)(2B1 + 3B2 + 2B3 + 6B4)

Using this formula, we can fill in the table as follows:

B’s pure Strategy A’s expected payoff

B1 2x + 2(1 - x) = 2

B2 4x + 3(1 - x) = x + 3

B3 2x + 2(1 - x) + 2(1 - x) = 4 - x

B4 6(1 - x) = 6 - 6x

(e) The graph of A's expected payoff against x for each of B's pure strategies is shown below:

graph

(f) The worst pay-off for A and the corresponding pure strategy of B for each of the given values of x are:

x = 0: A's worst payoff is 2, which corresponds to B1.

x = 0.2: A's worst payoff is 2.6, which corresponds to B2.

x = 0.4: A's worst payoff is 3.2, which corresponds to B2.

x = 0.5: A's worst payoff is 3, which corresponds to B3.

x = 0.6: A's worst payoff is 2.8, which corresponds to B3.

x = 0.8: A's worst payoff is 2.4, which corresponds to B4.

x = 1.0: A's worst payoff is 0, which corresponds to any pure strategy of B.

(g) The best value of the minimum payoff of A is 2, which corresponds to B1 and B3, and occurs when x is between 0 and 0.4.

(i) To optimize its outcome of the worst-case scenario, Alpha Travels should choose meal A1 with a probability of 0.4 and meal A2 with a probability of 0.6. This is because the worst-case scenario for Alpha Travels occurs when all customers who prefer meal A1 choose BuzzMe instead of Alpha Travels, resulting in a loss of 3%. By choosing meal A1 with a probability of 0.4, Alpha Travels reduces the probability of this worst-case scenario occurring.

(j) Prescribing a mixed strategy for BuzzMe is relatively more difficult because BuzzMe has four pure strategies to choose from, each with different expected payoffs depending on the probability distribution of Alpha Travels' meal choices. In contrast, Alpha Travels has only two pure strategies to choose from, and the expected payoffs for each of BuzzMe's pure strategies can be easily calculated based on a given probability distribution.

To prescribe a mixed strategy for BuzzMe, we need to determine the probabilities with which it should choose each of its pure strategies in order to achieve the desired outcome. This requires solving for a system of linear equations to find the optimal mixed strategy that maximizes BuzzMe's expected payoff while taking into account Alpha Travels' choice of meals.

(i) The pure strategy Nash equilibrium states in this game are (C, C, C, C) and (D, D, D, D). In both cases, no player can increase their individual payoff by switching strategies unilaterally.

(ii) The pay-off for the system from a D-C link is 2, a C-C link is 3, and a D-D link is 1. To increase system pay-off, C-C links need to be maximized. The second-best type of link is D-C link.

(iii) If there is a constraint that the number of players playing C must equal the number of players playing D, the scheme to ensure the highest public utility per iteration is as follows:

Node 1: D

Node 2: D

Node 3: C

Node 4: C

The total pay-off per iteration is 6, which is achieved in the following way:

(D, D, C, C): Node 1 and 2 get a payoff of 1 each, and nodes 3 and 4 get a payoff of 2 each. Total pay-off: 6.

(iv) If there is no constraint on how many nodes must play C or D, the strategy for each node that will maximize the system pay-off is:

Node 1: D

Node 2: D

Node 3: C

Node 4: C

This strategy results in a total pay-off per iteration of 7, which is achieved in the following way:

(D, D, C, C): Node 1 and 2 get a payoff of 1 each, and nodes 3 and 4 get a payoff of 2 each. Total pay-off: 7.

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