Please draw the graphs in the same format as in Figure $4.$ I drew some pictures, please modify them. Thank you so much. project. For an extended analysis, Figure $4$ shows the key performance results of the voting algorithms.

The performance parameters are the time to deliver a data to the end$-$user $($TTC$).$ the number of distinct

data proposals before effecting a data delivery, and the control message overhead expended to deliver

a data. One of the influential parameters is the fault severity of a mis$-$behaving voter: denoted as a

parameter $r,$ where $r$ is the probability that a faulty device does in fact behave incorrectly. For e$.$g$.,$

$r=0\mathrm{.}1$ means that a faulty device $x$ behaves incorrectly only for $10\%$ of the times $x$ participates in

the algorithm operations $-$ which means that $x$ behaves correctly for $90\%$ of the times. The results

are shown for a case of $r=0\mathrm{.}5.$ Empirically plot the results for two other cases: $r=0\mathrm{.}1$ and $r=0\mathrm{.}8.$

Show the plots on top of the results given for $r=0\mathrm{.}5.$

Figure $4$: Experimental performance results of replica voting algorithm

I use a software to draw the graphs, $r=0\mathrm{.}1$ : shown by the yellow line of each graph, $r=0\mathrm{.}8$ : shown by the red line

of each graph.

The explanation for $r=0\mathrm{.}1$ suggests that with a lower fault severity, the algorithm will perform more efficiently,

resulting in a lower Time to Consistency $($TTC$),$ fewer voting steps needed before successful data delivery, and

lower control message overhead. Therefore, in the chart, we would expect the performance line for $r=0\mathrm{.}1$ to

be below the line for $r=0\mathrm{.}5,$ indicating better performance.

For $r=0\mathrm{.}8,$ the explanation indicates that a higher fault severity leads to a worse performance. This is due to

the higher probability of failure, which increases the TTC$,$ the number of voting steps required, and the control

message overhead. Consequently, the performance line for $r=0\mathrm{.}8$ would be above the line for $r=0\mathrm{.}5$ on the

chart.