# Analytical Hierarchy Process for Construction Safety Management and Resource Allocation
## 4 Results and Discussion
### Case Study 2  
4.2.1. Case Study 2--Cost Analysis and Alternatives  
The second case study is an infrastructure project in southern Israel for building pump shafts for a desalination plant and connecting them to the sea. The estimated project duration is 36 months. The project's scope, including fees, is estimated at USD 11M. Additionally, a safety budget of about USD 200,000 was allocated for safety measures, including safety measures for water stay, which is 1.8% of the total project scope. The safety measures used in the project did not apply technological solutions. It is worth noting that compared to the first case study, the current case study is riskier and more complex and is of a larger scale. The safety budget allocated at the beginning of the project was much lower, which caused many work accidents at the project site. Due to the allocation of additional safety resources, risk prevention was improved, and the number of work accidents was notably lowered. The _TPB_ of the project is USD 11,000,000. A cost analysis of safety resources allocated is conducted for each alternative as follows:  
* Alternative A considers allocating an optimal safety investment ratio that is 1% of the total project cost. The cost to the owner, including safety investment according to alternative A, is therefore USD 11,110,000. * Alternative B considers allocating an initial safety investment ratio that is 0.5% of the total project cost. The cost to the owner, including initial safety investment according to alternative B, is therefore USD 11,055,000. Considering the optimal safety investment ratio of 1%, an upgraded safety investment is possible with the extra 0.5% available safety budget. However, due to the large number of accidents that occurred at the site, the safety budget was upgraded to a total safety budget of USD 200,000. This results in a supplement of the safety budget of USD 90,000 but without applying any advanced technological solutions. Allocating a supplement of USD 90,000 to the safety budget leads to a total safety budget of 1.8%. * Alternative C considers allocating a modest safety investment ratio that is 0.5% of the total project cost. The cost to the owner, including initial safety investment according to alternative B, is therefore USD 11,055,000. 4.2.2. Case Study 2--AHP for Decision-Making  
The three alternatives suggest allocating different _SIR_s of the total project scope:  
Alternative A: _SIR_(A) = 1% (50)  
Alternative B: _SIR_(B)= 1.8% (51)  
Alternative C: _SIR_(C) = 0.5% \begin{table}
\begin{tabular}{p{142.3 pt} p{142.3 pt}} \hline \hline
**Worker \#** & **Do You Have Any Suggestions for Improving Safety in Construction Sites?** \\ \hline Worker 31 & \\ \hline Worker 32 & \\ \hline Worker 33 & After the main contractor has fulfilled all requirements related to equipment and guidance, the constructor is allowed to impose fines on workers or subcontractors who violate the safety orders they have agreed to \\ \hline Worker 34 & Implement ongoing education programs for management and work teams. \\ \hline Worker 35 & \\ \hline \hline \end{tabular}
\end{table} 
Table 2: _Cont._According to Equation (19) the \(TSCR\) and \(TCAR\) values are as follows:  
\[\text{Alternative A }TSCR(A)=y(1.0\%)=1.89\%\to TCAR(\text{A})=1.89\%-1.0\%=0.89\% \tag{53}\] \[\text{Alternative B }TSCR(B)=y(1.8\%)=2.99\%\to TCAR(\text{B})=2.99\%-1.8\%=1.19\%\] (54) \[\text{Alternative C }TSCR(C)=y(0.5\%)=2.27\%\to TCAR(\text{C})=2.27\%-0.5\%=1.77\% \tag{55}\]  
Consequently, the comparison matrices are  
\[A_{C}= \begin{bmatrix}1&0.56&2\\ 1.8&1&3.6\\ 0.5&0.28&1\end{bmatrix} \qquad\qquad\qquad A_{R}= \begin{bmatrix}1&0.75&0.5\\ 1.34&1&0.67\\ 1.99&1.49&1\end{bmatrix}  \tag{56}\]  
Its column's sum normalizes each matrix:  
\[\text{column sum }AC= \begin{pmatrix}3.3&1.83&6.6\end{pmatrix} \quad\text{column sum }AR= \begin{pmatrix}4.33&3.24&2.18\end{pmatrix}  \tag{57}\]  
As follows:  
\[N_{C}= \begin{bmatrix}0.30&0.30&0.30\\ 0.55&0.55&0.55\\ 0.15&0.15&0.15\end{bmatrix} \qquad\qquad N_{R}=