Page 283 - Industrial Process Plant Construction Estimating and Man Hour Analysis
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262 Industrial process plant construction estimating and man-hour analysis


            12.9.3 Application of the range method for equipment
            12.9.3.1 Illustration: SCR—Exhaust stack silencer
            Data for input (Table 12.9.1):


              TABLE 12.9.1 Calculation of expected cost and variance for the installation
              of exhaust stack silencer probability that the cost will exceed $26,400
              L $                  M $                  H $
              4652                 4786                 4912
              4626                 4682                 4828
              4592                 4723                 5232
              6372                 5652                 5942
              6002                 6123                 6180
              Element         L $      M $     H $      E(Cr), $  Var (Cr), $
              Support brackets  4652   4786    4912     4784.69   1877.8
              Baffles         4626     4682    4828     4696.91   1133.4
              Wall bracket    4592     4723    5232     4785.91   11377.8
              Baffle guide    6372     5652    5942     5820.24   5136.1
              Perforated liner  6002   6123    6180     6112.24   880.1
                                                        $26,200   $20,405



                             PR mean cost > ECrފ ¼ 0:5 PZðÞ
                                ½
                                            ð
                                Z ¼ UL ECrÞ= var Crފ½
                                          ð
                                              ½
                                                  ð
               where Z ¼ value of the standard normal distribution, Appendix A
               UL ¼ upper limit of cost, arbitrarily selected
            Data for input:

                                              UL ¼ $26,400
                                             E(Cr) ¼ $26,200
                                           Var (Cr) ¼ $20,405
                                               Z ¼ 1.40
                                P(Z); Appendix A; 1.40 ¼ 0.4192
                                then, P(Cost > $26,400) ¼ 8.08%
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