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IENG 217 Cost Estimating for Engineers. Project Simulation. A 1 A 2. 10,000. Simulation. Let us arbitrarily pick a value for A 1 and A 2 in the uniform range (5000, 7000). Say A 1 = 5,740 and A 2 = 6,500. A 1 A 2. 5,740 6,500. 10,000. 10,000. Simulation. - PowerPoint PPT Presentation
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IENG 217Cost Estimating for
Engineers
Project Simulation
Simulation
A1 A2
10,000
Let us arbitrarily pick a value for A1 and A2 in the uniform range (5000, 7000). Say A1 = 5,740 and A2 = 6,500.
Simulation
A1 A2
10,000
5,740 6,500
10,000
NPW = -10,000 + 5,740(1.15)-1 + 6,500(1.15)-2
= (93.96)
Simulation
A1 A2
10,000
5,740 6,500
10,000
We now have one realization of NPW for a given realization of A1 and A2.
Simulation
A1 A2
10,000
5,740 6,500
10,000
We now have one realization of NPW for a given realization of A1 and A2. Choose 2 new values forA1, A2.
A1 = 6,820 A2 = 6,218
A1 A2
10,000
6,820 6,218
10,000
NPW = -10,000 + 6,820(1.15)-1 + 6,218(1.15)-2
= 632.14
Summary
A1 A2 NPW
5,740 6,500 (93.96)6,820 6,218 632.14
Choose 2 new values.
A1 = 5,273 A2 = 6,422
A1 A2
10,000
5,273 6,422
10,000
NPW = -10,000 + 5,273(1.15)-1 + 6,422(1.15)-2
= (558.83)
Summary
A1 A2 NPW
5,740 6,500 (93.96)6,820 6,218 632.145,273 6,422
(558.83)Choose 2 new values.
Summary
A1 A2 NPW
5,740 6,500 (93.96)6,820 6,218 632.145,273 6,422 (558.83) . . .6,855 5,947 457.66
Simulation
With enough realizations of A1, A2, and NPW, we can begin to get a sense of the distribution of the NPW.
-1,871 0 1,380 NPW
Freq.
Simulation
What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.
Simulation
What we now need is a formal method of selecting random values for A1 and A2 to avoid selection bias.
Recall the uniform
5,000 7,000
1/2,000
f(x)
Simulation
The uniform has cumulative distribution given by:
5,000 7,000
1
F(x)
F( x)
0 , x5,000
1 , x7,000
x 5,0002,000
Simulation
Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1 (Rand Function in Excel). Let P be a random variable uniformly from 0 to 1.
P U(0,1)
Simulation
Recall that 0 < F(x) < 1. Using modulo arithmetic, it is fairly easy to generate random numbers between 0 and 1. Let P be a random variable uniformly from 0 to 1.
P U(0,1)Algorithm: 1. Randomly generate P 2. Let P = F(x) 3. Solve for x = F-1(p)
Simulation
1. Randomly generate P U(0,1). P = .7
5,000 7,000
1
F(x)
F( x)x 5,0002,000
5,000 < x < 7,000
Simulation
1. Randomly generate P U(0,1). P = .72. Let P = F(x).
5,000 7,000
1
F(x)
F( x)x 5,0002,000
5,000 < x < 7,000
.7
Simulation
1. Randomly generate P U(0,1). P = .72. Let P = F(x).3. x = F-1(p).
5,000 7,000
1
F(x)
F( x)x 5,0002,000
5,000 < x < 7,000
.7
6,400
Formal Derivation
Recall, for
5,000 7,000
1
F(x)
F( x)x 5,0002,000
5,000 < x < 7,000. Then
P
Px
5 000
7 000 5 000
,
, ,
x 5 000
2 000
,
,
Formal Derivation
Solving for x = F-1(p),
5,000 7,000
1
F(x)
P
x
x P 5 000 2 000, ,
Note: 1. P = 0 x = 5,000
2. P = 1 x = 7,000
Class Problem
You are given the following cash flow diagram. The Ai are iid shifted exponentials with location parameter a = 1,000 and scale parameter = 3,000. The cumulative is then given by
7,000
A1 A2 A3
F x e x( ) ( , )/ , 1 1 000 3 000 , x > 1,000
Class Problem
You are given the first 3 random numbers U(0,1) as follows:
P1 = 0.8
P2 = 0.3
P3 = 0.5
You are to compute one realization for the NPW.MARR = 15%.
7,000
A1 A2 A3
F x e x( ) ( , )/ , 1 1 000 3 000
Class Problem
P1 ex 10003000
ex 10003000
1 P
x 10003000
ln(1 P)
x1,000 3,000 ln(1 P)
Class Problemx1,000 3,000ln(1 P)
A1 = 1,000 - 3000 ln(1 - .8)
= 5,828
A2 = 1,000 - 3000 ln(1 - .3)
= 2,070
A3 = 1,000 - 3000 ln(1 - .5)
= 3,079
Class Problem
7,000
5,8282,0703,079
NPW = -7,000 + 5,828(1.15)-1 + 2,070(1.15)-2 + 3,079(1.15)-3
= 1,657
