Variance than the average of two independent random

Info icon This preview shows pages 140–145. Sign up to view the full content.

View Full Document Right Arrow Icon
variance than the average of two independent random variable distributed according to N. Firstly, if 1 2 U U , N = 2 with probability !/2 and other is N = 2 + N a with probability ½. So, [ ] [ ] a N 2E 2 N E + = [ ] ( ) [ ] [ ] [ ] 2 a a 2 2 N E 1/2 N E 2 4 N 2 E 1/2 (4) 1/2 N E + + = + + = a [ ] [ ] ( ) [ ] [ ] ( ) 0.4997 8 - 4e - 14e N E - N E N Var Thus, 2e - 8 N E and 4 - 2e N E 2 2 a 2 a a 2 a a = = = = Now consider the random variable N and M, ( ) ( ) ( ) a a N Var N 4 Var M N Var = + = +
Image of page 140

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
So, ( ) ( ) 3.065 0.4997 1.5316 M N Var N N Var + + Therefore, the use of antithetic variable reduces the variance of the estimator by a factor of slightly more than 3. 21. Suppose we were interested in using simulation to compute θ =E [e u ]. Here, a natural variant to use as a control is the random number U. Discuss what sort of improvement over the raw estimator is possible. (20 marks) Solution θ = E[e U ] By using raw estimator, ( ) [ ] [ ] ( ) ( ) 0.2420 1 - e - dx e x e E - U E e Var 1 0 2 x 2 U 2U U = = = By using antithetic variable U, ( ) [ ] [ ] [ ] ( ) ( ) 0.14086 2 1 - e - 1 2 1 - e - dx e x e E U E - e U E U , e Cov 1 0 x U U U = = = = And then, ( ) [ ] [ ] ( ) ( ) ( ) ( ) ( ) [ ] ( ) ( ) ( ) ( ) ( ) 0.0039 0.2380 - 0.2420 0.14086 12 - e Var 1/2 - U c e Var Y Var Y X, Cov - X Var μ - Y c X Var have we 1/12 U E - U E U Var 2 U * U 2 y * 2 2 = = = + = + = = Therefore, the use of control variate U can lead to a variance reduction of up to 98.4 percent.
Image of page 141
Computer Integrated Manufacturing IT 4036 for Second Semester Sample Question
Image of page 142

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
1. Explain with the block diagram showing various functions of CAD and their relationships to design CAD and manufacturing operations. (20 marks) 2. Draw “Overview of CAM” and explain briefly the “Direct computer application” in manufacturing. (10 marks) 3. Sketch the scope of CAD/CAM. (10 marks) 4. Discuss briefly the “Direct application” in computer aided- manufacturing. (10 marks) 5. Explain the computerized elements of a CIM system. (20 marks) 6. Explain the Computer Aided Process Planning. (10 marks) 7. What are the benefits of Computer Aided Process Planning ? (10 marks) 8. Explain the fundamental concept in Material Requirements Planning. (10 marks) 9. What are the output reports and benefits of the MRP (Materials Requirements Planning)? (10 marks) 10. To illustrate how MRP works let us consider the requirements planning proceeded for one of the components of product PI. The components we will consider a C4. This part happens to be used also on one other product P2 (see figure). However, only one of item C4 is used on each P2 produced. The product structure of P2 is given in Figure 1. Component C4 is made out of raw materials M4. One unit of M4 is needed to produce 1 unit of C4. Find the ordering and manufacturing lead times and make the MRP computations? (20 marks) The current inventory order and status of item M4 is in Figure. There are no stocks or orders for any of the other items listed below: P1: assembly lead time = 1 week P2: assembly lead time = 1week S2: assembly lead time = 1week S3: assembly lead time = 1 week C4: manufacturing lead time = 2 weeks M4: ordering lead time = 3weeks
Image of page 143