IE383_Assignment5_Sol(updated)

# IE383_Assignment5_Sol(updated) - IE 383 Assignment 5...

This preview shows pages 1–3. Sign up to view the full content.

IE 383 Assignment 5 solution Problem 1 First determine the line yield of the subsystem X consisting of box 4 and 5. Here we need to pay attention that THE FACTOR FOR EACH LINE DENOTES THE PERCENTAGE OF MATERIAL COMING INTO THE BOX GOES OUT ALONG THE LINE, for example, 0.2 denotes 20% of material going into box 1 goes along the line. Thus if we combine box 4 and 5 the factor for the line going into box 4 denotes of the material flowing through box 4 and 5 how much goes through box 4 and the factor should be 0.3/0.5=0.6. As such, the factor for the line going into box 5 should be 1-0.3/0.5=0.2/0.5=0.4. LY(X) = (0.2/0.5) * (1-0.6) + (0.3/0.5) * (1-0.1)= 0. 7 Then, substitute 4, 5 with subsystem X and determine the line yield of subsystem Y. Workstation Scrap rate 1 0.24 2 0.15 3 0.2 4 0.1 5 0.6

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

View Full Document
Then substitute 1 and X with Y and calculate the line yield of the system. LY(S) =(0.5 * 0.4 + 0.5 * 0.8) *0.85= 0. 51 a) System line yield = 0. 51 b) Materials needed = 1000/line yield= 1960.78 Problem 2
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 09/23/2009 for the course IE 383 taught by Professor Leyla,o during the Spring '08 term at Purdue University-West Lafayette.

### Page1 / 3

IE383_Assignment5_Sol(updated) - IE 383 Assignment 5...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online