Material Management 2010 Mid-Term Exam-#2 Solution By: Prof. Y. Peter Chiu 5 / 2010

#1 (a) By SPT: 1 – 3 – 6 – 2 – 7 – 5 – 4 Total Flow time= 171 ∴ Mean Flow Time= 171 / 7 = #1 (b) By EDD: 1 – 6 – 3 – 2 – 5 – 7 – 4 Total Flow time= 173 Total Tardiness=12 ∴ Maximum Tardiness = 8

#1 (c) 1st by EDD: 1 – 6 – 3 – 2 – 5 – 7 – 4 According to Moore’s algorithm Delete Job #5 First #1 (c) 2nd iteration: 1 – 6 – 3 – 2 – 7 – 4 According to Moore’s algorithm According to Moore’s algorithm Done! ∴ 1 – 6 – 3 – 2 – 7 – 4 – 5 & Number of Tardy Job = 1 ; That is Job #5

By Chiu’s algorithm V 1 ={3,6,8} max D i → 6 V 2 ={3,5,8} max D i → 5 V 3 ={2,3,8} → 8 V 4 ={2,3,7} → 7 V 5 ={2,3,4} → 2 V 6 ={1,3,4} → 4 V 7 ={1,3} → 3 Last → 1  The Optimal sequence is #2

#3 Finite Production Rate Model

(a) (b) #4  (c)

# 5 W-82R X-37R P-64R $ $ $

The End