MGOC20H3 Study Guide - Final Guide: Admissible Set, Operations Management
Document Summary
Mgoc20 operations management: a mathematical approach l01, midterm solution. Q1. (a) p4 = . 75, h = . 75(. 10) = . 875/unit/year. = 140. 095, eoq4 is not admissible set q = 400. = 139. 17, eoq3 is not admissible set q = 200 (19. 00)(460) (1. 90)(200) = . 00 p2 = . 50, h = . 50(. 10) = . 95/unit/year. =137. 37, eoq2 is admissible so stop set q = 137. 37 (19. 55)(460) (1. 95)(137. 37) Therefore it is best to order 200 units each time since it has lowest cost. . 00 (b) # of orders = d/q* = 460/200 =2. 3, # of days between orders = q*/d = 200/460 = . 434 * 365 = 158. 7 days. E[l] = (5 + 2)/2 = 3. 5 weeks, var[l] = (5/50 2/50)2/12 = 0. 0003 years. 700 boxes lead time demand x is normally dist. with mean = e[x]= e[l]* e[d] = (3. 5/50)(5000) = 350 var[x] = e[l]*var[d] + e[d]2 * var[l] = (3. 5/50)(45)2 + 50002(0. 0003) = 7641. 75. From z-table, p[z 1. 91] = . 9719 close enough, solve.
