CH13 online show

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Published on March 22, 2009

Author: Alancai

Source: authorstream.com

Slide 1: CHAPTER 13: Material Requirements Planning (MRP) and Enterprise Resource Planning (ERP) Slides prepared by Romulus Cismaru University of Regina Example 1: Page 489 : Example 1: Page 489 Example 1 (Cont.) : Example 1 (Cont.) Using the information above to do the follows: a) Determine the quantities of B, C, D, E, and F needed to assemble one X. Solution to Example 1 : Solution to Example 1 a). Thus, one X will require B: 2 C: 1 D:6 F: 2 E: 28 (Note that E occurs in three places, with 2+2+24) Example 1 (Cont.) : Example 1 (Cont.) Using the information above to do the follows: a) Determine the quantities of B, C, D, E, and F needed to assemble one X. b) Determine the quantities of B, C, D, E, and F needed to assemble ten X's, if you have the following in inventory: Solution to Example 1 : Solution to Example 1 b). Thus, given the amounts of on-hand inventory, 10 Xs will require B: 16 C: 0 D: 40 F: 0 E: 116 (=16+100) MRP Calculations : MRP Calculations MRP processing takes the end item requirements specified by MPS and explodes them into time-phased requirements for fabrication or assembly of subassemblies Product Structure for “Awesome” A : Product Structure for “Awesome” A A simplified MRP for 50 “Awesome A” Speaker Kits : A simplified MRP for 50 “Awesome A” Speaker Kits 200 200 Format of MRP : Format of MRP Example 2 (p.494) : Example 2 (p.494) Shutter Frame (2) Wood selection (4) Slide 12: shutter LT=1 week Slide 13: Times 2 Times 2 Slide 14: Times 4 Times 4 400-70=330 Example 2—Page 497 : Example 2—Page 497 Consider the two product structure trees shown below: Develop a MRP for item D. Given that the master schedule calls for 80 units of A in week 4 and 50 units of C in week 5. There’s a beginning inventory of 110 units of D on hand and all items have lead times of one week. Solution - MPR for Item A : Solution - MPR for Item A 80 80 80 80 80 A LT=1 week 50 times 1 for D Solution MRP for Item C : Solution MRP for Item C 50 80 C LT=1 week 50 50 50 50 times 2 for D Solution -- MRP for Item D : Solution -- MRP for Item D D LT=1 week Item A times 1 80 50 Item C times 2 100 110 110 110 110 30 70 70 70 80 0 100-30=70 110-80=30 Example : Example We have 10 alpha and 20 B at hand Slide 20: 10 10 10 10 10 10 10 40 40 40 50-10=40 50 50 50 100 100 100 Slide 21: 20 20 20 20 20 20 40 50 100 20 20 20 50 50 50 100 100 100 C is used for making both Alpha & B : C is used for making both Alpha & B Slide 23: 40 50 100 40 100 Times 2 Add 200 50+200=250 250 40 40 40 100 100 100 250 250 250 100 100 100 40 40 40 Slide 24: 40 100 Times 2 200 40 40 40 100 100 100 200 200 200 EOQ lot sizing : EOQ lot sizing When there is a net requirement, we order EOQ amount Balance setup cost and inventory holding costs Example : Example Initial project on hand: 35 Total requirement =(35+30+40+0+10+40+30+0+30+55) =270 Slide 27: D=270 units S=100 H=$1*10weeks=10 Q*= (2*270*100)/10 =73 Slide 28: (0) 35-35 =0 73 73 73 73 0 0) 0 Part period method calculation : Part period method calculation Economic part period If we hold 10 units for 2 periods, it would be 10*2=20 part period (EPP)=setup cost/ holding cost Example 3 : Example 3 Setup cost =$80; holding cost =0.95 EPP=80/0.95=84.21 rounds to 84 Slide 32: If we place order of 60 at period 1 we would have to carry 100-60 =40 units for 1 period 40*1=40 40+0=40 If we place order at period 1 to satisfy demand in The period 1&2, we need to order 60+40=100 Slide 33: 120-100=20; this is the extra amount of inventory we need to carry compare to order 100 at period 1; This extra 20 units will be used at period 3, that means we carry them at period 1 and 2 86 is very close to EPP(84) Slide 34: 152 30 4 120 206 86 is more close to EPP (84) than 206 Slide 35: We will order 122 to cover the first 4 period at period 1 Slide 36: 140 exceeds 84, so we need to stop here, 140 is more close to 84 than 0. So, we order 100 at period 5 If LT=1 : If LT=1

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