2. Today’s Presentation It’s a day like all days – filled with those events which alter and change the course of history.

3. The Production Planning Problem – Homework #3 Cost per hour to operate line Hours available per month Fixed cost each month to reconfigure line for Product A Fixed cost each month to reconfigure line for Product B Factory 1 Line 1 $1324801030950 Factory 1 Line 2 $1306001200 Factory 2 Line 1 $13610009001200 Factory 2 Line 2 $13868014001500

4. More data Available labor hours per month Labor rate ($/hour) Factory Burden factor Available units of raw material per month Factory 12960$48.71100 Factory 24400$45.81500 Unit selling price Production line hours per unit Units raw material per unit of product Raw material cost per unit of raw material Labor hours per unit Minimum production per month Product A$4600810$10230120 Product B$60001213$62.504385 Determine how many units should be produced each month on each production line in order to maximize monthly profits.

5. Compute Unit Profit factory 1 factory 2 line 1 line 2 line 1 line 2 product Aproduct Bproduct Aproduct Bproduct Aproduct Bproduct Aproduct B XA11XB11XA12XB12XA21XB21XA22XB22 selling price46006000460060004600600046006000 line cost/hr132 130 136 138 line hr/unit8128 8 8 line unit cost10561584104015601088163211041656 raw mat unit cost10262.510262.510262.510262.5 raw mat unit/prod1013101310131013 raw mat cost/unit1020812.51020812.51020812.51020812.5 labor hr/unit3043304330433043 labor rate $/hr48 45 labor unit cost14402064144020641350193513501935 factory burden rate0.7 0.8 burden unit cost10081444.810081444.81080154810801548 total unit cost45245905.345085881.345385927.545545951.5 unit profit7694.792118.76272.54648.5 fixed cost10309501200 900120014001500

6. The Answer Product AProduct B Factory 1 Line 118.84 Factory 1 Line 250.0 Factory 2 Line 1123.5 Factory 2 Line 216.16 Maximum monthly profit ___$11,609.78

7. The Formulation Let X ijk = units produced product i (i = A,B) in factory j (j = 1,2) on line k (k = 1,2) Max 76XA11+94.7XB11+92XA12+118.7XB12+62XA21+72.5XB21+46XA 22+48.5XB22-1030ZA11-1200ZA12-900ZA21-1400ZA22 - 950ZB11-1200ZB12-1200ZB21-1500ZB22

8. Some of the constraints !Min production levels XA11+XA12+XA21+XA22>120 XB11+XB12+XB21+XB22>85 !Available labor hours 30XA11+30XA12+43XB11+43XB12<2960 30XA21+30XA22+43XB21+43XB22<4400 !Available raw material 10XA11+10XA12+13XB11+13XB12<1100 10XA21+10XA22+13XB21+13XB22<1500

9. More of the constraints !Production line hours 8XA11+12XB11<480 8XA12+12XB12<600 8XA21+12XB21<1000 8XA22+12XB22<680 ! !Fixed cost XA11-10000ZA11<0 XA12-10000ZA12<0 XA21-10000ZA21<0 XA22-10000ZA22<0 XB11-10000ZB11<0 XB12-10000ZB12<0 XB21-10000ZB21<0 XB22-10000ZB22<0

10. Inetger Solutions Best Integer Solution - OBJECTIVE FUNCTION VALUE = 11,478.10 VARIABLE VALUE XA11 0.000000 XA12 0.000000 XA21 122.000000 XA22 0.000000 XB11 18.000000 XB12 50.000000 XB21 0.000000 XB22 17.00

11. Continued Part Period Balancing (PPB) 10 Inventory holding cost = h (PP m ) Find m so that K h(PP m ) or PP m K / h We in the business call K / h “the economic part period factor” Order quantity = Q = D 1 + D 2 + … + D m

12. An Old Favorite 11 Wk1Wk2Wk3Wk4Wk5Wk6Wk7Wk8Wk9 Wk10 42 32122611245147638 PP m  K/h = 132 /.6 = 220 pp1 = 0 pp2 = 42 Pp3 = 42 + (2) 32 = 106 pp4 = pp3 + 3(12) = 142 pp5 = pp4 + 4(26) = 246 Q 1 = 42 + 42 + 32 + 12 + 26 = 154 pp1 = 0 pp2 = 45 pp3 = 45 + (2) 14 = 73 pp4 = 73 + 3(76) = 301 Q 6 = 112 + 45 + 14 + 76 = 247 Q 10 = 38 cost =3 x 132 + 328 x.60 = $724.20 Wk1Wk2Wk3Wk4Wk5Wk6Wk7Wk8Wk9 Wk10 42 32122611245147638 154 247 38 1127038260135907600

13. 12 No Fixed Cost I believe this problem can be solved as a transportation problem.

14. 13 The Transportation Problem x tj = number of units produced in month t to satisfy month j demands I’m going to need an example of this.

15. 14 The Example h = $2 per unit per month

16. 15 What can we conclude from all of this? Most heuristics outperform EOQ the Silver-Meal heuristic incurs an average cost penalty relative to Wagner-Whitin of less than 1 percent. Significant costs penalties using Silver-Meal will incur if demand pattern drops rapidly over several periods when there are a large number of periods having no demand

17. 16 Can we have some really neat homework problems? Huh? Text: Chapter 7: problems 13, 14, 17, 18, 19, 22

18. 17 Safety Stock When demand or lead-time is random (or both), then safety stock may be established as a “hedge” against uncertain demands. For the deterministic case: R = D L For the stochastic case: R = LTD avg + s where LTD = a random variable, the lead-time demand, LTD avg = average lead-time demand and s is the safety stock. Uncertain Lead-time Demands

19. 18 Safety Stock based on Fill Rate Fill rate criterion: set s = z STD where STD = standard deviation of the lead-time demand distribution then R = LTD avg + z STD LTD LTD avg Shortage probability Pr{LTD > R) = p R s

20. 19 Compute the variability coefficient, v = variance of demand per period square of average demand per period If V <.25, use EOQ with D avg else use a DLS method But that is only a “rule of thumb.” But I need to know when demands are lumpy, don’t I?