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Reorder point Lead time Order qty., q NO BACKLOGGING PURCHASE Inv. I.

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Presentation on theme: "Reorder point Lead time Order qty., q NO BACKLOGGING PURCHASE Inv. I."— Presentation transcript:

1 Reorder point Lead time Order qty., q NO BACKLOGGING PURCHASE Inv. I

2 PRODUCTION q Inv. II

3 q b BACKLOGGING Inv. III

4 b Inv. IV

5 WITHOUT BACKLOGGING C = unit cost (Rs/piece) C 1 = i X C = carrying cost (Rs/unit/time) C 2 = shortage/backlogging cost (Rs/unit/time) C 3 = order cost (Rs/order)

6 C = unit cost (Rs/piece) C 1 = i X C = carrying cost (Rs/unit/time) C 2 = shortage/backlogging cost (Rs/unit/time) C 3 = order cost (Rs/order)

7 C = unit cost (Rs/piece) C 1 = i X C = carrying cost (Rs/unit/time) C 2 = shortage/backlogging cost (Rs/unit/time) C 3 = order cost (Rs/order) b WITH BACKLOGGING

8 C = unit cost (Rs/piece) C 1 = i X C = carrying cost (Rs/unit/time) C 2 = shortage/backlogging cost (Rs/unit/time) C 3 = order cost (Rs/order) b

9 SENSITIVITY STUDIES ON CLASSICAL LOT-SIZE MODEL q ANNUAL COSTS LOT SIZE q q1q1 q*q* q2q2 TC TC min Total cost Carrying cost Order cost Inv. Level Average Inventory = q/2 Sensitivity Q = bq*, b > 0 b TC/TC min

10 Total annual cost = Annual usage  )( EOQ

11 OPERATING AT A LOT-SIZE of 1000 rather than EOQ of 700 is WARRANTED HERE ANNUAL COSTS q ,700 TAC(Rs 5) 1000 TAC(Rs 4.85) TAC(Rs 4.75) 25,098 24,995

12 = = )(. XX )Rs( EOQ = = )(. XX )Rs( EOQ IN THIS CASE A LOT SIZE OF 112 RESULTS IN MINIMUM COST ANNUAL COSTS q EOQ =

13 DETERMINISTIC SINGLE ITEM MODEL t1t1 t2t2 tptp t3t3 t4t4 t I max 0 -b Rate of fall, d Rate of rise p-d

14 COSTS/CYCLE Notice that t1t1 t2t2 t3t3 t4t4

15 AVERAGE ANNUAL COSTS K (q, b) K (b, q) = Substituting for t, (t 1 + t 4 ), (t 2 + t 3 ) & I max in terms of q, b we obtain K (b, q) =

16 OPTIMAL RESULTS Annual cost is K (b, q) The solution of these simultaneous equations yields the optimum values q* and b* as follows: and

17 U3U3 U2U2 U1U1 Amt. of inventory on hand LT 2 Amt. of inventory on hand Reorder level, R Safety stock (s s) Avg. lead time usage (U) Amt. used during Lead time Amt. of inventory on hand Q LT 1 LT 3 Q Order qty, Q Time

18 COMPUTATIONS FOR R Z Probability of stock out

19 Similar data on demands for last six months yield EXAMPLE (p305, ch. 10) d = 40 units/day Var (d) = 30 (units/day) 2 LT = = days Var (LT) = (7 – 14.83) 2 + (12 – 14.83) 2 + … 6 -1 = (day) 2 (contd.)

20  EXAMPLE (contd.) Units demanded per lead time ,u  Units per lead time Desired SO/yr = 0.33 (as stated earlier) P = desired probability of stockout per order cycle From tables Z = 1.39 SS = 1.39 (237.5) = R = = Z Order cycles/yr = 4 (given) = n

21 IMPROVING RELIABILITY OF LEAD TIME Safety stock = 1.39 (21.09) = 29 ( compared to 330 earlier) R = 622 (compared to 923 earlier)  Inventory lowered by 301 units  Annul savings = Rs 1 X 301 = Rs 301 Thus it is worthwhile to improve reliability of lead time If var (LT) = 0 Then var (U) = 30 X =  u =  = units per lead time (compared to the original 237.5)

22 Lead time Reorder point Avg. Lead time consumption Reserve stock Safety stock Demand uncertainties Lead time uncertainties   Stock Avg. demand for maximum delay Probability of delay a)Avg. demand during avg. lead time (buffer) b)Variations in demand during avg. lead time, depending on service level(reserve stock) c) Avg. demand during delivery delays (safety stock) )kx( L D 

23 Q system units 1414 x).( x,x qty Order  Safety Reserve Buffer point reorder  units x 52 20,000 time lead during demand avg. Buffer  time lead during demand of deviation std. x k stock Reserve  units ) x 4( 1.64  delay of yprobabilit x delay max during demand Avg. stock Safety  units.x.x ) 3 x 20,000 (  units point Reorder  time S 2062 Stock level Q = 1414 Lead time = 4 weeks 1.64  Mean,  0.95

24 P system wks.3.7 weeks52 x, years,Demand EOQ periodReview  Can be rounded off to either 3 or 4 weeks depending on cost consideration 3 weeks 1,154 Rs.xxx ,000 cost carrying inventory Annual   Total inv. Cost = = Rs Rs 100 Rs x 3 52 cost ordering Annual 

25 4 weeks 1300 Rs 100 x 4 52 cost ordering Annual  1,154 Rs.xxx 13 20,000 cost carrying inventory Annual   Total inv. Cost = = Rs 2840  Review period is 4 weeks Desired inventory level = Buffer + Reserve + Safety = 3668 units x 52 20,000 Buffer  units x 50 x 8 Reserve  units x 3 x )(20,000/52 Safety  P system

26 SAFETY STOCK DETERMINATION Chosen reorder level Distribution of lead time demand Prob. Of stockout


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