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 b0.50.80.91.01.11.21.52.0 TC/TC min 1.2501.0251.0061.0001.0051.0171.0831.29

10. Total annual cost = Annual usage 700 1 4950002  )( EOQ

11. OPERATING AT A LOT-SIZE of 1000 rather than EOQ of 700 is WARRANTED HERE ANNUAL COSTS q 7002500 25,700 TAC(Rs 5) 1000 TAC(Rs 4.85) TAC(Rs 4.75) 25,098 24,995

12. 79 220 52502 2 = = )(. XX )Rs( EOQ 112 120 52502 1 = = )(. XX )Rs( EOQ IN THIS CASE A LOT SIZE OF 112 RESULTS IN MINIMUM COST ANNUAL COSTS q EOQ =79 112100

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 = 7 + 12 +25 + 16 + 14 + 15 6 = 14.83 days Var (LT) = (7 – 14.83) 2 + (12 – 14.83) 2 + … 6 -1 = 34.97 (day) 2 (contd.)

20.  EXAMPLE (contd.) Units demanded per lead time 523739756.,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) = 330.1 R = 593.3 + 330.1 = 923.4 Z 0.083 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 14.83 = 444.9  u =  444.9 = 21.09 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  1020 100000202 Safety Reserve Buffer point reorder  units 1540 4 x 52 20,000 time lead during demand avg. Buffer  time lead during demand of deviation std. x k stock Reserve  units 164 50) x 4( 1.64  delay of yprobabilit x delay max during demand Avg. stock Safety  units.x.x ) 3 x 20,000 ( 3583101154310 52  units 2062 358 164 1540 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  00020 1414 00020 1414 Can be rounded off to either 3 or 4 weeks depending on cost consideration 3 weeks 1,154 Rs.xxx 17.33 20,000 cost carrying inventory Annual  20010 2 1  Total inv. Cost = 1733 + 1154 = Rs 2887 1733 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  20010 2 1  Total inv. Cost = 1300 + 1154 = Rs 2840  Review period is 4 weeks Desired inventory level = Buffer + Reserve + Safety = 3668 units 3080 8 x 52 20,000 Buffer  units 230 1.64 x 50 x 8 Reserve  units 350 0.31 x 3 x )(20,000/52 Safety  P system

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