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Inventory Mgmt

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Chapter Sixteen: Inventory Management
PROBLEM SUMMARY
1. EOQ model 2. EOQ cost analysis (problem 1) 3. EOQ model 4. EOQ model 5. Noninstantaneous receipt model 6. Shortage model 7. EOQ model and reorder point 8. EOQ model and reorder point 9. Noninstantaneous receipt model 10. Noninstantaneous receipt model 11. EOQ model and reorder point 12. Noninstantaneous receipt model 13. Shortage model 14. Shortage model 15. Shortage model 16. Quantity discount model 17. EOQ model 18. EOQ model 19. Noninstantaneous receipt model 20. EOQ model 21. EOQ model analysis 22. Noninstantaneous receipt model analysis 23. Carrying cost determination 24. Quantity discount model 25. Quantity discount model 26. Quantity discount model 27. Quantity discount …show more content…

d = 180 lbs./day r = 250 lbs./day

( )

(

)

≅ 149,138 + 149,138 = $298,276

D = 65,700 Co = $125 243

12.

Operates 360 days/\year 12 converters 5 tons coal/day/converter D = (5 tons)(12 converters)(360 days) = 21,600 tons/year Co = $80 Cc = 20% of average inventory level Cc = (.20)($12) = $2.40 a) Q = 2CoD = 2(80)(21,600) = Cc 2.4 1, 440, 000 = 1,200 tons b) TC = Cc Q + Co D Q 2 = (2.4)

c) N = D = 10, 000 = 3.95 orders / year Q 2, 529.8 Tb = T = N 250 3.59 = 63.3 days

d) Q = 2,529.8, R = 60 The number of operating days to receive the entire order is Q = 2,529.8 = 42.16 days R 60

14.

D = 3,700 Co = $420 Cc = $1.75 Cs = $4 Q = = 2Co D  Cs + Cc  Cc  Cs    2(420)(3, 7000) 1.75 + 4 1.75 4

(1,200 ) +  (80)(,21,600)  2 1 200

= 1,440 + 1,440 = $2,880

(

)

= 1,598 tires c) R = L D = 5(21,600) = 300 tons 360 360 13. D = 10,000 logs/year T R = 250 days/year = 60(250) = 15,000/year Cc  S = Q C +C   c s = (1,598) = 486.3 TC = Cs S2 C (Q − S)2 + c + Co D Q 2Q 2Q

. (1.17575 4) +

Co = $1,600 Cc = $15 a) Q = 2CoD 2(1,600)(10, 000) = D 10, 000 (15) 1− Cc 1− R 15, 000

( )

(

)

=

(4)(486.3)2 (1.75)(1,598 − 486.3)2 + 2(1,598) 2(1, 598) 700 ( 3,,598 ) 1

= 6, 400, 000 = 2,529.8 logs b) TC = Co D + Cc Q 1− D 2 Q R

+ (420)

( ) =

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