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Problem 13-14 A jewelry firm buys semiprecious stones to make bracelets and rings. The supplier quotes a price of $8.20 per s
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Answer #1

a) Order quantity of 600 minimizes total cost.

Annual demand, D = 18*111 1998
Order cost, Co $49
Quantity 1-399 400-599 600+
Cost, C $9.10 $8.60 $8.20
Holding cost , Cc $3.00 $3.00 $3.00
EOQ = sqrt(2DCo/Cc) 255.48 255.48 255.48
Considered Order quantity - Q for calculation 255.00 400.00 600
EOQ within range so Q = 255 Consider min 400 to get price discount Consider min 600 to get price discount
Order size 1-399 400-599 600+
Product cost = D*Cost, Different Costs $18,181.8 $17,182.8 $16,383.6
Holding cost = (Q/2) * Cc, Differenct Q as mentioned above $382.50 $600.00 $900.00
Ordering cost = (D/Q) * Co $383.93 $244.76 $163.17
Total cost = Sum of above 3 $18,948 $18,028 $17,447
Lowest Cost

b) Optimal order size is still 600 stones

Annual demand, D = 18*111 1998
Order cost, Co $49
Quantity 1-399 400-599 600+
Cost, C $9.10 $8.60 $8.20
Holding cost , Cc = 31%*C $2.82 $2.67 $2.54
EOQ = sqrt(2DCo/Cc) 263.46 271.01 277.54
Considered Order quantity - Q for calculation 263.00 400.00 600
EOQ within range so Q = 263 Consider min 400 to get price discount Consider min 600 to get price discount
Order size 1-399 400-599 600+
Product cost = D*Cost, Different Costs $18,181.8 $17,182.8 $16,383.6
Holding cost = (Q/2) * Cc, Differenct Q as mentioned above $370.96 $533.20 $762.60
Ordering cost = (D/Q) * Co $372.25 $244.76 $163.17
Total cost = Sum of above 3 $18,925 $17,961 $17,309
Lowest Cost

c) Reorder quantity = Lead-time * average demand = 4*18 = 72 stones

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