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Cost minimization, Linear trend equation, Maximization

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Part 1: Inventory manager for Itex received wheels bearing from Wheel - Rite a small producer of metal parts , Unfortunately , Wheel -rite can only produce 500 wheel bearings from wheel rite each year. Since Itex operates 200 working days each year., its average daily demand for wheel bearings is 50. The ordering cost for Itex is $40 per order and the carrying cost is 60 cents per wheel bearing per year. How many wheel bearings should Itex order from wheel Rite at one time? Wheel -rite has agreed to ship the maximum number of wheel bearings that it produces each day to Itex when an order has been received.

Part 2: National Scan, Inc., sells radio frequency inventory tags (RFID). Monthly sales for a seven-month period were as follows:

Month Sales (000 units)
Feb..................19
Mar..................18
Apr...................15
May..................20
Jun...................18
Jul....................22
Aug..................20
What is a linear trend equation?

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Solution Summary

How to minimize the cost for Itex by choosing the optimal number?

How to construct the linear trend equation?

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Part 1: Inventory manager for Itex received wheels bearing from Wheel-Rite a small producer of metal parts , Unfortunately , Wheel-rite can only produce 500 wheel bearings from wheel rite each day. Since Itex operates 200 working days each year., its average daily demand for wheel bearings is 50. The ordering cost for Itex is $40 per order and the carrying cost is 60 cents per wheel bearing per year. How many wheel bearings should Itex order from wheel Rite at one time? Wheel-rite has agreed to ship the maximum number of wheel bearings that it produces each day to Itex when an order has been received.

Carrying cost per bearing per year is 0.60. Daily carrying cost is 0.6/365 = 0.001643835.

Let Itex order x bearings at a time. Then it has to place n orders where n = 10000/x.

Cost of each order is 40.

The total cost of ...

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