Chapter 10

Chapter 10 - Chapter 10 Managing Economies of Scale in the...

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Chapter 10: Managing Economies of Scale in the Supply Chain: Cycle Inventory Exercise Solutions 1. The economic order quantity is given by hC DS 2 . In this problem: D = 109,500 (i.e., 300 units/day multiplied by 365 days/year) S = \$1000/order H = hC = (0.2)(500) = \$100/unit/year So, the EOQ value is 1480 units and the total yearly cost is \$147,986 The cycle inventory value is EOQ/2 = 1480/2 =740 Worksheet 10.1 provides the solution to this problem. 2. (a) If the order quantity is 100 then the number of orders placed in a year are: D/Q = 109500/100 = 1095. So, 1095 orders are placed each year at a cost of \$1000/order. Thus, the total order cost is \$1,095,000. Cycle inventory = Q/2 = 100/2 = 50 and the annual inventory cost is (50)(0.2)(500) = \$5,000 (b) If a load of 100 units has to be optimal then corresponding order cost can be computed by using the following expression: order per S S hC DS Q 57 . 4 \$ ) 109500 )( 2 ( ) 500 )( 2 . 0 ( ) 100 ( ) 500 )( 2 . 0 ( ) 109500 )( 2 ( 100 2 2 = = = = This analysis is shown in worksheet 10-2. 3. (a) We first consider the case of ordering separately: For supplier A: Order quantity (Q) = ) 5 )( 2 . 0 ( ) 100 400 )( 20000 ( 2 + = 4,472 units/order 1

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Total cost = order cost + holding cost = (20000/4472)(500) + (4472/2)(0.2)(5) = \$4,472 Similarly, for suppliers B and C the order quantities are 1768 and 949 and the associated total costs are \$1,414 and \$949, respectively. So, the total cost is \$6,835 (b) In using complete aggregation, we evaluate the order frequency (n*) as follows: So, n* of the case is = * 2 S hC D hC D hC D C C B B A A + + S* = 400 + 3(100) = \$700 So, n* = ) 700 ( 2 ) 5 )( 2 . 0 ( 9000 ) 4 )( 2 . 0 ( 2500 ) 5 )( 2 . 0 ( 20000 + + = 4 orders/year For supplier A: Q = D/n = 20000/4 = 5000 units/order Total cost = order cost + holding cost = 4(500) + (5000/2)(0.2)(5) = \$4,500 Similarly, for suppliers B and C the order quantities are 625 and 225 and the associated total costs are \$650 and \$513, respectively. So, the total cost is \$5,663. Worksheet 10-3 provides the solution to this problem 4. (a) This is a quantity discount model and the decision is to identify the optimal order quantity in the presence of discounts. We evaluate the order quantities at different unit prices using the economic order quantity equation as shown below: For, price = \$1.00 per unit Q= EOQ = 984 , 30 ) 1 )( 2 . 0 ( ) 12 )( 400 )( 20000 ( 2 = Since Q > 19,999 We select Q = 20,000 (break point) and evaluate the corresponding total cost, which includes purchase cost + holding cost + order cost 2
Total Cost = ( 29 ( 29 400 20000 ) 12 )( 20000 ( ) 98 . 0 ( 2 . 0 2 20000 ) 98 . 0 )( 12 )( 20000 ( + + = \$ 241,960 Similarly we evaluate the EOQs at prices of p = 0.98 (Q = 31298) and p = 0.96 (Q = 31623, which is not in the range so use Q = 40001). The corresponding total costs are \$241,334 and \$236,640. So, the optimal value of Q = 40001 and the total cost is \$236,640

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This note was uploaded on 09/17/2009 for the course INDUSTRIAL 0906547 taught by Professor Khaldontahboub during the Three '09 term at ADFA.

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Chapter 10 - Chapter 10 Managing Economies of Scale in the...

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