Question

Steps and equations used would be very helpful please !

Practice Problem (EOQ) Name Kathoarius Seta ar costs is 20% of A food processor uses approximately 27,000 glass jars a month for its fruit juice product. Each jar costs 510.8. Because of storage limitations,a lot size of 4,000 jars has been used. Annuai cost is 20% of a VE 2 1,000 x 12 334,000 jar's price, and reordering cost is60 ber order The company operates an average of 20 days a month Answer the following questions, showing calculation steps: a) Mow many times a year an order size of 4000 is placed? b) What ig the total cost of Q-4000 (Include ordering, holding and purchase cost) 160 31-1000 x60- 2. c) What is optimal order quantity and its total cost? d) What penalty is the company incurring by its present order size? The manager would prefer ordering 10 times each month but would have to justify any change in order size. One possibility is to simplify order processing to reduce the ordering cost. What ordering cost would enable the manager to justify ordering every other day (i.e., 10 times a month) e)

Answer 1

Given Data:

                Monthly Demand (d) = 27000 units

                Unit cost (P) = $10.8

                Lot size/Order size (Q) = 4000

                Annual holding cost(H) = 20% of unit cost = 0.2 X 10.8 = $2.16

                Cost per Order(S) = $60

a. The number of times the order size of 4000 has to be placed

To calculate that first we need to calculate the yearly demand (D) = 27000 X 12 = 3,24,000 units

So, no of orders = Total demand/ Order size = 324000/4000 = 81

So, 81 times we have to place an order with lot size 400

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b. Total Annual Cost for order size = 4000

                                                              =   Purchase cost + Ordering cost + Holding cost

=   unit cost * Demand + No of orders * Ordering cost + Average        inventory *Annual Holding Cost

                                                         =     10.8 * 324000 + (324000/4000) * 60 + (4000/2) * 2.16

                                                                =     3499200 + 4860 + 4320

                                                                =     $3508380

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c. Optimal order quantity can be calculated by applying EOQ formula

= (2DS/H)

= (2* 324000 * 60/2.16)

= 4242.64 = 4243 (Rounded up)

                Total cost with this order size = 10.8* 324000 + (324000/4243)* 60 + (4243/2)* 2.16

                                                                                = 3499200 + 4581.66 + 4582.44

                                                                                = $3508364.1

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d. Penalty the company getting = difference between total cost = 3508380 – 3508364 = $16

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e. Order will be 10 times a month so, no of orders annually = 12*10= 120

Ordering cost, we need to calculate with 120 orders

Hence Q we need to calculate first => 324000/Q =120

• Q = 324000/120 = 2700

To keep the total cost optimal i.e. $3508364

TC=3508364 = 10.8* 324000 + 120 * S + 2700/2 * 2.16

Solving the above equation, we can get S = $52

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f. Quantity Discount

     All units quantity discount model

                Q<4999 P= $10.8

                5000<Q<6000 P =$10

               Q>= 6000 P =$9.5

Compare total cost to get the optimum quantity of order size

EOQ at each level we need to calculate

EOQ 1 = (2* 324000*60)/10.8*0.2 = 4243 its valid for < 4999

TC1 = $ 3508364 ( Already calculated above )

   EOQ2 = (2*324000*60)/10*0.2 = 4409 which is not valid for 5000 to 6000 order quantity

EOQ 3= (2*324000*60)/9.5*0.2 = 4523.62 Also not valid for order >= 6000

So The order size should not be changed from current