Assignment 02 Solution

ASSIGNMENT 2 (Operation Research) Maximum Marks: 40 Due Date:

Question 1 The demand for a product is 600 units per week, and the items are withdrawn at a constant rate. The setup cost for placing an order to replenish inventory is $25. The unit cost of each item is $3, and the inventory holding cost is $0.05 per item per week. (a) Assuming shortages are not allowed, determine how often to make a production run and what size it should be. (b) If shortage are allowed but cost $2 per item per month, determine how often to order and what size the order should be. Solution: * (a) Here size of order = Q 

2C2 D = 774.5967 C3

How often to order for time between order = t *  Q* / D = 1.291 (b) Size of order =

Q* 

2 C2 D C3

C3 C4 C4

= 812.4039

Time between order = t *  Q* / D = 1.3540 Question 2 Neon lights on the U of A campus are replaced at the rate of 100 units per day. The physical plant orders the Neon lights periodically. It cost $100 to initiate a purchase order. A neon light kept in storage is estimated to cost about $0.02 per day. The load time between placing and receiving an order is 12 days. Determine the optimal Inventory policy for ordering the neon lights. Solution: D = 100 units per day C2 = $100 per day C3 = $0.02 per day L = 12 days 2C2 D Q*  C3 

2 100  100  1000 neon lights 0.02

The associate cycle length is T = Q/D = 10 days Because the lead time L = 12 days exceeds the cycle length t (=10 days). We must compute L e. The no. of integer cycles included in L is n = (largest integer  12 /10 ) Thus, Le  L  nt  2 days The reorder point thus occurs when the inventory level drops to Le D  2 100  200 neon lights The inventory policy for ordering the neon lights is order 100 units whenever the inventory order drops to 200 units. The daily inventory cost associated with the proposal inventory policy is C  Q  * 2  C3   Q /D  2  20 / day Question 3 Find the optimal order quantity for a product for which the price-breaks are as follows: Items q 0  q < 100 100  q < 200 200  q

Price/Unit Rs. 20 Rs. 18 Rs. 16

The monthly demand for the product is 600 units. The storage cost is 15% of unit cost and the cost of ordering is Rs. 30 per order. Solution: EOQ at Rs. 20 = EOQ at Rs. 18 = EOQ at Rs. 18 =

2  600  30  109.54 20  0.15 2  600  30  115.47 18  0.15 2  600  30  122.47 16  0.15

The EOQ at Rs. 20 per item = 109.54. But the price per item is Rs. 20 only if the items are ordered in the range of 0 to less then100. This is therefore an infeasible solution. Similarly the EOQ at Rs. 16 per item is 122.47. This price is valid only for items ordered in the range 200 or more. This is also an infeasible solution.

Question 4 A company stocks an item that is demanded 1000 units per month and the shortages are allowed. If the unit cost is Rs 20 per unit, the cost of making one purchase is Rs 1000, the holding cost for one unit is Rs 40 per year and the cost of one shortage is Rs 150 per year, determine: (a) The economic purchase quantity. (b) The time between orders. (c) The number of orders per year. (d) The optimum shortages. (e) The maximum inventory. (f) The time of items being held. (g) The optimum annual cost. Solution: Let D  1000 units / month  12000 units / year C1  20 / unit , C2  1000 C3  40 units / year , C4  150 / year The economic purchase quantity. 2C2 D C3  C4 Q C3 C4 = 871.7798 The time between orders. t *  Q* / D = 0.0726 The number of orders per year  1/ t *  13.774 The optimum shortages C3 2C2 D S*  C4  C3  C4  = 183.53 The maximum inventory. I max  Q*  S *  688.24 The time of items being held t1  I max / D  0.0574 / years The optimum annual cost. Annual Cost = Item cost + Ordering cost + holding cost + shortage cost Now Item Cost = 688.24 Ordering cost = 1000

C3 I max t1  65.84 per order 24 Shortage cost = 65.949 Annual Cost = 1820.02 Holding cost =

So