5351: IE Advanced Production and Inventory Control Homework #1
1. A company XYZ purchases air filters that are used at the rate of 350 per year. The cost of each filter is $30 and the cost of placing each order is $10. The annual inventory carrying cost rate is 18%. Shortage cost consists of two components: 1. Fixed cost of $0.30 per unit 2. Variable cost of $5 per unit short per year a) Find both the optimal order quantity and the optimal shortage b) The cycle length and the total annual cost
2. A gift shop sells Little Lentils Cuddly animal dolls stuffed with dried lentils- at a very steady pace of 10 per day, 310 days per year. The wholesale cost of the dolls is $5, and the gift shop uses an annual interest rate of 20% to compute the holding costs. a) If the shop wants to place an average of 20 replenishment orders per year, what order quantity should it use? b) If the shop orders dolls in quantities of 100, what is the implied fixed order cost? c) If the shop estimates the cost of placing a purchase order to be $10 , what is the optimal order quantity?
3. The El Paso Company produces a chemical compound that is used in beauty products. The compound can be produced at a rate of 10,000 pound per day. Annual demand for the compound is 0.6 million pounds per year. The fixed cost of setting up for a production run of the chemical compound is $1,500, and the variable cost of production is $3.50 per pound. The company uses an interest rate of 22% to account for the cost of capital, and the costs of storage and handling of the chemical amount to 12% of the value. Assume that there are 250 working days in a year. a) What is the optimal size of the production run for this particular compound? b) What proportion of each production cycle consists of uptime and downtime? c) What is the average annual cost of holding and setup attributed to this item? d) If the compound sells for $3.90 per pound, what is the annual profit the company is realizing for this item?
P K c I
= 10,000/day 6 6 = .6 x 10 /year = .6 x 10 /250 = 2400/day = 1500 = 3.50 = .22 + .12 = .34
h = Ic(1 - /P) = (.34)(3.50)(1 - 2400/10,000) = .9044 a) b) Q=
2 K (2)(1500)(2400)(250) = = 44.612 h' lbs .9044
c) d)
T1 = Q/P = 44,612/10,000 = 4.46 days T = Q/ = 44,612/2400 = 18.59 days T2 = T - T1 = 14.13 Up time = 4.46 days or 4.46/18.59 = 24 (24% of cycle is up time) Down time = 14.13 days (76% of cycle is down time) Annual holding and set-up cost = 2 K h' = (2)(1500)(600, 000)(.9044) = $40,347.49 annually. If the compound sells for $3.90 per lb., then the annual profit exclusive of holding and set-up cost is (3.90 - 3.50)(600,000) = $240,000.00
Hence the profit generated from this item is $240,000.00 - 40,347.49 = $199,652.51 annually.
4. Hervis Rent-a-Car has a fleet of 2,500 Rockets serving the Los Angeles area. All Rockets are maintained at a central garage. On the average, eight Rockets per month require a new engine. Engines cost $850 each. There is also a $120 order cost (independent of the number of engines ordered). Hervis has an annual holding cost rate of 30% on engines. It takes two weeks to obtain the engines after they are ordered. For each week a car is out of service, Hervis loses $40 profit. a) What are the optimal ordering quantity and the optimal shortage b) At what point should Hervis reorder more engines? c) How many days after receiving an order does Hervis run out of engines? d) How long is Hervis without any engines per cycle? D = 8 x 12 = 96; Co = $120; Ch = .30(850) = $255;
Cb = 40 x 52 = $2080 Q * = 2DCo/Ch (Ch + Cb)/Cb = 2(96)(120)/255 x (255+2080)/2080 = 10.07 10 S * = Q *(Ch/(Ch+Cb)) = 10(255/(255+2080)) = 1.09 1 Demand is 8 per month or 2 per week. Since lead time is 2 weeks, lead time demand is 4. Thus, since the optimal policy is to order 10 to arrive when there is one backorder, the order should be placed when there are 3 engines remaining in inventory. How many days after receiving an order does Hervis run out of engines? How long is Hervis without any engines per cycle? Inventory exists for Cb/(Cb+Ch) = 2080/(255+2080) = .8908 of the order cycle. (Note, (Q *-S *)/Q * = .8908 also, before Q * and S * are rounded.) An order cycle is Q */D = .1049 years = 38.3 days. Thus, Hervis runs out of engines .8908(38.3) = 34 days after receiving an order. Hervis is out of stock for approximately 38 - 34 = 4 days.