CHAPTER 12. (The interpretation of the symbols used in the equations is given in page 3)

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1 CHAPTER 12 The equations needed: (The interpretation of the symbols used in the equations is given in page 3) 1. ATI = t D ws = I average weekly CU 2. I turn = CU I average 5. ds = I average daily CU 3. ws = 52 I turn Example 12.1 The XYZ Company has an annual demand of 30,0 units. The annual cost of goods sold is BD,227,000 and the average inventory is BD 479,000. Find the a. average transportation inventory given that the company sends some of its finished goods from its factory to one of its distribution centers via road freight which takes days transit. b. Inventory turnover c. Weeks of supply d. Days of supply, assuming that the company operates 20 days a year. e. Annual holding costs of inventory if the capital cost, storage cost and risk costs are 5%, 2%, and 3% per unit of inventory, respectively. Solution a. ATI = td = x30,0 = 504 units b. I turn = CU =,227,000 = 13 I average 479,000 c. WS = I average = 479,000 = 4 weeks of supply weekly CU,227, Alternatively: WS = 52 = 52 = 4 weeks of supply I turn 13 e. Days of supply = I average = 479,000 = 20 days of supply daily CU,227, days a year f. Annual holding costs = (5+2+3)% x I average = 10% x 479,000 = 47,900 [Inventory turnover is the no. of times a company sold and replaced its inventory during an accounting period. The higher the inventory turnover the most efficient the inventory system is]

2 The equations of the EOQ model: (The interpretation of the symbols used in the equations is given in page 3) 1. Q = 2DS H 5. I max = Q 2. TC =Annual ordering costs + Annual holding costs. I min = 0 = DS Q + Q 2 x H 7. R = d x L 3. N = D Q 8. t cycle = Q d 4. I average = Q 2 Example 12.2 on the EOQ model The cost per unit of a major material of a company is BD 80. The annual demand for this material is 3,000 units and the ordering cost is BD 135 per order. The annual holding cost is 15% of the cost per unit of the material, the lead time is 2 days and the company operates 240 days a year. a. Find the i. EOQ. EOQ = 2DS H = 2x3,000x x80 ii. Annual total costs of inventory. = 972, = 900 TC = DS Q + QH 2 = 3,000x x12 2 = 5, ,400 = 10,800 iii. The time, in days, of the inventory cycle d = D = 3,000 = 150 (the daily demand) no.ofoperating days in a year 240 t cycle = Q d = = days iv. Reorder point. R = d ı = 150 x 2 = 300 v. Number of orders. N = D = 3,000 = 40 times a year Q 900

3 b. Show the costs and quantity curves of the EOQ model for this problem. TC 10, (EOQ = I max ) Holding costs (QH/2) I average = 450 5,400 R= (EOQ) Ordering costs (DS/Q) I min = 0 days L=2 t cycle = The costs curve. The quantity curve Example 12.3 The given diagram shows the cost curves of the EOQ model. Answer the following questions assuming the ordering cost per order is BD 200. a. For Q = 1250, find the annual total costs of inventory and the annual holding cost per unit. b. For Q = 1425, find the annual holding costs and the annual ordering costs. 12,500 Solution: a. For Q = Annual total cost of inventory = annual ordering costs + annual holding costs. TC = 12, ,500 = 25,000 TC Annual holding costs = QH 2 12,500 = 1250xH 2 25,215 25,000 H = 12,500x2 1,250 = 20 14,250 12,500 b. For Q = 1,425 10,95 Annual holding costs = QH 2 = 1,425x20 2 = 14,250 Annual ordering costs = DS Q. We need to find D from the EOQ situation

4 For EOQ, the annual ordering costs = DS Q 12,500 = Dx200, D = 12,500x1,250 = 78,125 units 1, Therefore, the annual ordering costs for the quantity of 1,425 = 78,125x200 = 10,95 1,425 [Note that TC for the quantity of 1,425 = annual ordering costs + annual holding costs] = 10,95 +14,250 = 25,215

5 The equations of the EPQ model: (The interpretation of the symbols used in the equations is given in page 3) 1. Q = 2DS H(1 d p ) 7. P = d + I build 2. Q = d run + I max 8. TC = Annual ordering costs + Annual holding costs 3. I max = Q (1- d p ) = DS + I max H Q 2 4. P = Q t 9. I average = I max 2 5. d = d run t 10. R = d x L. I build = I max t 11. N = D Q Example 12.4 on the EPQ model The CBC Company has a yearly demand of 42,000 units of a certain material for their operations. The company has the capacity to produce 200 units from this material per day. The cost of setting up each production run is BD 19, the annual holding cost is BD 7 per unit of inventory and the company operates 300 days a year. a. Calculate the i. EPQ Q = 2DS H(1 d p ) Demand rate per day, d = d = 140 = 0.7 p 200 D = 42,000 = 140 no.of operating days in the year 300 Q = 2x42,000x19 7(1 0.7 ) = 2,800 ii. Annual total inventory costs TC = DS + I max Q 2 x H I max = Q (1- d ) = 2,800 (1 0.7) = 840 p TC = 42,000x19 2, x7 2 = 2, ,940 = 5,880

6 iii. Amount of material used to meet immediate demand. d run = Q - I max (from equation 2) = 2, = 1,90 iv. Number of production runs per year. N = D = 42,000 = 15 times a year Q 2,800 v. Length (time) of the production run in days. t = Q P = 2, = 14 days (from equation 4) [Note that t cycle = Q d = 2, vi. Reorder point, if the lead time is 2 days. = 20. This is not the same as t] R = dl = 140x2 = 280 units. This means that when the inventory level reaches 280 units, an order for a new production run will be made. b. Show the curves of the EPQ model EPQ = 2,800 d run =190 I max = 840 R=280 t= Time Example 12.5 on the EPQ model Refer to the given EPQ diagram to answer the following questions: 4,500 a. Find the EPQ, daily I build, p, d run, and R if the lead time is 2 days. EPQ = 4,500 (from the diagram) Daily I build = I max t P = Q = 4,500 t 20 = = 45 (equation ) = 225 (equation 4) 900 d run = Q - I max = 4, = 3,00 (equation 2)

7 R= dl. We need to find d; the daily demand rate. It can be found as d = d run t = 3,00 20 = 180 (equation 5). 20 days [Alternatively, d = p - I build = = 180 (equation 7)] R = 180x 2 days = 30 This means that 4,500 units are produced during the production run of EPQ=4, days; 3,00 of them go to meet immediate demand and 900 units go to build up inventory. During the non-production phase, the company uses material to meet demand from the inventory. When inventory level d run = 3,00 reaches 30 units a new quantity of 4,500 is ordered. Every day of the production phase (20 days), 225 units are produced, 180 units of them go to meet immediate demand and 45 units go to inventory. I max = 900 b. Suppose D, H, and S are the same for the EOQ and the EPQ models. R=30 Will the two models produce the same order quantity (Q)? We can answer this question using data from example 12.4 where t=20 D = 42,000, S = 19, H = 7, d = 140, p = 200 Using the EOQ model The order quantity, Q = 2DS H Using the EPQ model = 2x42,000x19 7 = DS The order quantity, Q = = H(1 d p ) 2x42,000x19 7(1 140 = 2, ) The order quantity using the EPQ model is always higher than that of the EOQ model (2,800 > )

8 CHAPTER 1 Example 1.1 on the critical path method 1.1 The following data describe the activities of a small project. a. Draw the project network and determine the possible paths in it. b. Identify the critical path. c. Find the completion time of the project. Activity : S T W U V Y Immediate Predecessor : -- S S T, W W U, V Activity time (or duration) in days : Solution a. The project network The possible paths in the network are STUY = = 18 days SWUY = = 22 SWVY= = 20 b. The critical path is the longest path in the network: SWUY c. The completion time of the project is 22 days. Notes:

9 1. In the forward pass, we calculate ES and EF. ES is the earliest start time for the activity and EF is the earliest finish time for the activity. EF = ES + D, where D is the duration or the activity time. In the forward pass the longest EF is chosen. 2. In the backward pass, we calculate LS and LF. LS is the latest start time for the activity while LF is the latest finish time of the activity. LS = LF D. In the backward pass, the shortest LS is chosen. 3. The slack time of an activity is S where S =LS ES or S = LF EF. 4. Activities T and V are non-critical because they have non-zero slack. 5. The path or route that connects the critical activities is the critical path which is the longest path in the network.. The activities that fall on the critical path are critical and have zero slack time. They must be monitored carefully since any delay in them can cause a delay to the whole project. 7. The length of the critical path determines the completion time of the project. 8. There could be more than one critical path in a network. Example 1.2 on PERT You are given the following data about a small project. Find the critical path and the completion time of the project: Activity : K L M N P Immediate predecessor : - K K L, M N, M Optimistic time (weeks) : Most likely time (weeks) : Pessimistic time (weeks) : Solution Before drawing the network, we calculate the expected time t e for each activity using the following formula: t e = t o+(4xt m )+t p t e (K) = 8+(4X10)+12 t e (L) = 3+(4X5)+7 = 5 = 10 weeks t e (M) = 8+(4X8)+8 = 8 t e (N) = 2+(4X)+1 = 7 t e (P) = 1+(4X5)+9 = 5 The possible paths in the network are KLNP = = 27

10 KMNP = = 30 KMP = = 23 The critical path is KMNP and the completion time of the project is 30 weeks.. Example 1.3 on Gantt chart The earliest-start Gantt chart for the project in example 1.1 is as follows: Time in days Activity S T W U V Y a. i. How many activities that must be completed before activity V? 3 activities: S, T, and W ii. How many activities are parallel to activity W? 1 activity: T, they have common time. iii. How many activities are parallel to activity Y? Zero activities. There is no common time with any other activity. iv. How many activities are critical? 4 Activities: S, U, W, Y. Activities S and Y are critical because they are the first and last activities. Activities U and W are critical because they have no gaps before or after them. Gaps mean slack times. b. Identify the critical path. SUWY c. i. How much is the slack time of activity V? 2 days. ii. How much is the slack time of activity W? Zero, because it is critical. iii. How much is ES and EF of activity T? ES = 3, and EF = 7 iv. How much is LS and LF of activity U?

11 LS = 11, LF = 15. The same as ES and EF, respectively, because for critical activities like U, ES = LS, and EF = LF. v. How much is the completion time of the project? 22 days Note: for activity T, LF = 11, LS = LF D = LF 4 = 7 and for activity V, LF = 15, LS =LF 2 = 13 END First written in September, 201 Updated in January, 2017

MSA 640 Homework #2 Due September 17, points total / 20 points per question Show all work leading to your answers

Name MSA 640 Homework #2 Due September 17, 2010 100 points total / 20 points per question Show all work leading to your answers 1. The annual demand for a particular type of valve is 3,500 units. The cost