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CDAE 266 - Class 23 Nov. 14 Last class: Result of Quiz 6

CDAE 266 - Class 23 Nov. 14 Last class: Result of Quiz 6 4. Queuing analysis and applications Project 3 Today: 4. Queuing analysis and applications Problem set 4 Project 3 5. Inventory analysis and applications Next class:

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CDAE 266 - Class 23 Nov. 14 Last class: Result of Quiz 6

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  1. CDAE 266 - Class 23 Nov. 14 Last class: Result of Quiz 6 4. Queuing analysis and applications Project 3 Today: 4. Queuing analysis and applications Problem set 4 Project 3 5. Inventory analysis and applications Next class: 5. Inventory analysis and applications Reading:

  2. CDAE 266 - Class 23 Nov. 14 Important dates: Problem set 4, due today Project 3, Saveway Supermarket (page 4-20) due Thursday, Nov. 16 Final exam: 8:00-11:00am, Thursday, Dec. 14

  3. 4. Queuing analysis and applications 4.1. One example 4.2. Basic queuing situations 4.3. Importance of queuing analysis 4.4. Single-server queuing models 4.5. Multiple-server queuing models 4.6. Cost analysis of queuing models 4.7. How to use Excel to solve queuing models?

  4. 4.5. Multiple-server queuing models 4.5.1. Assumptions: Structure: multiple-server & single-stage Discipline: FIFO 4.5.2. Information we need:  = mean customer arrival rate  = mean service rate per server s = number of servers 4.5.3. Condition:  < s 

  5. 4.5. Multiple-server queuing models 4.5.3. Queuing results (pages 4-13 and 4-14) (1) Probability that there are a particular number of customers in the system (2) Average number of customers in the system (L) (3) Average number of customers waiting (Lq) (4) Average time a customer spends in the system (W) (5) Average waiting time of each customer (Wq) (6) Server utilization rate (ρ)

  6. The formulas of the multiple-server queuing models are too complex -- that is why we have a computer program!

  7. 4.6. How to solve queuing models using Excel? 4.6.1. Excel worksheet 4.6.2. Enter the data:  = mean customer arrival rate  = mean service rate per server s = number of servers 4.6.3. Results

  8. 4.7. Cost analysis of queuing models: An example A company has a central parts cage for mechanics to pick up supplies. Mechanics arrive at the cage at an average rate of 25 per hour. The only clerk Helen working in the cage can complete a mechanic's part request in an average of 2 minutes. The cost of mechanics’ time is $20 per hour and the cost of the clerk’s time is $12 per hour. The manufacturer wants to evaluate three alternative options to see if the cost can be reduced: (1) To rent an electric-controlled storage bin system at a cost of $15 per hour and the system can double Helen’s speed (2) To add another clerk (Amber) with the same wage rate and same check out speed (3) To replace Helen by Ashley who can check out one person in one minute and her wage rate is $24 per hour.

  9. 4.7. Cost analysis of queuing models: An example Hourly cost of the current system: = Cost of Helen + cost of mechanics’ time = $12 + cost per mechanic x number of mechanics = $12 + W x $20 x  = $12 + 0.2 x 20 x 25 = $112 Hourly cost of the storage bin system: = Cost of Helen + rent + cost of mechanics’ time = $12 + $15 + cost per mechanics x number of … = $12 + $15 + W x $20 x  = $12 + $15 + ? x $20 x 25 = $ Hourly cost of the 2nd option: = Cost of 2 clerks + cost of mechanics’ time = $24 + cost per mechanic x number of mechanics = $24 + W x $20 x  = $24 + ? x 20 x 25 = $

  10. 4.7. Cost analysis of queuing models: An example Hourly cost of the 3rd option: = Cost of Ashley + cost of mechanics’ time = $24 + cost per mechanic x number of mechanics = $24 + W x $20 x  = $24 + ? x 20 x 25 = $

  11. Problem 20-13:  = mean customer arrival rate = 1.5 per hour  = mean service rate = 2 per hour (a) L = Lq = Wq = W = p = (b) (1) Hourly queuing system cost = $2 x Wq x 1.5 = 2 x 1.5 x 1.5 = $4.5 (2) Gross profit = 1.5 x 20 – the cost = 30 – 4.5 = $25.5 (c)  = 1.9 per hour,  = 2 per hour (1) Profit per hour = 1.9 x 20 – $2 x Wq* x 1.9 = 38 – 2 x 9.5 x 1.9 = 1.9 (2) Number of hours to break even = $1000 / gross profit per hour = 1000 / 1.9 = 526.3 hours What does Wq = 9.5 hours mean? Problem set 4

  12. Project 3 Saveway Supermarket What is the business decision problem? What information do we have? What information does the business need from us? How to analyze the problem? How to present our analysis and results?

  13. Project 3 --------------------------------------------------------------------------------------------------------- Labor Construction Goodwill Total  S cost Rent cost cost cost ---------------------------------------------------------------------------------------------------------Current Option 1 Option 2 Option 3a Option 3b Options 1+3a Options 1+3b Options 2+3a Options 2+3b ---------------------------------------------------------------------------------------------------------

  14. 5. Inventory analysis and applications 5.1. Basic concepts 5.2. Inventory cost components 5.3. Economic order quantity (EOQ) model 5.4. Inventory policy with backordering 5.5. Inventory policy and service level 5.6. Production and inventory model

  15. 5.1. Basic concepts -- Inventories: parts, materials or finished goods a business keeps on hand, waiting for use or sell. -- Inventory policy (decision problems): Q = Inventory order quantity? R = Inventory reorder point (R = level of inventory when you make the order)? -- Optimal inventory policy: Determine the order quantity (Q) and reorder point (R) that minimize the inventory cost. -- SKU: Stock-keeping units

  16. 5.1. Basic concepts -- Lead time (L): the time between placing an order and receiving delivery -- Inventories models: (1) Economic order quantity (EOQ) model (2) Backordering model (3) Production and inventory model

  17. 5.2. Inventory cost components: -- Inventory ordering and item costs -- Ordering costs (telephone, checking the order, labor, transportation, etc.) -- Item cost (price x quantity) -- Inventory holding costs (interest, insurance, storage, etc.) -- Inventory shortage costs (customer goodwill and satisfaction costs)

  18. 5.3. The economic order quantity (EOQ) model 5.3.1. Assumptions: -- One item with constant demand (A) -- Lead time = 0 -- No backordering -- All the cost parameters are known

  19. 5.3. The economic order quantity (EOQ) model 5.3.2. A graphical presentation 5.3.3. Mathematical model -- Variable definitions: k = fixed cost per order A = annual demand (units per yr.) c = price h = annual holding cost per $ value T = time between two orders

  20. A graphical presentation of the EOQ model: The constant environment described by the EOQ assumptions leads to the following observation THE OPTIMAL EOQ POLICY ORDERS THE SAME AMOUNT EACH TIME. Q Q Q This observation results in the inventory profile below:

  21. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Assumptions: Constant demand No backordering

  22. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Objective: choose order quantity (Q) to minimize the total annual inventory cost What is the reorder point (R)?

  23. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Total annual inventory cost = annual ordering costs + annual holding cost + annual item costs (a) Annual order costs Annual demand = A Quantity of each order = Q Number of orders per yr. = A/Q Fixed cost per order = k Annual ordering costs = k (A/Q)

  24. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Total annual inventory cost = annual ordering costs + annual holding cost + annual item costs (b) Annual holding costs Average inventory = Q/2 Annual holding cost per unit = hc Annual holding cost = (Q/2)*hc

  25. Holding Costs (Carrying costs) Cost of capital Storage space cost Costs of utilities Labor Insurance Security Theft and breakage Annual unit holding cost h = Annual holding cost rate (cost per dollar value) c = Unit value (price) h * c

  26. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Total annual inventory cost = annual ordering costs + annual holding cost + annual item costs (c) Annual item cost Average demand = A Cost per unit (price) = c Annual item cost = Ac

  27. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Total annual inventory cost = annual order costs + annual holding cost + annual item costs -- Total annual relevant (variable) cost: -- Examples

  28. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Derive the optimal solution: (1) A graphical analysis: the sum of the two costs is at the minimum level when the annual holding cost is equal to the annual ordering cost: => => =>

  29. 5.3. The economic order quantity (EOQ) model 5.3.3. Mathematical model -- Derive the optimal solution: (2) A mathematical analysis: At the minimum point of the curve, the slope (derivative) is equal to zero: => =>

  30. 5.3. The economic order quantity (EOQ) model 5.3.4. Reorder point if lead time > 0 If L > 0, R= L x A Note that L and A must have consistent units 5.3.5. Examples (1) Liquor store (pp. 209-211) Additional question: If the sale price is $3 per case, what will be the total “gross” profit per year?

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