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Chapter 16

Chapter 16. Management of Short-term Assets. Preliminary Definitions. Working capital: Current (short-term) assets Net working capital: Current assets minus current liabilities Working capital policy: Management of current assets and current liabilities. The Operating Cycle.

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Chapter 16

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  1. Chapter 16 Management of Short-term Assets

  2. Preliminary Definitions • Working capital: Current (short-term) assets • Net working capital: Current assets minus current liabilities • Working capital policy: Management of current assets and current liabilities

  3. The Operating Cycle • Time it takes to produce output, sell it, and collect payment • Timing of payments • Varies among industries • Affects the need for short-term finance

  4. Financing and Working Capital Policy • Matching sources and uses of funds • Basic principle: do not use short-term sources to finance long-term assets • Cost structure of funds: • Long-term sources tend to be more expensive than short-term source

  5. Cost Structure Illustrated by a Yield Curve

  6. Aggressive Working Capital Policy • Use of more short-term sources • Increases profitability • Increases risk • But also increases the need to roll over or refinance short-term debt

  7. Inventory Cycle • Inventory is • acquired • drawn down to some level • replenished • and the cycle is repeated • The safety stock

  8. The Inventory Cycle $ B A T1 T2 Time

  9. The Economic Order Quantity (EOQ) • Ordering costs versus carrying costs • Fewer orders reduce ordering costs but increases carrying costs

  10. The Economic Order Quantity (EOQ) • The EOQ minimizes the sum of • Ordering costs (e.g., shipping and processing costs) plus • Carrying costs (e.g., warehouse expense, insurance, and interest expense)

  11. The Economic Order Quantity (EOQ) • EOQ = (2SF/C).5 • EOQ = ((2 x 10,000 x $50)/$10).5 = 316

  12. The Economic Order Quantity (EOQ) $ Costs A Total Costs E F Total Carrying Costs D $3,180 Total Ordering Costs B C 316 Order Size

  13. The Inventory Cycle Combined with the EOQ Units of Inventory 366 Level of Inventory Safety Stock 50 0 5 10 15 Days

  14. The Inventory Cycle Combined with the EOQ • Maximum inventory: EOQ plus the safety stock • Minimum inventory: Safety stock • Average inventory: EOQ/2 + safety stock

  15. EOQ Model • A very simple model • Assumes • sales occur evenly • no quantity discounts • individual items are identifiable • Highlights the trade off between ordering costs and carrying costs

  16. Just-in-Time inventory management • Designed to minimize the amount of inventory • Requires • accurate forecasts • excellent timing • dependable shipping • flexible schedules • frequent communication

  17. Management of Accounts Receivable • Credit policy encompasses • Who will receive credit • Terms of credit • Collections

  18. Decision to Grant Credit • Trade-off between additional sales versus additional costs • Additional costs • Cost of goods sold • Credit and collection costs • Bad debt expense • Carrying costs

  19. Analyzing Accounts Receivable • Receivables turnover: Credit sales/Accounts receivable • Aging schedules: Determine payment patterns

  20. Cash Management • Faster collections • Slower disbursements • Investing short-term funds

  21. Facilitating Faster Collections • Electronic funds transfers • Lockbox

  22. The Flow of Checks and Funds

  23. Money Market Instruments and Yields the Instruments • U.S. Treasury bills • Commercial paper • Negotiable certificates of deposit • Banker's acceptances • Eurodollar CDs • Repurchase agreements (REPOs)

  24. Yields: the Discount Yield • Depends on • The principal amount • The amount of the discount • The time to maturityYd = Par value - Price x ________360_______ Par value Number of days to maturity

  25. Yields: the Discount Yield • Yd = Par value - Price x ________360_______ Par value Number of days to maturity • $10,000 - $9,791 x _360_ = $10,000 180 4.18%

  26. Yields: the Discount Yield • Weaknesses • Use of 360 days • Return based on par value and not price

  27. Alternative Calculation: Simple Interest • Yd = Par value - Price x _ 365______ Price Number of days to maturity • $10,000 - $9,791 x _365_ = $9,791 180 4.33%

  28. Alternative Calculation: Compound Interest • Price(1 + i)t = Face Value • Example • price: $9,791face value: $10,000time period: 180 days$9,791(1 + i)0.49315 = $10,000i = (1.02135)2.0278 - 1 = 4.38%

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