NewsvendorProblem

NewsvendorProblem - Single Period “Newsvendor” Problem...

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Unformatted text preview: Single Period “Newsvendor” Problem IE 251 1 Probabilistic Demand: Single Period Problem Consider a single ordering/production decision, Q, to be made at the beginning of a period Demand, which is probabilistic, occurs during the period Relevant costs are incurred (inventory holding, variable purchasing or production costs, fixed costs (when exists), backordering or lost sales costs) at the end of the period Single period model is a building block for more complicated models under uncertainty Direct applications: style goods and perishable items (dated publications, garment manufacturing, fresh produce, blood, medical supplies) 2 Probabilistic Demand: Single Period Problem inventory Q inventory End of period 0 Backorder or lost sales Total demand within the period 3 Model Environment Relatively short selling season (weeks, 2 months,…) with a well-defined beginning and end When the total demand in the period exceeds the stock available, there is an associated underage cost, cu – cost of lost profit and/or goodwill – expediting or acquiring the material at a higher unit value When the total demand is less than the stock available, overage cost occurs, co 4 Simple Case: The Newsvendor Problem Zero initial inventory at the beginning of the period Only overage and underage costs We are interested in the “expected cost” Expected Underage Cost: cu {1× P( X = Q + 1) +2 × P( X = Q + 2) + ...} = cu ∞ ∑ (i − Q )P( X = i) i =Q +1 5 Simple Case: The Newsvendor Problem Expected Overage Cost co {Q × P( X = 0) + (Q − 1) × P( X = 1) + L +1× P( X = Q − 1)} Q = co ∑ (Q − i)P( X = i) i =0 Total Expected Cost: Q C(Q ) = co ∑ (Q − i)P( X = i) + cu i =0 ∞ ∑ (i − Q )P( X = i) i =Q +1 6 Marginal Analysis From the graph we see that q* is the smallest value of q for which C (q*+1) – C (q*) ≥ 0. If C(q) is a convex function of q we can minimize expected cost by finding the smallest value of q that satisfies the inequality. Determine q* by repeatedly computing the effect of adding a marginal unit to the value of q. Method is called marginal marginal C(q) q*–1 q* q q*+1 analysis. 7 The Concept of Marginal Analysis Marginal analysis: finding the expected profit of ordering one more unit. one Probability of not selling Your Last item in stock P(X < Q) Probability of Selling everything P(X ≥ Q) μ Q 8 Single Period Inventory Model Marginal Analysis Goal in finding the proper Q: E (revenue on last sale) = E (loss if last paper not sold) P(X ≥ Q)Cu = P(X < Q)Co ⎡1 − P(X < Q)⎤ Cu = P(X < Q)Co ⎣ ⎦ Cu P(X < Q) = Cu + Co Critical Ratio where: Cu = unit contribution from newspaper sale ( opportunity cost of underestimating demand)…underage cost is avoided by Qth newspaper Co = unit loss from not selling newspaper (cost of overestimating demand)… adding Qth newspaper adds “exposure” to overage cost X = demand Q = newspaper stocked 9 Critical fractile for the newsvendor problem When the demand is a discrete random variable, the cu condition PX < (Q ) = cu + c o may not be satisfied at equality (jumps, due to discreteness). The appropriate condition to use: Q * = min{Q : PX ≤ (Q ) ≥ Q cu } cu + co C(Q+1)-C(Q) ≥0 thus… cuPX≥(Q+1) ≤ coPX<(Q+1) 10 Single-Period & Discrete Demand: Lively Lobsters Lively Lively Lobsters (L.L.) receives a supply of fresh, live lobsters from Maine every day. Lively earns a profit of $7.50 for every lobster sold, but a day-old lobster is worth only $8.50. Each lobster costs L.L. $14.50. – unit cost of a L.L. stockout Cu = 7.50 = lost profit – unit cost of having a left-over lobster Co = 14.50 - 8.50 = cost – salvage value = 6. – target L.L. service level CR = Cu/(Cu + Co) = 7.5 / (7.5 + 6) = .56 11 Lively Lobsters Demand follows a discrete (relative frequency) distribution: Result: order 25 Lobsters, because that is the smallest amount that will serve at least 56% of the demand on a given night. Cumulative Relative Relative Frequency Frequency (cdf) (pmf) D emand 19 0.05 0.05 20 0.05 0.10 21 0.08 0.18 22 0.08 0.26 23 0.13 0.39 24 0.14 0.53 25 0.10 0.63 26 0.12 0.75 27 0.10 … 28 0.10 … 29 0.05 … * pmf = prob. mass function Probability that demand will be less than or equal to x P(D < 1 9 ) P(D < 2 0 ) P(D < 2 1 ) P(D < 2 2 ) P(D < 2 3 ) P(D < 2 4 ) P(D < 2 5 ) P(D < 2 6 ) P(D < 2 7 ) P(D < 2 8 ) P(D < 2 9 ) 12 Notation: Profit Maximization Q = order quantity v = unit cost A = ordering cost p = selling price p-v = profit per item sold g = salvage value (g < v) if g < 0, then storage, holding, or removal cost v - g = loss per unsold item B = goodwill cost (lost sales penalty) X = a random variable, the demand for the item p(x) = probability function for X. 13 Model formulation: Profit Maximization For x ≤ Q: Profit = px + g(Q - x) - A - vQ For x > Q: Profit = pQ - B(x - Q) - A - vQ Q E ⎡ profit ⎤ = ∑ ⎡ px + g(Q − x) − A − vQ ⎤ p(x) ⎣ ⎦ ⎣ ⎦ x =0 + ∞ ∑ x =Q +1 Q ⎡ pQ − B(x − Q ) − A − vQ ⎤ p(x) ⎣ ⎦ = ∑ ⎡(p − g) x + gQ ⎤ p(x) + ⎣ ⎦ x =0 ∞ ∑ x = Q +1 ⎡(p + B) Q − Bx ⎤ p(x) − A − vQ ⎣ ⎦ 14 Newsvendor Problem Interpreting as Simple Newsvendor Problem: – Cost of overage: » Purchase cost - Salvage Value = v - g – Cost of underage: » Lost profit: Selling price – Purchase cost = p - v » Or, p – v + (any additional penalty, B) – Critical ratio is then: » cu / (cu+co) = (p-v+B)/(p-g+B) »For the discrete demand case take the smallest value of Q* that satisfies the condition P(D ≤ Q*) ≥ (p - v + B) / (p - g + B) 15 Single Period Inventory Model Newsvendor Problem Example D = newspapers demanded p(D) = probability of demand Q = newspapers stocked P = selling price of newspaper, $10 v = cost of newspaper, $4 g = salvage value of newspaper, $2 Cu = unit contribution: p-v = $6 Co = unit loss: v-g = $2 16 Single Period Inventory Model Expected Value Analysis p(D) D 6 7 Stock Q 8 9 10 .028 .055 .083 .111 .139 .167 .139 .111 .083 .055 .028 2 3 4 5 6 7 8 9 10 11 12 4 12 20 28 36 36 36 36 36 36 36 2 10 18 26 34 42 42 42 42 42 42 0 8 16 24 32 40 48 48 48 48 48 -2 6 14 22 30 38 46 54 54 54 54 -4 4 12 20 28 36 44 52 60 60 60 $34.43 $35.77 $35.99 Expected Profit $31.54 $35.33 17 THE SENTINEL NEWSPAPER Management at Sentinel wishes to know how many newspapers to put in its new vending machine location. Data – – – – – Unit selling price is $0.30 Unit production cost is $0.38. Advertising revenue is $0.18 per newspaper. Unsold newspaper can be recycled and net $0.01. Unsatisfied demand costs $0.10 per newspaper. Demand distribution is Discrete Uniform from 30 through 49 newspapers. 18 SOLUTION Input to the optimal order quantity formula p = 0.30 v = 0.20 [0.38-0.18] g = 0.01 B = 0.10 p+ B - v The optimal service level = p+ B - g = 0.30 + 0.10 - 0.20 = 0.513 0.30 + 0.10 - 0.01 19 Finding the optimal order quantity Q* 1.0 P(D ≤ 39) = 0.50 P(D ≤ 40) = 0.55 0.513 0.55 0.50 30 Q**= 40 Q = 40 39 40 49 20 Example Demand for a critical component purchased by the Air Force for use in the B-2 bomber is Poisson with an estimated mean of 12 over the life of the bomber fleet. The manufacturer, Lockheed-Grumman, sells the part for $5,000. The unit production cost is $3,750. Some of the parts found in the component may be salvaged for $900. The production line for this component will be shut down in several months in order to make room for a new product line. How many units should Lockheed-Grumman produce in order to maximize expected profits? Assume A = 0. 21 Example Solution CR = p − v 5000 − 3750 = = .305 5000 − 900 p−g X Pr{X≤x} Q* is smallest Q such that: Q 8 9 10 11 .155 .242 .347 .462 ∑ p( x) ≥ .305 x =0 Q* = 10 22 Expected Profit ∞ E[P ] = ( g − v ) Q + ( p − g ) μ − ( p − g + B ) ∑ ( x − Q )p(x) x =Q = ( 900 − 3750 )(10 ) + ( 5000 − 900 )(12 ) ∞ − ( 5000 − 900 ) ∑ ( x − 10 ) e−12 (12 ) x =10 x x! x ⎡∞ e−12 (12) ⎤ ⎥ = −28,500 + 49,200 − ( 4100 ) ⎢ ∑ ( x − 10 ) x! ⎥ ⎢ x =10 ⎣ ⎦ Q x −12 ⎡ e (12 ) Q e −12 12 ⎛ ( ) ⎞⎤ ⎢μ ⎟⎥ = −28,500 + 49,200 − ( 4100 ) + ( μ − Q ) ⎜1 − ∑ ⎜ x =0 ⎟⎥ Q! x! ⎢ ⎝ ⎠⎦ ⎣ = −28,500 + 49,200 − ( 4100 )( 2.56 ) = $10,204 23 Newsvendor : Continuous Demand Case If demand is continuous, we’ll have: Q ∞ 0 Q C(Q ) = co ∫ (Q − x)fX (x)dx + cu ∫ (x − Q )fX (x)dx – Where fX(x) is the density function for X Check for the properties of the function: Q ∞ dC(Q ) = co ∫ fX (x)dx − cu ∫ fX (x)dx dQ 0 Q = co PX ≤ (Q ) − cu (1 − PX ≤ (Q )) d 2C(Q ) = (co + cu )fX (Q ) > 0 dQ 2 24 Newsvendor : Continuous Demand Case Liebnitz’s Rule to take the derivative of an integral d ⎡b(Q)⎤ d ⎡a(Q)⎤ d dg(x,Q) ⎣ ⎦ ⎣ ⎦ ∫Q) g(Q,x)dx = dQ g(Q,b(Q)) − dQ g(Q,a(Q)) + a(∫Q) dQ dx dQ a( b(Q ) b(Q ) Since the second derivative of C(Q) is positive, C(Q) is convex dC(Q ) = co PX ≤ (Q ) − cu (1 − PX ≤ (Q )) dQ = (cu + co )PX ≤ (Q ) − cu ⇒ PX ≤ (Q *) = =0 cu cu + c o 25 An example: Demand in the period is uniformly distributed over [50,150]. cu=$10, co=$7 PX ≤ (x) = x − 50 for 50 ≤ x ≤ 150 100 Q − 50 10 = 100 17 10 ⇒ Q * = 50 + 100 ≅ 109 17 Solve PX ≤ (Q ) = 26 Example The Hotel Holiday rents rooms for $100 a night. The Operations Research Institute is holding a conference at the Hotel Holiday in two months. They have negotiated a special rate of $70 a night for those attending the conference. Empty rooms cost the hotel $25. The demand for regular rooms is exponential with a mean of 140 per night. How many rooms should be kept at the regular rate in order to maximize profit? All rooms at the lower rate can be rented. 27 Example Solution CR = Q Cu 30 = = .5455 Cu + Co 30 + 25 Q ∫ f(x)dx = ∫ 0 e 0 Q − 140 e − x μ μ dx = 1 − e − Q μ = .5455 = 1 − .5455 Q * = −140ln (.4545 ) = 110.384 → 110 28 Continuous Demand case: Max Profit For the continuous case: E[ profit] = ∫ Q 0 ⎡(p − g) x + gQ ⎤ f(x)dx ⎣ ⎦ ∞ + ∫ ⎡(p + B)Q − Bx ⎤ f (x) dx − A − vQ ⎦ Q⎣ Q Q = gQ ∫ f (x)dx + (p − g)∫ xf (x)dx 0 0 ∞ ∞ Q Q +(p + B)Q ∫ f (x)dx − B ∫ xf (x)dx − A − vQ 29 Continuous Demand case: Max Profit ∞ ∞ 0 Q = gQ ∫ f (x)dx − gQ ∫ f (x)dx ∞ ∞ 0 Q +(p − g)∫ xf (x)dx − (p − g)∫ xf (x)dx ∞ ∞ Q Q +(p + B)Q ∫ f (x)dx − B ∫ xf (x)dx − A − vQ ∞ = gQ + (p − g)μ + ( p + B − g)∫ Qf ( x)dx Q ∞ −(p + B − g)∫ xf (x)dx − A − vQ Q 30 Continuous Demand case: Max Profit ∞ = gQ + (p − g)μ + (p + B − g)∫ Qf (x)dx Q ∞ −(p + B − g)∫ xf (x)dx − A − vQ Q ∞ = (g − v ) Q + (p − g)μ − A − (p + B − g)∫ (x − Q )f (x)dx Q 31 Continuous Demand case: Max Profit ∞ d {.} = (g − v ) − (p + B − g)∫ − f (x)dx = 0 Q dQ ∞ v−g f (x)dx = ∫Q Solving: Critical p+B−g ratio Loss per unsold item v−g v−g = p + B − g (p − v ) + B + (v − g) Item profit + goodwill PX ≤ (Q * ) = p−v +B p−g+B 32 Example The historical demand for Christmas wreaths sold during the annual Boy Scout fundraising drive has the following PDF: ⎧1 ⎪ f ( x) = ⎨ 50 ⎪0 ⎩ 50 ≤ x ≤ 100 elsewhere Wreaths sell for $35 each and cost the Boy Scouts $20. A goodwill cost of $2 is assumed as a matter of policy by the National Boy Scout Fund Raising Committee. It cost $1.00 per wreath for disposal. 33 Example Solution 50 1 − F (Q ) = Q 100 100 − Q 20 − (−1 ) 21 v −g = = = 50 ) p + B − g 35 + 2 − (−1 38 ⎛ 21 ⎞ 100 − Q = 50 ⎜ ⎟ ⎝ 38 ⎠ ⎛ 21 ⎞ Q * = 100 − 50 ⎜ ⎟ = 72.368 ≈ 72 ⎝ 38 ⎠ 34 Expected Profit E ⎡Profit ⎤ = (g − v ) Q + (p − g)μ ⎣ ⎦ ∞ − A − (p + B − g)∫ (x − Q )f (x)dx Q = ( −1 − 20 ) 72 + ( 35 + 1) 75 − A − ( 35 + 2 + 1) ∫ 100 72 = −1512 + 2700 − ( 38 ) (100 − 72 ) (x − 72) dx 50 2 − A = 890.08 − A 100 = −1512 + 2700 − 38(7.84) − A = 890.08 − A Order Q if A − Bμ < 890.08 Otherwise, order nothing 35 Normally Distributed Demand The optimality equation can be written as: Φ( Q −μ σ )= cu = Pu≤(k) cu + co When the demand is normally distributed, expected cost E(Q) can be written as: E(Q ) = co (Q − μ) + (co + cu )σ Gu ( k ) Expected profit equation may be written as: E ⎡Profit⎤ = (g − v) Q + (p − g)μ ⎣ ⎦ − A − (p + B − g)σ Gu (k) 36 Normally Distributed Demand Example: X~N(100, 36), μ = 100, σ = 6 Let’s find the optimal order quantity and optimal expected cost for cu=$10 and c0=$2.5. k is the 10/12.5 percentile of standard normal From Table B.1 of appendix B, for Pu≥(k)=1-0.8=0.2, k is approximately 0.84. We also read Gu(k) as 0.1120. Q*=100+(0.84)(6)=105.04 E(Q*)=2.5(5.04)+12.5(6)(0.1120)=$21 37 Example 2 Demand for the UD daily newspaper at the campus newsstand is normal with a mean of 100 and a standard deviation of 15. Newspapers sell for 25 cents and costs 10 cents to produce. Goodwill costs are assumed to be zero and there is a 2 cents salvage value. 38 Example 2 Solution ∞ ∫ f(x)dx = Q v−g .10 − .02 = = .3478 p + B − g .25 − .02 Pu>(k) = .3478 k = .39 100 Q Q* = μ + k σ = 100 + 15(.39) = 105.85 ≈ 106 39 Multi-Item Newsvendor Problem n: number of different items fi(x)=probability density function for item i Fi(x)=cdf for item i cio: overage cost for item i ciu: underage cost for item i W: available budget We want to minimize C(Q 1,..., Q n ) = subject to n Qi ∞ i =1 0 Qi ∑ { cio ∫ (Q i − x ) fi ( x )dx + ciu ∫ ( x − Q i ) fi ( x )dx } n ∑vQ i =1 i i ≤W 40 Multi-Item Newsvendor Problem Very similar to multi-item EOQ under constraints First check unconstrained optimum. Qi ∂C = −ciu + (ciu + cio ) ∫ fi (x)dx = 0 ∂Q i 0 ⇒ Fi (Q i ) = ciu ciu + cio If ∑ v i Q i ≤ W then we are done. Otherwise, form the Lagrangian function. 41 Multi-Item Newsvendor Problem n Qi ∞ i =1 0 Qi L(Q1,...,Q n, M) = ∑{cio ∫ (Qi − x)fi (x)dx + ciu ∫ (x − Qi )fi(x)dx } + M(∑viQi − W) Taking derivatives with respect to (Q1,…, Qn, M) : −ciu + (ciu + cio )Fi (Q i ) + Mv i = 0 for i = 1,2,...,n ∑v Q i i −W = 0 Usually requires numerical methods to solve. 42 Multi-Item Newsvendor Problem p v g B x hat sigma Item 1 Item 2 Item 3 Item 4 Total 50.3 40 32 6.1 35.1 15 28 4.8 25 12.5 15.1 2 9.7 0 10.3 1.5 900 800 1200 2300 122 200 170 200 M p(k) k Q G(k) E[P(Q)] Qv 0.509306 0.50930629 0.50930629 0.50930629 0.7993 0.3687 0.9986 0.9365 -0.8392 0.3353 -2.9783 -1.5264 798 867 694 1995 0.9514 0.2535 2.9787 1.5540 $ 10,652 $ 18,438 $ (2,442) $ 2,104 $ $ 27,996 $ 13,006 $ 19,423 $ 9,575 $ W = Budget $ 70,000 Goal Seek to set Qv total to W by changing M. 28,752 70,000 43 Multi-Item Newsvendor Problem M 1.667 1.567 1.467 1.367 1.267 1.167 1.067 0.967 0.867 0.767 0.667 0.567 0.467 0.367 0.267 0.167 0.067 -0.033 -0.133 Qv $ 70,000 $ 33,572 $ 40,760 $ 41,875 $ 42,631 $ 43,235 $ 43,756 $ 44,228 $ 44,668 $ 45,090 $ 45,502 $ 56,801 $ 62,006 $ 77,137 $ 82,319 $ 85,993 $ 89,289 $ 92,574 $ 96,191 $ 101,116 E[P(Q)] $ 28,752 $ (5,699) $ 7,611 $ 9,307 $ 10,380 $ 11,177 $ 11,812 $ 12,339 $ 12,787 $ 13,174 $ 13,511 $ 21,456 $ 24,615 $ 32,291 $ 34,477 $ 35,647 $ 36,363 $ 36,746 $ 36,803 $ 36,373 44 Multiple Period Problems Difficulty: Technically, Newsvendor model is for a single period. Extensions: But Newsvendor model can be applied to multiple period situations, provided: – demand during each period is independent and identically distributed – amount ordered simply the difference between Q* and the present backorder or positive inventory level Key: make sure co and cu appropriately represent overage and shortage cost. 45 Example Scenario: – – – – GAP orders a particular clothing item every Friday mean weekly demand is 100, std dev is 25 wholesale cost is $10, retail is $25 holding cost has been set at $0.5 per item per week (to reflect obsolescence, damage, etc.) Problem: how should they set order amounts? 46 Example (cont.) Newsvendor Parameters: co = $0.5 cu = $15 15 = 0.9677 0.5 + 15 ⎛ Q − 100 ⎞ Φ⎜ ⎟ = Pu ≤ (k) = 0.9677 ⎝ 25 ⎠ Every Friday, they should order-up-to 146, that is, if Q − 100 = 1.85 there are x on hand, then 25 order 146-x. Q = 100 + 1.85(25) = 146 47 Solution: CR = Newsvendor-like Problem Problem Condo Construction Company is bidding on an important construction job. The job will cost $2 million to complete. One other company is bidding for the job. Condo believes that the opponent’s bid is equally likely to be an amount between $2 million and $4 million. If Condo wants to maximize expected profit, what should its bid be? 48 Newsvendor-like Problem Problem Let B = random variable representing bid of Condo’s opponent b = actual bid of Condo’s opponent q = Condo’s bid If b > q, Condo outbids the opponent and earns a profit of q – 2,000,000. If b < q, Condo is outbid by the opponent and earns nothing. 49 Newsvendor-like Problem Problem Let E(q) be Condo’s expected profit if it bids q. Then E(q) = (0)∫ q 2,000,000 f (b)db +(q − 2,000,000)∫ 4,000,000 q f (b)db Since f(b) = 1/2,000,000 for 2,000,000 ≤ b ≤ 4,000,000, we obtain E(q) = (q − 2,000,000)(4,000,000 − q) 2,000,000 50 Newsvendor-like Problem Problem To find the value of q maximizing E(q), we find E ′(q) = −(q − 2,000,000) + (4,000,000 − q) 6,000,000 − 2q = 2,000,000 2,000,000 Hence, E ′(q) =0 for q=3,000,000. We know that E (q) is a concave function of q, and 3,000,000 does indeed maximize E (q). Hence, Condo should bid $3 million. Condo’s expected profit will be $500,000. 51 Impact of Supply Chain Contracts on Profitability: Buyback Contracts Buyback Contracts Specifies the parameters within which a buyer places orders and a supplier fulfills them Example parameters: quantity, price, time, quality Double marginalization: – buyer and seller make decisions acting independently instead of acting together – gap between potential total supply chain profits and actual supply chain profits results Buyback contracts can be offered that will increase total supply chain profit 52 Returns Policy: Buyback Contracts A manufacturer specifies a wholesale price and a buyback price at which the retailer can return any unsold items at the end of the season Results in an increase in the salvage value for the retailer – which induces the retailer to order a larger quantity The manufacturer is willing to take on some of the cost of overstocking (market risk) because the supply chain will end up selling more on average Manufacturer profits and supply chain profits can increase 53 Returns Policy: Buyback Contracts An example with high demand variability v = manufacturer production cost = $50 c = wholesale price to retailer = $250 p = retail selling price = $300 sR = retail clearance price = $100 Retailer : Pu ≤ (k) = cu (300 − 250) = = .25 cu + co (300 − 250) + (250 − 100) k = −.6745 μ = 5000 σ = 3000 Q * = 5000 − .6745(3000) = 2977 = Manufacturer's Sales 54 Returns Policy: Buyback Contracts Retailer’s profit: E(PR (k)) = ( p − s)μ − (c − s)Q * −( p − s)σ x Gu ( k ) where Q * = μ + kσ E (PR (−.6745)) = (300 − 100)(5000) − (250 − 100)(2977) −(300 − 100)(3000)(.8236) = 59,334 Manufacturer’s profit E(PM (Q *) = (c − v )Q * = (250 − 50)(2977) = 595,400 Manufacturer makes significantly more profit, but faces no market risk 55 Returns Policy: Buyback Contracts Independent decisions by Manufacturer and Retailer Inputs Mean Demand, Mu Standard Deviation of demand Retailer's Cost, c = Retailer's Sale price, p = Retailer's Salvage value, s=b = Retailer's Return cost = Mfg's Cost, v Mfg's Sale price, c Mfg's Buyback Price, b Mfg's Salvage Value, sm $ $ $ $ $ $ $ $ 5000 3000 250 300 100 50 250 - Intermediate Calculations Cost of Understocking, Cu Cost of Overstocking, Co $ $ 50 150 Outputs Order size, O* overstock understock Retailer's Expected Profit Manufacturer's E(Profit) Total Supply Chain Profit = 2977 447 2471 $ 59,334 $ 595,306 56 $ 654,640 Returns Policy: Buyback Contracts Product Availability maximizing supply chain profit – solving single period problem from a supply chain perspective – shift some overstock risk to manufacturer so that retailer will order more – assume that manufacturer can salvage overstock for $20/unit ( sM ) co = v − s M = 50 − 20 = 30 cu = p − v = 300 − 50 = 250 CR = cu 250 = = .8929 → k* = 1.2419 cu + co 280 Q * = 5000 + (1.2419)(3000) = 8726 57 Returns Policy: Buyback Contracts Define – b = price manufacturer will pay to buy back remaining retailer inventory at the end of the selling period – sM = salvage value of any items the manufacturer buys back Expected manufacturer’s profit under buyback = Q * (c − v) − (b − sM )E[Retailer overstock] Q* E[Retailer overstock] = ∫ (Q * −x)f(x)dx = (Q * −μ)Pu≤(k) + σ fu (k) 0 = σ (k + Gu (k)) Q * = μ + kσ 58 Returns Policy: Buyback Contracts Determining buyback price that maximizes supply chain profit – Retailer CR must match Supply Chain CR CR* = .8929 = cu 300 − 250 = cu + co (300 − 250) + (250 − b) b = 244 b must be greater than sR to provide any incentive for the retailer to order a larger amount 59 Buyback Contracts “Breakeven” buyback price Inputs Mean Demand, Mu Standard Deviation of demand Retailer's Cost, c = Retailer's Sale price, p = Retailer's Salvage value, s=b = Retailer's Return cost = Mfg's Cost, v Mfg's Sale price, c Mfg's Buyback Price, b Mfg's Salvage Value, sm $ $ $ $ $ $ $ $ 5000 3000 250 300 100 50 250 - $ $ $ $ $ $ $ $ Intermediate Calculations Cost of Understocking, Cu Cost of Overstocking, Co $ $ 50 150 $ $ Outputs Order size, O* overstock understock Retailer's Expected Profit Manufacturer's E(Profit) Total Supply Chain Profit = 2977 447 2471 $ 59,334 $ $ 595,306 $ $ 654,640 $ 5000 3000 250 300 116 50 250 116.00 20.00 50 134 3177 500 2323 66,898 587,480 60 654,378 Buyback Contracts Inputs Mean Demand, Mu Standard Deviation of demand Retailer's Cost, c = Retailer's Sale price, p = Retailer's Salvage value, s=b = Retailer's Return cost = Mfg's Cost, v Mfg's Sale price, c Mfg's Buyback Price, b Mfg's Salvage Value, sm Less demand uncertainty …less benefit from buybacks $ $ $ $ $ $ $ $ $ $ 50 150 $ $ Outputs Order size, O* overstock understock Retailer's Expected Profit Manufacturer's E(Profit) Total Supply Chain Profit = Buyback Contracts $ $ $ $ $ $ $ $ Intermediate Calculations Cost of Understocking, Cu Cost of Overstocking, Co Buyback price optimizing Supply Chain profit 5000 3000 250 300 100 50 250 - 5000 3000 250 300 244 50 250 244.00 20.00 2977 8726 447 3880 2471 154 $ 59,334 $ 219,002 $ 595,306 $ 876,010 61 $ 654,640 $ 1,095,012 50 6 Mean Demand, Mu Standard Deviation of demand Retailer's Cost, c = Retailer's Sale price, p = Retailer's Salvage value, s=b = Retailer's Return cost = Mfg's Cost, v Mfg's Sale price, c Mfg's Buyback Price, b Mfg's Salvage Value, sm $ $ $ $ $ $ $ $ 5000 300 250 300 100 50 250 - Intermediate Calculations Cost of Understocking, Cu Cost of Overstocking, Co $ $ 50 $ 150 $ Outputs Order size, O* overstock understock Retailer's Expected Profit Manufacturer's E(Profit) Total Supply Chain Profit = 4798 5373 45 388 247 15 $ 230,933 $ 246,900 $ 959,531 $ 987,601 $ 1,190,464 $ 1,234,501 $ $ $ $ $ $ $ $ 5000 300 250 300 244 50 250 244.00 20.00 50 6 62 Buyback Contracts Restricted manufacturer margin… buyback contract can not be justified Mean Demand, Mu Standard Deviation of demand Retailer's Cost, c = Retailer's Sale price, p = Retailer's Salvage value, s=b = Retailer's Return cost = Mfg's Cost, v Mfg's Sale price, c Mfg's Buyback Price, b Mfg's Salvage Value, sm $ $ $ $ $ $ $ $ 5000 3000 250 300 100 200 250 - Intermediate Calculations Cost of Understocking, Cu Cost of Overstocking, Co $ $ 50 $ 150 $ $ $ $ 2977 447 2471 59,334 $ 148,827 $ 208,161 $ Outputs Order size, O* overstock understock Retailer's Expected Profit Manufacturer's E(Profit) Total Supply Chain Profit = $ $ $ $ $ $ $ $ 5000 3000 250 300 160 200 250 160.00 20.00 50 90 3902 727 1825 93,305 93,305 186,611 63 ...
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This note was uploaded on 10/24/2011 for the course IE 351 taught by Professor Wil during the Spring '08 term at Lehigh University .

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