Prob_Inventory_Class_A (1)

Prob_Inventory_Class_A (1) - Managing Class A Inventories...

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Unformatted text preview: Managing Class A Inventories IE 251 1 Base Stock Policies If there is no significant setup/order cost, or If there is no limit on the number of replenishments that can be placed in a year, or If demand is low (slow moving items) Then, order one at a time! Different names for the same policy – (s,Q) policy with lot/order size Q=1 – One-for-one replenishment policy – (S-1,S) policy, S=base-stock level=s+1 Note that – Inventory position is always S=s+1 2 1 Base Stock for Slow-Moving / Expensive Items Conditions for Q = 1 : TRC (Q ) = ∴ TRC (1) = Q 2 AE ( D ) vr + vr 2 Q + AE ( D ) TRC(2) = vr + AE ( D ) 2 ambivalent between Q=1 & Q=2 when vr 2 vr 2 + AE ( D ) = vr + = AE ( D ) 2 AE ( D ) 2 → AE ( D ) = vr ∴ Q = 1 if E( D ) ≤ vr A 3 Base Stock for Slow-Moving / Expensive Items B2 criterion / Q = 1 : – all on-order stock arrives by t+L » net stock = n = S-xL » PNS(n) = P{xL=S-n} – for Poisson demand » P{a particular demand is back ordered} = PN S ≤ (0) = P x L ≥ ( S ) 4 2 Example with Poisson Demand Demand for a specific model of Kenmore refrigerator at a Sears store is Poisson with mean of 5 per month. The lead time is 2 months and the Sears store manager uses a base stock policy with a reorder point of 12. Evaluate the fill rate. Probability Probability 0.14 0.12 0.1 0.08 0.06 0.04 0.02 0 Probability 0 10 20 30 LT Demand 5 Example with Poisson Demand PxL ≤ (S − 1) = PxL ≤ (s) = fill rate = P2 Suppose the manager’s goal is to achieve 90% fill rate. What is the required reorder point? Demand 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 Probability 4.54E-05 0.000454 0.00227 0.007567 0.018917 0.037833 0.063055 0.090079 0.112599 0.12511 0.12511 0.113736 0.09478 0.072908 0.052077 0.034718 0.021699 0.012764 0.007091 Cum Prob 4.54E-05 0.000499 0.002769 0.010336 0.029253 0.067086 0.130141 0.220221 0.33282 0.45793 0.58304 0.696776 0.79156 0.864464 0.916542 0.95126 0.972958 0.985722 0.992813 6 3 Expected Shortage Shortage/Stockout occurs when LT demand > S=s+1 b(S) = Expected backorder level at any given time = ∑ ( x − S)p x ≥S xL (x) = ∞ ∑ x = S +1 Px L ≥ ( x ) b(S) = μ L p x L (S − 1) + ( μ L − S)Px L ≥ (S) for S = 15 : = (10)(.052077) + (10 − 15)(1 − .916542) = .10348 S S S n =0 n =0 j =0 I = ∑ nPNS (n) = ∑ nPxL (S − n) = ∑(S − j)PxL ( j) = S − DL + b(S) E[back order cost / yr] ETRC(S) = rv(S − DL) + B2vDPxL ≥ (S) + (rv + B3v)b(S) 7 Base Stock for Slow-Moving / Expensive Items Indifference analysis to determine S ETRC(S) = Ivr + Cs S setup cost for every unit of demand thus, it is ignored = vr ∑ (S − j)PxL ( j) + B2vDPxL ≥ (S) j =0 Ambivalent between S and S+1 : ETRC(S) = ETRC(S + 1) ... after algebraic reduction ... PxL (S) r = PxL ≤ (S) DB2 8 4 Base Stock for Slow-Moving / Expensive Items B2 criterion / Q = 1 Example: Item 1 Demand 20 Lead Time Order Unit Carry Mean Dem. Std Dev of Weeks Cost cost Charge during LT LT Dem. 3.5 $ 3.20 $ 400.00 0.24 1.346153846 Q= 1 Decision Rule Q? Demand 0 1 2 3 4 5 6 PxL ( s + 1) 1 Slow Moving Poisson Q=1 s 5 Q 1 x hat L 1.346153846 1.8 B2 r/D*B2 0.006666667 EOQ 1 PxL ≤ ( s + 1) s=S-1 Probability Cumulative 0.26024 0.2602 0.35032 0.6106 0.23579 0.8464 0.10580 0.9522 0.03561 0.9878 0.00959 0.9974 0.00215 0.9995 = r DB2 Target RHS s 0 1 2 3 4 5 6 ratio 0.5738 0.2786 0.1111 0.0360 0.0096 0.0022 0.0004 Ratio≤RHS for first time 9 (s,Q) Policy for Slow Moving Items B2 criterion / Q > 1 : – An indifference analysis analogous to the Q = 1 case (see Problem 8.2) leads to : Example : Q ∑P (s + j) ∑P (s + j) j =1 Q j =1 xL xL ≤ = Lead Time Order Unit Carry Mean Dem. Std Dev of Weeks Cost cost Charge during LT LT Dem. 3.5 $ 3.20 $ 40.00 0.24 2.355769231 Probability Cumulative s ratio Slow Moving Poisson Q>1 0.09482 0.0948 0 0.2447 s 6 0.22338 0.3182 1 0.1585 Q 5 0.26311 0.5813 2 0.0894 x hat L 2.355769231 0.150 B2 0.20661 0.7879 3 0.0734 r/D*B2 0.045714286 0.12168 0.9096 4 0.0559 0.05733 0.9669 5 0.0472 0.02251 0.9894 6 0.0429 0.00758 0.9970 7 0.0404 sums 0.41571 4.6509 0.08938 0.0001 Item 1 r D B2 Demand 35 EOQ 5 10 5 (s,Q) Policy for Slow Moving Items B1 criterion / Q ≥ 1 : – Cycle stock holding cost and setup cost not a function of s, thus ignored in ETRC… D B P (s) Q 1 xL ≥ x =0 Indifferent when ETRC(s) = ETRC(s + 1) s ETRC(s) = vr ∑ (s − x)PxL ( x) + ⇒ PxL (s + 1) PxL ≤ (s) = Qvr DB1 11 (s,Q) Policy for Slow Moving Items Example: Item 1 Demand 35 Lead Time Order Weeks Cost 3.5 $ 3.20 $ Slow Moving Poisson Q>=1 s 4 Q 5 x hat L 2.355769231 10 B1 Qvr/D*B1 0.137142857 PxL ( s + 1) PxL ≤ ( s ) = Qvr DB1 Unit Carry Mean Dem. Std Dev of cost Charge during LT LT Dem. 40.00 0.24 2.355769231 Probability Cumulative 0.09482 0.0948 0.22338 0.3182 0.26311 0.5813 0.20661 0.7879 0.12168 0.9096 0.05733 0.9669 0.02251 0.9894 0.00758 0.9970 0.00223 0.9992 s 0 1 2 3 4 5 6 7 8 ratio 2.3558 0.8269 0.3554 0.1544 0.0630 0.0233 0.0077 0.0022 0.0006 EOQ 5 Calc. s 1 1 1 1 0 0 0 0 0 12 6 (s,Q) Policy Determination ETRC( s , Q ) = AE (D ) + E ( D )v + rv Q E (D) ⎡Q ⎤ ⎢ 2 + s − E ( D )L ⎥ + B 2 Q vb( s) ⎣ ⎦ ∞ where loss integral b( s) = ∫ ( x − s) f ( x )dx s 2 E ( D )( A + B 2 vb( s)) AE (D ) rv B 2 E ( D )b( s )v ∂ ETRC =− + − = 0.......Q = 2 2 2 rv ∂Q Q Q B E ( D )v ∂ b( s) ∂ ETRC = rv + 2 =0 Q ∂s ∂s ∞ ∂ b( s) = ∂s ∂ ∫ ( x − s) f ( x )dx s ∂s ∞ = − ∫ f ( x )dx = −(1 − F ( s)) s rQ F ( s) = 1 − B2 E (D) 13 (s,Q) Policy Determination E (D )B1 AE (D ) ⎛ Q ⎞ + ⎜ + kσ L ⎟ vr + Pu ≥ (k) Q Q ⎝2 ⎠ ∂ ETRC(Q , k) AE (D ) vr E (D )B1 =− + − Pu ≥ (k) = 0 ∂Q 2 Q2 Q2 E (D )B1 ∂ ETRC(Q , k) = σ L vr + ( − fu (k)) = 0 ∂k Q 2 Qv σ L r 1 = fu (k) = e − k /2 E (D )B1 2π ETRC(Q , k) = B1 criterion / Normal demand: ⎡ E (D )B1 ⎤ 2 ln ⎢ ⎥ ....similarly, solvin g for Q... ⎢ 2π Qv σ L r ⎥ ⎣ ⎦ vr AE (D ) E (D )B1 = + Pu ≥ (k) 2 Q2 Q2 Q2 AE (D ) E (D )B1 = + Pu ≥ (k) 2 vr vr B Q = EOQ 1 + 1 Pu ≥ (k) ...solve for k and Q iteratively 14 A k= 7 (s,Q) Policy Determination B1 criterion / Normal demand Example : Item 1 B1 30 Demand 6000 k 0.76 Sequential (s, Q) EOQ s Order Cost Holding Cost Shortage Cost Total Cost Stockout/year Lead Time Weeks 4.333 $ SS 15 447 515 $ 1,342 $ 1,433 $ 90 2864.489 3.0 Order Cost 100.0 $ SS$ 458 Unit cost 30.00 p(k) 0.2227 Carry Mean Dem. Std Dev of Charge during LT LT Dem. 0.2 500 20 f(k) 0.2981 Simultaneous (s, Q) Iteration 1 2 Q 447.214 461.91 k 0.7632 0.7196 p(k) 0.22267 0.2359 Q s Order Cost Holding Cost Shortage Cost Total Cost Stockout/year 463 514 $ 1,296 $ 1,475 $ 92 2862.953 3.1 G(k) 0.1282 3 462.77 0.7170 0.2367 P1 0.7773 4 462.82 0.7169 0.2367 EOQ 447.21 P2 0.994 5 462.82 0.7169 0.2367 Percent Savings 0.05% $ 1.54 15 (s,Q) Policy Determination The Example with an enhanced “A” profile : Item 1 Demand 50000 Lead Time Order Unit Weeks Cost cost 3 $ 10.0 $ 300.00 Simultaneous (s, Q) Iteration 1 2 Q 129.099 300.51 k 0.8451 0.0000 p(k) 0.19902 0.5000 S equential (s, Q) EOQ s Order Cost Holding Cost Shortage Cost Total Cost Stockout/year 129 3,223 $ 3,873 $24,156 $17,112 45140.79 77.1 3 449.07 0.0000 0.5000 4 449.07 0.0000 0.5000 Carry Mean Dem. Std Dev of Charge during LT LT Dem. 0.2 2,885 400 Q s Order Cost Holding Cost Shortage Cost Total Cost Stockout/year 449 2885 $ 1,114 $13,470 $12,361 26944.39 55.7 EOQ 129.10 B1 222 Percent Savings 67.53% $ 18,196.41 16 8 Order Point, Order Level Model (s,S) Inventory Level S s Place order Lead Time Safety Stock (SS) Receive Time order Allows variable Q if there is batch demand Equivalent to (s,Q) if demand is one unit at a time TRC(s,S) < TRC(s,Q) – but more computational effort Dominant control system in practice 17 Order Point, Order Level Model (s,S) Inventory Level S s undershoot Place order Lead Time Safety Stock (SS) Receive Time order S-s ≈ Q – the replenishment size is at least as large as an EOQ S and Q can be adjusted for demand “undershoot” of s for a more refined model needed for “A” items 18 9 Undershoot-based (s,S) Model Let – Z = undershoot random variable – X = total lead time demand ~ N(µ,σ) – X’ = Z+X …. assumed to be approximately Normal – Pz(zo) = P{undershoot = zo} – Pt(to) = P{transaction size = to} – E(t) = expected transaction size From renewal theory… 1∞ – Pz(zo) = ∑ P (t ) E(t) to =zo +1 t o 19 Undershoot-based (s,S) Model ⎤ 1 ⎡ E(t 2 ) − 1⎥ ⎢ 2 ⎢ E (t) ) ⎥ ⎣ ⎦ 2 ⎤ ⎛ E (t 2 ) ⎞ 1 ⎡ 4 E(t 3 ) ⎢ var(Z ) = − 3⎜ ⎟ − 1⎥ 12 ⎢ E(t) ⎥ ⎝ E(t) ⎠ ⎣ ⎦ ...if Z and X are independent random E(Z ) = Developing Undershoot Adjustment variables... E( X ') = E(Z ) + E( X ) var( X ') = var(Z ) + var( X ) ...thus... Q let I ≈ + s − E( X ') 2 and Q ' = S − s + E(Z ) = Q + E(Z ) assume X ' is distributed as N( μ ,σ ) SS = kσ x ' and s = E( X ') + kσ x ' 20 10 Undershoot-based (s,S) Model Undershoot adjusted ETRC function AD D ⎛Q ' ⎞ + + kσ x ' ⎟ vr B1Pu ≥ (k) + ⎜ Q' Q' ⎝2 ⎠ ∂ETRC(k, Q ) ∂ETRC(k, Q ) = =0 ∂Q ∂k Solving... (Q ' = Q + E(Z)) ETRC(k, Q ) = Q = EOQ 1 + B1 P (k) − E(Z) A u≥ ⎡ 1 ⎛ B1 ⎞ ⎤ EOQ 2 k = 2ln ⎢ ⎥ ⎜⎟ ⎣ 2 2π ⎝ A ⎠ (Q + E(Z))σ x ' ⎦ 21 Undershoot-based (s,S) Model Example: Normal Demand / Discrete Transaction Distribution Lead Time Order Unit Carry Mean Dem. Std Dev of Item Demand Weeks Cost cost Charge during LT LT Dem. EOQ 1 5200 2 $ 10.0 $ 83.20 0.02 200 7.07107 250.00 E(t) 24 Simultaneous (s, S) E(t^2) 700 B1 = $20 E(t^3) 22800 trans Prob t^2 t^3 E(z) 14.08333 10 0.3 100 1000 E(x') 214.08 20 0.2 400 8000 var(z) 103.91 30 0.3 900 27000 var(x') 153.91 40 0.2 1600 64000 sigma(x') 12.41 Iteration Q k p(k) s S 1 250 2.01 0.0220 239 480 2 241.35 2.03 0.0211 3 241.14 2.03 0.0211 4 241.14 2.03 0.0211 Sequential (s, S) s S 216 466 22 11 Optional Replenishment Model (R,s,S) Inventory Level S s R R Time R (s,S), (R,S) hybrid control system An option exists at each review to not reorder S-s ≈ Q – the replenishment size is at least as large as an EOQ TRC(R,s,S) best among control systems reviewed – but, computationally most difficult 23 Optional Replenishment Model (R,s,S) Revised Power Approximation (B3 shortage cost) – Regression used against known optimal policies on: b ⎛A⎞⎛ σ R +L 2 ⎞ ˆR1− b ⎜ Q p = ax ⎟ ⎟ ⎜1 + ˆ xR2 ⎠ ⎝ vrR ⎠ ⎝ c – Regression used to fit parameters of sp , also – Power Approximation estimates of the Order Quantity and Reorder Point: .116 .506 ⎛ σ L2 ⎞ 1 + R +2 ⎟ ⎜ ˆ xR ⎠ ⎝ ⎛ .183 ⎞ + σ R +L ⎜ + 1.063 − 2.192 z ⎟ ⎝z ⎠ ⎛A⎞ ˆ Q p = 1.3 x R .494 ⎜ ⎟ ⎝ vrR ⎠ ˆ s p = .973 x R + L where z = Q pr σ R + L B3 ˆ x R = DR ˆ x R + L = D(R + L) 24 12 Optional Replenishment Model (R,s,S) Initial estimates are used if it appears that order quantity will typically cover more than 1 review period if Qp ˆ xR > 1.5 → s = sp and S = sp + Qp Otherwise, a pure periodic review system may be more attractive ˆ So = xR+L + kσ R+L such that Pu≥(k) = Final model specification { s = min s p , So } { r B3 + r S = min s p + Q p , So } Note : if s = S = So …… (R, s, S) becomes (R, S) 25 Optional Replenishment Model (R,s,S) Example: Example: E(Dt)=100/wk v=$5 L=3 wk = 49 B3=$0.1/$/wk R=1 wk A=$10 r=$0.005/$/wk σt2 Not dependent on form of distribution .506 ˆ Q p = 1.3x R .494 ⎛A⎞ ⎜ vr ⎟ ⎝⎠ .116 ⎛ σ R +L 2 ⎞ ⎜1 + ⎟ ˆ xR2 ⎠ ⎝ .506 .494 ⎛ 10 ⎞ = 1.3 (100 ) ⎜ ⎜ 5 (.005 ) ⎟ ⎟ ⎝ ⎠ z= Q pr σ R +LB3 = .116 ⎛ 4(49) ⎞ ⎜1 + ⎟ ⎜ (100 )2 ⎟ ⎝ ⎠ (262.76)(.005) = .969 (14)(.1) = 262.76 26 13 Optional Replenishment Model (R,s,S) ⎛ .183 ⎞ ˆ + 1.063 − 2.192 z ⎟ s p = .973xR + L + σ R + L ⎜ ⎝z ⎠ ⎛ .183 ⎞ = .973 (100 + 300 ) + (14 ) ⎜ + 1.063 − 2.192 (.969 ) ⎟ = 377 ⎝ .969 ⎠ Step2 : Q p 262.76 = = 2.63 > 1.5 ˆ xR 100 Thus... s = s p = 377 S = s p + Q p = 377 + 263 = 640 27 Optional Replenishment Model (R,s,S) Example 2: Example v=$50 E(L)=2 mo. A=$50 B3=$20/$/yr var(L)=1 r=$0.40/$/yr σ R2 = E(Dt)=10,000/yr R=1 mo. σt=2000 (2000)2 = 333,333 12 σL2 = E ( L) σt 2 + E(Dt )2 var(L) 2 ⎛2⎞ ⎛ 10000 ⎞ ) = ⎜ ⎟(2000)2 + ⎜ ⎟ (1 ⎝ 12 ⎠ ⎝ 12 ⎠ ,361 = 1 ,111 σ R+L2 = σ R2 + σL2 = 333,333 + 1 ,111 = 1 ,361 ,694,444 → σR+L = 1302 28 14 Optional Replenishment Model (R,s,S) Step 1: ˆ Q p = 1.3 x R .494 ⎛A⎞ ⎜ vr ⎟ ⎝⎠ .506 ⎛ σ R +L 2 ⎞ ⎜1 + ⎟ ˆ xR2 ⎠ ⎝ .116 ⎛ ⎜ ⎜ 1 + 1694444 2 ⎜ ⎛ 10000 ⎞ ⎜ ⎜ 12 ⎟ ⎜ ⎝ ⎠ ⎝ = (1.3)(27.725)(5.59)(1.154) = 232.55 ⎛ ⎞ .494 ⎜ ⎟ ⎛ 10000 ⎞ ⎜ 50 ⎟ = 1.3 ⎜ ⎟ ⎝ 12 ⎠ ⎜ 50 ⎛ .4 ⎞ ⎟ ⎜ ⎜ 12 ⎟ ⎟ ⎝ ⎝ ⎠⎠ z= Q pr σ R + L B3 = .506 ⎞ ⎟ ⎟ ⎟ ⎟ ⎟ ⎠ .116 (232.55)(.4) = .0598 (130 2 )(20) ⎛ .183 ⎞ ˆ s p = .973 x R + L + σ R + L ⎜ + 1.063 − 2.192 z ⎟ ⎝z ⎠ ⎛ 10000 ⎞ ⎛ .183 ⎞ = .973 ⎜ ⎟ + (1302 ) ⎜ .0598 + 1.063 − 2.192 (.0598 ) ⎟ = 7630.3 ⎝4⎠ ⎝ ⎠ 29 Optional Replenishment Model (R,s,S) Step 2: Qp ˆ xR = 232.55 < 1.5 833.3 Thus, we need Pu ≥ (k) = .4 r = = .0196 → k = 2.06 r + B3 20 + .4 So = x R +L + kσ R +L = 2500 + (2.06)(1302) = 5182 { S = min {s } s = min s p , So = min {7630,5182} = 5182 p } + Q p , So = min {7630 + 233,5182} = 5182 So, (R, s, S) reduces to (R, S) ˆ Notice that Q = E(Dt )R = 833 >> Q p = 233 30 15 Optional Replenishment Model (R,s,S) Fill Rate (P2) Service Criterion – Expected Order Size may be larger than S-s because system only viewed at the end of review periods » do not know when inventory position crosses s – Nominal order size 1 2 AD + σ L2 S − s = EOQ or P2 vr – Expected “undershoot” of s by end of review period σ R 2 + μR 2 E[U ] = 2 μR – Thus, E[Q] = Expected order size = S - s + E[U] 31 Optional Replenishment Model (R,s,S) Fill Rate (P2) Service Criterion – Expected shortage per replenishment cycle ⎛ s − μ R +L ⎞ ⎛ s − μL ⎞ 2 ⎟ − σ L Ju ⎜ ⎟ ⎝ σ R +L ⎠ ⎝ σL ⎠ 2 μR σ R + L 2 Ju ⎜ E [ short ] = – Using fill rate definition: E [ short ] = 1 − P2 E [Q ] – Rearranging after substitution: σR+L 2 ⎡ ⎛ s − μR+L ⎞ ⎛ s − μL ⎞ σR2 + μR2 ⎤ 2 Ju ⎜ ⎟ − σL Ju ⎜ ⎟ = 2(1 − P2)μR ⎢S − s + ⎥ 2μR ⎦ ⎝ σ R+L ⎠ ⎝ σL ⎠ ⎣ – Solve for s 32 16 Optional Replenishment Model (R,s,S) Simplified fill rate model – Second term on LHS of complex model may be dropped if R > .5L and P2 > .9 with little impact on accuracy ⎡ ⎛ s − μR+L ⎞ σ R2 + μR2 ⎤ σ R+L Ju ⎜ ⎟ = 2(1 − P2 )μR ⎢S − s + ⎥ 2μR ⎦ ⎝ σ R +L ⎠ ⎣ s − μR+L let k = 2 σ R +L – Ju (k) can be solved for and an appropriate k found in the Normal Tables, then s = μR +L + kσ R +L 33 Optional Replenishment Model (R,s,S) Fill Rate Example Fill A=$25 v=$10 r=$.25/$/yr μt=25000/yr R=2 weeks L=2 weeks P2=.99 σt=509 S − s = EOQ = 2(25)(25000) = 707 (10)(.25) 2 ˆ (25000) = 962 xR+L = 1924 52 2 2 σR = σL = (509) = 100 σ R+L = 2(100) = 141.42 → σ R +L = 20000 52 ˆ ˆ x R = xL = 34 17 Optional Replenishment Model (R,s,S) Use simplified formula: R>.5L , P2>.9 2(1 − P2)μR ⎡ σ R2 + μR2 ⎤ s − μR+L Ju ( k) = let k = ⎢S − s + ⎥ 2 2μR ⎦ σ R+L σ R+L ⎣ 2(1 − .99)(962) ⎡ 10002 + 9622 ⎤ = ⎢707 + ⎥ = 1.1479 → k = −.575 20000 2(962) ⎦ ⎣ ˆ s = xR+L + kσ R+L = 1924 + (−.575)(141.42) = 1842.68 → 1843 S = s + EOQ = 1843 + 707 = 2550 Complex formula returns the same result. 35 Optional Replenishment Model (R,s,S) Adjusting the example data so that simple formula conditions violated… A=$25 v=$10 r=$.25/$/yr μt=25000/yr R=2 weeks L=10 weeks P2=.65 σt=2280 Recalculating… σR = σ R+ L = 2 (2280) = 447.21 52 12 (2280) = 1095.45 52 36 18 Optional Replenishment Model (R,s,S) 2(1 − P2 )μR ⎡ σ 2 + μR 2 ⎤ s − μR + L S−s+ R let k = ⎢ ⎥ 2 2 μR ⎦ σ R +L σ R +L ⎣ 2 2 ( 447.21) + ( 962 ) ⎤ = .7246 → 2(.35)(962) ⎡ ⎢707 + ⎥ = 2 2(962) ⎥ (1095.45) ⎢ ⎣ ⎦ 12 ˆ s = x R +L + kσ R +L = (25000) + (−.242)(1095.45) = 5504 52 S = s + EOQ = 5504 + 707 = 6211 Ju ( k ) = Complex formula result… s = 5350 , S = 6057 If P2=.99 and L=10 weeks… – Simple: (s, S) = (7454, 8161) (s, S) = (7441, 8149) – Complex: k = −.242 37 19 ...
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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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