The News Vendor Problem, Part 2

Basic Probability Distributions Part 2 The News Vendor Problem The expected number of participants in a conference is normally distributed with mean of 500 and standard deviation of 150. The participants spend one night in the hotel and the cost is paid by the conference. The hotel has offered a rate of $200 per room if a block of rooms is reserved (non-refundable) in advance. The rate in the conference day is $300. All rooms will be single occupied. How many rooms should we reserve in the non-refundable block to minimize our expected total cost. Service level (Probability of demand not exceeding what we have ordered) SL* = Cu/(Cu+Co) Co = 200, Cu = 300-200 = 100 SL* = Cu/(Cu+Co) = 100/(100+200) = 1/3 =NORM.INV(1/3,500,150) =436 21 100,000 in One Stock Suppose there is a stock with its return following Nomal distribution with mean of 4% and standard deviation of 4%. Therefore if we invest one dollar in this stock, our investment after one year will have pdf of N(1.04, 0.04). What will be mean and standard deviation of our investment if we invest 100,000. =NORM.INV(probability, mu, sigma) x:N(1.04,.04) 1.00 1.03 0.99 1.08 1.05 1.06 1.06 1.02 1.01 1.07 y=100,000x 100296.9134 103355.9772 99461.79536 107857.9764 104640.5643 106107.4342 105839.8813 102089.6817 101131.6528 107078.3105 =NORM.INV(probability, 5%, 5%) =NORM.INV(rand(), 5%, 5%) $1 Mean 1.04 Var 0.0016 StdDev 0.0402 CV 0.0386 $100,000 Ratio 104008 100000 16121032 4015 100000 0.0386 22 y = Kx µ(y) = k µ(x), σ(y) = Kσ(x) A random variable x with mean of µ , and standard deviation of σ is multiplied by a constant K and generates the random variable y=Kx. Then Mean (y) = K*Mean(x) StdDev (y) = |K|*StdDev(x) If $1 after one year has mean of $1.04 and StdDev of 0.04 Then 100,000 after one year has mean of $(100,000)1.04 and StdDev of 100,000(0.04) 23 90000 in One Stock or 10000 in Each of 10 stocks Suppose there are 9 stocks and with high probability they all have Normal pdf return with mean of 5% and standard deviation of 5%. These stocks are your only options and no more information is available. You have to invest $100,000. What do you do? =NORM.S.INV(rand()) =z = 0.441475 X= µ + sz = X = 5%+ 0.441475(5%) X= 5%+2.21% = 7.21% =NORM.INV(probability, mu, sigma) =NORM.INV(probability, 5%, 5%) =NORM.INV(rand(), 5%, 5%) = -8.1% 24 100,000 vs. 10(10,000) investment in N(5%, 5%) 90000 4306 4742 693 11964 1941 67 -2280 5914 1943 6565 3682 10000 864 422 891 -165 464 785 505 -55 1468 882 1410 10000 388 1038 585 1400 1406 -417 -106 1434 521 49 919 10000 513 316 -561 895 1459 1089 150 629 172 124 836 10000 1016 -667 674 861 855 917 770 740 771 1244 919 10000 385 619 757 297 200 582 700 538 459 1123 950 10000 452 681 715 204 665 -605 519 1156 1014 590 1955 10000 1097 599 736 194 1197 1119 855 -264 610 715 1649 10000 448 751 462 934 -828 264 587 1232 262 -118 962 Alternatives Min= Max= Mean= Variance StdDev= VTM CV Mean/SrdDev 10000 -126 -3 236 -641 906 781 148 860 1350 -93 1091 90000 5036 3757 4494 3978 6326 4516 4128 6269 6626 4517 10692 90000 -8198 20162 4568 20979621 4580 4592.29 1.00 0.9974 9(10000) -63 9698 4569 2325737 1525 509.02 0.33 2.9960 25 Risk Aversion Individual 20000 15000 10000 5000 0 -5000 -10000 -15000 26 y = x1+x2+…+xk µ(y) = k µ(x), VAR(y) = KVAR(y) A random y is equal to summation of K random variable x each with mean of µ and standard deviation of σ. y = x1+x2+….xK Mean (y) = K*Mean(x) VAR (y) = K*VAR(x) σ y = Kσ x 27 ROP; total demand during lead time is variable If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute safety stock. =NORM.INV(.9,200,25) Safety Stock 232.0388 32.03879 This Problem: If average demand per day is 20 units and standard deviation of demand per day is 5, and lead time is 16 days. Compute ROP at 90% service level. Compute safety stock. If we can transform this problem into the previous problem, then we are done, because we already know how to solve the previous problem. 28 μ and σ of demand per period and fixed L If Demand is variable and Lead time is fixed L: Lead Time R: Demand per period (per day, week, month) R: Average Demand per period (day, week, month) σR: Standard deviation of demand (per period) LTD: Average Demand During Lead Time LTD = L × R σLTD: Standard deviation of demand during lead time 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐿𝐿𝜎𝜎𝑅𝑅 29 μ and σ of demand per period and fixed L If demand is variable and Lead time is fixed L: Lead Time = 16 days R: Demand per day R: Average daily demand =20 σR: Standard deviation of daily demand =5 LTD: Average Demand During Lead Time LTD = L × R = 16 × 20 = 320 σLTD: Standard deviation of demand during lead time 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 = 𝐿𝐿𝜎𝜎𝑅𝑅 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 = 16 5 = 20 30 μ and σ of demand per period and fixed LT R: demand rate per period (a random variable) R: Average demand rate per period σR: Standard deviation of the demand rate per period L: Lead time (a constant number of periods) LTD: demand during the lead time (a random variable) LTD: Average demand during the lead time σLTD: Standard deviation of the demand during lead time 31 μ and σ of demand per period and fixed LT A random variable R: N(R, σR) repeats itself L times during the lead time. The summation of these L random variables R, is a random variable LTD If we have a random variable LTD which is equal to summation of L random variables R LTD = R1+R2+R3+…….+RL Then there is a relationship between mean and standard deviation of the two random variables LTD = LR 2 2 σ LTD = Lσ R σ LTD = Lσ R 32 Now It is Transformed The Problem originally was: If average demand per day is 20 units and standard deviation of demand is 5 per day, and lead time is 16 days. Compute ROP at 90% service level. Compute safety stock. We transformed it to: The average demand during the lead time is 320 and the standard deviation of demand during the lead time is 20. Compute ROP at 90% service level. Compute safety stock. Which is the same as the previous problem: If average demand during the lead time is 200 and standard deviation of demand during lead time is 25. Compute ROP at 90% service level. Compute safety stock. 33 ROP; Variable R, Fixed L e) If demand per day is 20 units and lead time is 16 days and standard deviation of lead time is 4 days. Compute ROP at 90% service level. Compute Isafety. What is the average demand during the lead time What is standard deviation of demand during lead time 34 μ and σ of L and Fixed R If Lead time is variable and Demand is fixed L: Lead Time L: Average Lead Time σL: Standard deviation of Lead time R: Demand per period LTD: Average Demand during lead time LTD = L × R σLTD: Standard deviation of demand during lead time 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 =R𝜎𝜎𝐿𝐿 35 μ and σ of L and Fixed R If Lead time is variable and Demand is fixed L: Lead Time L: Average Lead Time = 16 days σL: Standard deviation of Lead time = 4 days R: Demand per period = 20 per day LTD: Average Demand During Lead Time LTD = 16 × 20 = 320 σLTD: Standard deviation of demand during lead time 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 =R𝜎𝜎𝐿𝐿 𝜎𝜎𝐿𝐿𝐿𝐿𝐿𝐿 =20(4) =80 36 μ and σ of lead time and fixed Demand per period L: lead time (a random variable) L: Average lead time σL: Standard deviation of the lead time RL R: Demand per period (a constant value) LTD: demand during the lead time (a random variable) LTD: Average demand during the lead time σLTD: Standard deviation of the demand during lead time R L 37 μ and σ of demand per period and fixed LT A random variable L: N(L, σL) is multiplied by a constant R and generates the random variable LTD. If we have a random variable LTD which is equal to a constant R times a random variables L LTD = RL Then there is a relationship between mean and standard deviation of the two random variables RL LTD = LR 2 σ LTD = R 2σ L2 σ LTD = Rσ L R L 38 Variable R fixed L………….Variable L fixed R LTD = LR 2 σ LTD = Lσ R2 σ LTD = Lσ R R R R R R + R + R + R + R R RL LTD = LR 2 σ LTD = R 2σ L2 σ LTD = Rσ L R L L 39 μ and σ of L and Fixed R LDT = 320 σLTD = 80 SL = 90% z =1.28 ROP = LTD +zσLTD ROP = 320 +1.28(80) ROP = 320 + 102.4 Isafety = 102.4 40 Comparing the two problems I. Average demand per day is 10 units and standard deviation of demand per day is 3 units. The lead time is 5 days II. Demand per day is 10 units. The average lead time is 3 days and standard deviation of lead time is 1 days. 1 2 3 4 5 1 2 3 4 5 41 Service Level, Fill Rate Within 100 time intervals, stockouts occur in 20. Service Level = 80/100 = 80%. Probability of stock-out = 20%. Risk = # of stockout intervals/ Total # of intervals Service Level = 1- (# of stockout intervals/ Total # of intervals) Suppose that in each time interval in which a stockout occurred, the number of units by which we were short. Suppose that cumulative demand during the 100 time intervals was 15,000 units and the total number of units short in the 20 intervals with stockouts was 1,500 units. Fill rate = (15,000-1,500)/15,000 = 13,500/15,000 = 90%. Fill Rate = Expected Sales / Expected Demand Fill Rate = (1- Expected Stockout )/ Expected Demand 42 2 20 20 0 0 0 Demand During the Lead Time 43