Re-Order Point Problems Set 2: NVP The first two problems are also included in the lecture slides. Since they have not been recorded, I use them along with the exercises.

Optimal Service Level: The News Vendor Problem An electronics superstore is carrying a 60” LEDTV for the upcoming Christmas holiday sales. Each TV can be sold at $2,500. The store can purchase each unit for $1,800. Any unsold TVs can be salvaged, through end of year sales, for $1,700. The retailer estimates that the demand for this TV will be Normally distributed with mean of 150 and standard deviation of 15. How many units should they order? Note: If they order 150, they will be out of stock 50% of the time. Which service level is optimal? 80%, 90%, 95%, 99%?? Cost =1800, Sales Price = 2500, Salvage Value = 1700 Underage Cost = Marginal Benefit = p-c = 2500-1800 = 700 Overage Cost = Marginal Cost = c-v = 1800-1700 = 100 Optimal Service Level = SL* = P(LTD ≤ ROP) = Cu/(Cu+Co) Or in NVP Terminology SL* = P(R ≤ Q*) = Cu/(Cu+Co)

Optimal Service Level: The Newsvendor Problem Underage cost = Marginal Benefit =Cu = 2500-1800 = 700 Overage Cost = Marginal Cost = Co = 1800-1700 = 100 SL* = Cu/(Cu+Co) SL* = 700/800 = 0.875 LTD =N(150,15) ROP = LTD + Isafety = LTD + zσLTD = 150+1.15(15) Isafety = 17.25 = 18 ROP = 168 Risk = 12.5% Probability of excess inventory 0.875 Probability of shortage 0.125 1.15

The Newsvendor Problem Terminology In the NVP terminology, the following notations is used R: demand during the single period of study R =N(R, σR) Q* : What we order (only once) P(R ≤ Q*) = Cu/(Cu+Co) = 0.875 Compute z value related to Cu/(Cu+Co) = 0.875 probability. z (0.875) = 1.15 z = (Q* - R)/σR 1.15 = (Q*-150)/15 Q* = 168

Optimal Service Level: The Newsvendor Problem Demand for a product in the upcoming period is normally distributed with mean of 4000 and standard deviation of 1000. Unit Revenue = Sales Price = p = 30. Unit purchase cost = c = 10. Salvage value = v = 6. Goodwill cost = g = 1 Goodwill cost is the present value of all the future lost sales due to creating bad reputation. LTD = N(4000,1000) Overage Cost = Marginal Cost = Co = c-v = 10-6 = 4 Underage Cost = Marginal Benefit = p-c+g = 30-10 +1 = 21 SL*= Cu/(Cu+Co) SL*= 21/25 = 0.84  Z(0.84) =

Optimal Service Level: The Newsvendor Problem Probability of excess inventory 0.84 Probability of shortage 0.16 0.99 ROP = LTD + Isafety = LTD + zσLTD ROP = 4000+0.99(1000) ROP = 4999 Risk = 16%

Problem 7.4- The News Vendor Problem Johnson Electronics sells electrical and electronic components through catalogs. Catalogs are printed once every two years. Each printing run incurs a a variable production cost of $5 per catalog. Annual demand for catalogs has normal distributed; mean of 16,000 and standard deviation of 4,000. On average, each customer ordering a catalog generates a $35 sales revenue. Assuming that Johnson wants only one printing run in each two-year cycle (and leftovers of the previous cycle are worthless), how many catalogs should be printed in each run? Is this an EOQ problem? We do a run every two years. That would mean, in a deterministic EOQ setting, that Q must equal two years of average demand; 32000 units. This is indeed a periodic inventory system, and the period is 2 years. The question is whether 32000 is the best quantity we can print every two years? Or we should print more. If we produce 32,000 the probability of stockout is 50%. So this is an ROP problem when we want to have Isafety for 2 years.

Problem 7.4- The News Vendor Problem What the appropriate safety stock (or service level) should be. Q = 2 year demand S =25000 C = 5 R/yr = N(16000,4000) Sales price P = 35 Every catalog fetches sales of 35 and costs 5 to produce. The underage cost Cu = p – c = 35 - 5= 30. What is my overage cost? Each catalog costs $5, if not used, salvages at the end of two year period. The overage cost = Co = 5-0 =5 The optimal service level (or critical fractile): SL* = Cu/(Cu+Co) = 30/(30+5) = 0.857. Z(0.857) = 1.07. L = 2  σLTD= √2(4000) = 5657

Problem 7.4- The News Vendor Problem The total average demand for 2 years is R ×L = 16000(2) = 32000 Q = 2 year demand S =25000 C = 5 R/yr = N(16000,4000) Sales price P = 35 The optimal ROP = LTD + z σLTD ROP = 32000 + 1.07(5657) ROP = 38,053 catalogs. If we order 38,053, there is 85.7% probability that we are not out of stuck in the nest two years. The Risk is 14.3%. The inventory cost is minimize. How many catalog should be distributed to break-even the fixed cost of 25,000 Marginal profit = 30, Q = 25000/30 = 833.3 =834

Problem Game- The News Vendor Problem Daily demand for your merchandise has mean of 20 and standard deviation of 5. Sales price is $100 per unit of product. You have decided to close this business line in 60 days. Your supplier has also decided to close this line immediately, but has agreed to provide your last order at a cost of $60 per unit. Any unsold product will be disposed at cost of $10 per unit. How many units do you order LTD = R ×L =20 ×60 = 1200. Should we order 1200 units or more or less? It depends on our service level. Underage cost = Cu = p – c = 100 – 60 = 40. Overage cost = Co = 60-0+10 =70 SL = Cu/(Cu+Co) = 40/(40+70) = 0.3636. Due to high overage cost, SL*< 50%. Z(0.3636) = ？

The standard Normal Distribution F(z) Given a 36.36% SL 36.36% Probability Compute z

The standard Normal Distribution F(z) Given a 36.36% SL 0.3636% Probability 0.6364 Probability 0.3636% Probability Compute z Compute z Then put a - sign in front of z

z Value using Table Go to normal table, look inside the table. Find a probability close to 0.6364. Read its z from the corresponding row and column. 0.6364 Probability Normal table 0.05 z Second digit after decimal The table will give you z Up to the first digit after decimal Z = 0.35 Probability 0.3 Z = - 0.35

Problem Game- The News Vendor Problem L = 60  σLTD = √60(5) = 38.73 The optimal ROP = LTD + z σLTD ROP = 1200 -0.35(38.73) ROP = 1200-13.55 ROP = 1186