Chapter 7: managing flow variability: safety inventoryProblem 7.1Problem 7.2Problem 7.3Problem 7.4Prob(R Q*) = .Problem 7.5Problem 7.6Another approach to get the critical fractile probability SL uses the newsvendor solution directly:Problem 7.7 (This is an advanced problem)We are concerned about the overbooking problem; that is, how many seats to overbook. The randomness in demand arises from uncertain cancellations, which are uniformly distributed between 0 and 20. One can think of this question as asking “what is the optimal service level of cancellations?”If I overbook by 1 additional unit, thenThus optimal service level is MB/(MB+MC) = 600/(600+250) = 70%. The optimal overbooking quantity is determined by Prob(RQ) = MB/(MB+MC) = 0.7. For uniform distribution between 0 and 20, Prob(RQ)=Q/20. Thus Q = 20 * 0.7 = 14 seats so that the optimal overbooking level is 14 seats.Problem 7.8Problem 7.9MBPF Ch7 solutions. Last updated: March 12, 2013CHAPTER 7: MANAGING FLOW VARIABILITY: SAFETY INVENTORYProblem 7.1 [a] Given quantities: mean weekly demand = 400; standard deviation of weekly demand = 125; replenishment lead time = 1 week and reorder point (ROP) = 500 units. We computethe average demand during leadtime to be 400 units. Thus the safety stock, Isafety = 100 units; to find the service level provided, we need to find the area under the normal curve to the left of the reorder point (ROP) = 500. Let the demand during lead time be LTD. The cycle service level Prob( LTD  ROP) = Prob( LTD  R + Isafety) = 0.7881.So the cycle service level is 78.81%.[b] The standard deviation of lead time demand, LTD = 125 units. For each service level the z-value can be read from the standard normal table. The safety inventoryIsafety = z x LTD Finally, ROP = 400 + Isafety.Cycle Service Level 80% 90% 95% 99%z = 0.842 1.282 1.645 2.326Isafety = 105 160 205 290ROP = 505 560 605 690Problem 7.2[a] Average weekly demand (R) = 1000Standard deviation of weekly demand (R) = 150.Lead time (L) = 4 weeks.Standard deviation of demand during lead time (LTD ) = RLs= 300.Current reorder point (ROP) = 4,200.Average demand during lead time (LTD) = L x R= 4,000.Current level of safety stock (Isafety)= 200.Current order quantity (Q) = 20,000Average inventory (I) = Isafety + Q/2 = 200 + (20,000/2) = 10,200.H = rC = 0.25 x 9.99 = 2.4975  2.50Average time in store (T) = I/R = 10,200/1,000 = 10.2 weeks.Annual ordering cost = S x R/Q = $100 x 2.5 = $250.Annual holding cost = H x I = 2.5 x 10,200 = $25,500.[b] We use the EOQ formula to determine the optimal order quantity.H = $1 * 25%/year = $2.50/yearR = 1,000 /week = 50,000/yearS = $100.61Thus, the economic order quantity isQ = √2 SRH=√2× 100×50 , 0002. 50=√4 ,000 , 000=2 ,000 To determine the safety inventory, Isafety, for a 95% level of service, we first observe that the z-value = 1.65. Then Isafety = z x LTD = 1.65 x 300 = 495.Average inventory (I) = Isafety + Q/2 = 495 + (2,000/2) = 1,495.Average time in store (T)= I/R = 1.495 weeks.[c] If lead time (L) reduces to 1 week, then standard deviation of demand during lead time (LTD) = 150. Safety stock for 95% level of service = 1.65 x 150 = 247.5.Average inventory = 247.5 + (2,000/2) = 1,247.5.Average time in store = 1.25 weeks.Problem 7.3(a) The optimal order quantity of planters for HG isQ¿=√2 RSH=√2×1500×52×100002. 5=24 , 980(b) If the delivery lead time from Italy is 4 weeks and HG wants to provide its customers a cycle service level of 90%, Safety stock = NORMSINV(.9)*sqrt(4)*800 = 2050(c) Quantify the impact of the change.Additional transportation cost per year = 1500*52*.2 = $15,600Savings in holding cost = NORMSINV(.9)*800*(sqrt(4)-sqrt(1))*10*0.25 = $2562.5Thus Fastship should not be used.(One could be more precise and compare the total costs under current shipping with that with Fastship. The latter has slightly higher unit holding cost H, which also will slightly increase the cycle stock, in addition to the transportation cost. Given that even at the old holding cost, transportation increased cost exceed holding cost savings, the above answer is sufficient to draw the correct conclusion.) Problem 7.4First, is this an EOQ problem? Well, notice that the question dictates that we do a run every two years. That would mean, in a deterministic EOQ setting, that Q must equal twoyears of mean demand, i.e., 32000. Hence, this question does not give us the freedom to change when we do a run (which is what EOQ is all about).Thus, the question is whether 32000 is the best quantity we can print every two years? This thus asks about what the appropriate safety stock (or service level) should be. We know that this is answered by newsvendor logic. Answer these two questions:1. What is my underage cost (cost of not having enough)? I.e., if I were to stock onemore unit, how much could I make? Every catalog fetches sales of $35.00 and costs $5.00 to produce. Thus, the net marginal benefit of each additional unit (MB), or the underage cost, is p – c = $35 - $5= $30. 2. What is my overage cost? I.e., if I had stocked one less unit, how much could I have saved? The net marginal cost of stocking an additional unit (MC) = c – v = $5 – 0 = $5. Now, we can figure out the optimal service level (or critical fractile): SL = 30/(30+5) = 0.857.The last step is to convert the SL into a printing quantity. Recall that total average demand for 2 years (R) = 32,000 with a standard deviation of 5656.86. The optimal printing quantity, Q* is determined such thatProb(R ¿ Q*) = ( )300.85730 5MBMB MC= =+ +.The optimal order quantity Q* = R + z  where z is read off from the standard Normal tables such that area to the left of z is 0.857. That is, z = 1.07. This gives Q* = 38,053 catalogs. It can be verified that the optimal expected profit (when using Q* = 38053) is larger than $25,000, the fixed cost of producing the catalog.Problem 7.5The revenue per crate, p = $120.00, variable cost, c = $18.00, and salvage value, v = – $2.00. The marginal benefit of stocking an additional crate (MB) = p – c = $120 – $18 = $102. The marginal cost of stocking an additional unit (MC) = c – v = $18 + $2 = $20. ThenMB/(MB+MC) = 102/(102+20) = 0.836.The probability density of demand and its cumulative probability is listed below.Demand 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15Frequency 0 0 0 1 3 2 5 1 6 7 6 8 5 4 1 3Prob. 0 0 0 0.02 0.06 0.04 0.1 0.02 0.12 0.13 0.12