Chapter 7: managing flow variability: safety inventory




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MBPF Ch7 solutions. Last updated: August 28, 2011

Chapter 7: managing flow variability: safety inventory

7.1 Objective

In the previous chapter on inventory, we focused on economies of scale as the major driver for inventory. The purpose of this chapter is to introduce the notion of safety inventory as a buffer against stochastic variability in supply / demand and discuss various levers for reducing it.

The chapter is covered over two classes each of duration 100 minutes. In the first class, we first motivate the need for forecasting as a way of estimating demand. We emphasize the four key characteristics of forecast without getting into any details of forecasting methodologies. This is key since the strategies for managing inventories critically exploit these characteristics. The first two characteristics of a forecast emphasize the need to estimate the variability of demand in addition to its mean. Building up on the examples of economic order quantity model in chapter 6, we discuss the notion of stock-outs (when demands become uncertain), introduce the cycle service level measure, and derive the safety stock as a buffer against uncertain demand to provide a certain service level. The key determinants of safety stock – demand variability and the mean and the variability of replenishment lead time - are emphasized. These are positioned as primary levers. Subsequently, the third characteristic of forecast (the aggregation principle) is used to discuss the concept of physical centralization of stocks as a way to reduce safety stock without affecting the cycle service level. Other manifestations of this principle in terms of virtual centralization, substitution, specialization, and component commonality are discussed qualitatively. Discussion of periodic review models can also be introduced here. Either this can be done soon after the basic model for continuous review is done (so there is a direct contrast) or after completing the discussion of centralization concepts.

In the second class, we emphasize the last characteristic of forecasts (the effect of forecast horizon on the accuracy of forecasts) as a lever to reduce safety stock investment and build flexibility to react to market changes through postponement or delayed differentiation. We use the Hewlett Packard and the Benetton case. (While discussing the Benneton case, for lack of time, we ignore the U.S. entry issue). We also introduce the classical newsvendor problem in this class.

Thus the supply chain module occupies three 100 minutes classes, one based on chapter 6 and two based on chapter 7.

7.2 Additional Suggested Readings

We continue with the Hewlett-Packard case to illustrate the notion of safety stock and the concept of centralization (across DCs in Europe).

  • “Hewlett-Packard: DeskJet Printer Supply Chain (A)”. Stanford Case 1993. Authors: Laura Kopczak and Hau L. Lee.

Suggested assignment questions (continued from chapter 6):

    1. Are there any other factors that need to be considered when deciding on the inventory stocking policy of different types of printers in Europe? How would you come up with appropriate levels of safety stock?

    2. How would you recommend that Hewlett-Packard structure its supply chain to best match supply with demand? Do you find it worthwhile for DCs to start supporting manufacturing?

  • Benetton (A), Harvard Business School case # 9-685-014.

Suggested questions:

    1. Summarize the important elements of Benetton’s marketing, logistics, manufacturing and financial strategies. Identify the presence or absence of interdependencies among these functional strategies.

    2. Is being a Benetton retailer a worthwhile business? Explain.

    3. How does Benetton gain advantage over its European competition?

  • Palu Gear, Author: Jan van Mieghem (email: permissions@vanmieghem.us).

This case can be used to all of the concepts in Chapters 6 and 7 including EOQ, centralization, as well as the newsvendor model. The questions are part of the case.

7.3 Solutions to the Problem Set

Problem 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 compute the 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 inventory

Isafety = z x LTD Finally, ROP = 400 + Isafety.


Cycle Service Level

80%

90%

95%

99%

z =

0.842

1.282

1.645

2.326

Isafety =

105

160

205

290

ROP =

505

560

605

690



Problem 7.2


[a] Average weekly demand (R) = 1000

Standard deviation of weekly demand (R) = 150.

Lead time (L) = 4 weeks.

Standard deviation of demand during lead time (LTD ) = = 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) = 20,000

Average inventory (I) = Isafety + Q/2 = 200 + (20,000/2) = 10,200.


Average 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 = $0.25 x 10,200 = $2,550.


[b] We use the EOQ formula to determine the optimal order quantity.

H = $1 * 25%/year = $.25/year

R = 1,000 /week = 50,000/year

S = $100.

Thus, the economic order quantity is


Q = = 6,325 units.

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 + (6,325/2) = 3,657.5.

Average time in store (T)= I/R = 3.6575 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 + (6,325/2) = 3,410.

Average time in store = 3.41 weeks.

Problem 7.3


  1. The optimal order quantity of planters for HG is





  1. 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


  1. Quantify the impact of the change.


Additional transportation cost per year = 1500*52*.2 = $15,600

Savings in holding cost = NORMSINV(.9)*800*(sqrt(4)-sqrt(1))*10*0.25 = $2562.5

Thus 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 7.4



First, 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 two years 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 one more 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 that

Prob(R  Q*) = .

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 7.5



The 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) = pc = $120 – $18 = $102. The marginal cost of stocking an additional unit (MC) = cv = $18 + $2 = $20. Then

MB/(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

15

Frequency

0

0

0

1

3

2

5

1

6

7

6

8

5

4

1

3

Prob.

0

0

0

0.02

0.06

0.04

0.1

0.02

0.12

0.13

0.12

0.15

0.1

0.08

0.02

0.06

Cumulative

















































Prob.

0

0

0

0.02

0.08

0.12

0.21

0.23

0.35

0.48

0.6

0.75

0.85

0.92

0.94

1


The optimal order quantity is the smallest number of crates such that cumulative probability is at least 0.836. From the table this gives the number of crates to be 12.

Problem 7.6




How many crews should the city assign to trash collection? For simplicity, you may treat the number of crews as a continuous variable. For example, 4.1 crews would be a perfectly acceptable answer.


One solution approach (starting from the basics):

Note that the marginal cost of scheduling one more ton = $125/ton.

This only has value if demand exceeds current planned schedule, in which case it saves $650. In other words, the expected marginal revenue is $650 Prob(R>Q).

At optimality, marginal cost equals marginal revenue:

 Prob(R>Q) = 125/650 or SL = 525/650 = 80.77%  z = .87  Q = 35 tons + .87*9 tons = 42.8 tons = 8.56 crews.


Another approach to get the critical fractile probability SL uses the newsvendor solution directly:

Here we are stocking up on local trash collection capacity.

Cost of overstocking by 1 ton = MC = 625/5 = $125

Cost of understocking by 1 ton = additional cost of using outside trash pick up = MB = $650-$125 = $525

Thus: SL = Prob(RQ) = MB /( MB + MC) = 525/(125 + 525) = 0.8077


appropriate # of crews = 8.56 crews


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, then


  • If there are more cancellations than “stocked”, we are fine: there are sufficient seats for every passenger who shows up. The net benefit is that we sold one more ticket at $600. (This is the “underage cost”; i.e., cost of having more cancellations than “stocked.”)

  • If there are fewer cancellations than “stocked,” there are insufficient seats for those passengers that have a ticket and show up. The net cost of this is that we must compensate the “bumped” customer (who had a reservation but did not get a seat) by $250 (the $600 earned from the additional ticket is spent on getting another ticket on another flight).

Thus 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 7.8




[a] To compute the optimal order quantity at each store we use the EOQ formula.

Assume 50 sales weeks/year.

H = 25%/year * $10 = $2.5/year

R = 10,000 /week = 500,000/year

S = $1000. Thus,

Q = EOQ = = 20,000 units.

The replenishment lead time (L) = 1 week.

Standard deviation of demand during lead time at each store (LTD) = 2,000.

Safety stock at each store for 95% level of service (Is) = 1.65 x 2,000 = 3,300.

Reorder point (ROP)= + Is = 10,000 + 3,300 = 13,300.

Average inventory across four stores (Id)

= 4 x (Is + Q/2) = 4*(3,300+(20,000/2)) = 53,200.

Annual order cost for all four stores = 4 x S x R/Q = 4 x 1,000 x25= $100,000.

Annual holding cost for all four stores = H x Id = $133,000.

Average time unit spends in store (T) = Id / 4 x R = 53,200/40,000 = 1.33 weeks.


[b] To compute the optimal order quantity at centralized store observe that this store

faces a cumulative average weekly demand = 4 x 10,000 = 40,000. This gives an annual demand of 2,000,000 units.

Q = EOQ = = 40,000 units.


Standard deviation of demand during lead time at central store (LTD)

= = 4,000.

Safety stock at central store for 95% level of service = 1.65 x 4,000 = 6,600.

Reorder point (ROP) = 40,000 + 6,600 = 46,600.

Average inventory in central store (Ic) = 6,600+(40,000/2) = 26,600.

Annual order cost for central store = S x R/Q = $1,000 x 50 = $50,000.

Annual holding cost for central store = H x Ic = $66,500.

Average time unit spends in store (T) = Ic / 4 x R = 26,600/40,000 = 0.67 week .

Problem 7.9





  1. Given that each outlet orders independently and gets its own delivery, the optimal order size at each outlet is

sqrt(2RS/H) = sqrt(2*4000*50*900/(.20*200)) = 3,000


Average cycle stock at each outlet = Q/2 = 1,500. Total cycle stock (and hence average inventory) across all outlets = 4 × 1,500 = 6,000.


  1. On average, each unit spends

T = I / R = (Q/2) / R = 1,500 / 4000 weeks = 3/8 weeks = .375 weeks in the Hi-Tek system before being sold


  1. With a fixed cost of $1,800, the new order quantity with centralized purchasing is


Q =  (2x4,000x4x50x1800)/40 = 8,486


This quantity is split into four and shipped to each outlet. So each outlet received 8,486/4= 2,122 units per shipment (rounded up to make whole number of units). So cycle stock at each outlet is 2,122/2 = 1,061 units.


Total average inventory across all four outlets will be four times the cycle stock in each outlet = 4×1,061 = 4,244.


Problem 7.10


Mean demand, 1000/day with a daily standard deviation 150.

Annual unit holding cost, H = 0.25×$20/unit/year = $5.00 / unit /year.

Review period, T = 2 weeks and replenishment leadtime, L = 1 week.


  1. Average weekly demand, R = 7×1000 = 7,000; weekly standard deviation of demand =

Standard deviation of demand during review period and replenishment leadtime



For a 98% service level z = NORMSINV (0.98) = 2.054 and safety stock



OUL = R×(Tr+L) + Isafety=7000×3+1413=22,413 units

Average order quantity, Q = R× Tr = 14,000 and therefore cycle stock = 14,000/2 = 7,000.

Average inventory, I = Q/2 + Isafety=7,000+1,413 = 8,413.

Total average annual holding cost = H×I = 5.00 × 8,413 = $42,065 per unit per year.

  1. If review period, T, is reduced 1 week, then,

Standard deviation of demand during review period and replenishment leadtime



For a 98% service level z = NORMSINV (0.98) = 2.054 and safety stock



OUL = R×(Tr+L) + Isafety=7000×2+1154=15,154 units

Average order quantity, Q = R×Tr = 7,000 and therefore cycle stock = 7,000/2 = 3,500.

Average inventory, I = Q/2 + Isafety=3,500+1,154 = 4,654.

Total average annual holding cost = H×I = 5.00 × 4,654 = $23,720 per unit per year.

Total savings in holding costs =$ 42,065 – $2327 = $18,795 per year.

Of course, now we order twice as frequently. So any associated costs related to placing orders needs to be balanced off against the savings in inventory holding costs of a shorter review period.


Problem 7.11

[Same data as in Problem 7.8 but with periodic review]

Review period length Tr = 2 weeks

[a] Assume 50 sales weeks/year.

H = $10 * 25%/year = $2.5/year

R = 10,000 /week = 500,000/year

Review period, Tr = 2 weeks

So, average order quantity Q = R×Tr = 10,000×2 = 20,000 units

The replenishment lead time (L) = 1 week.

Standard deviation of demand during review period and lead time at each store.



Safety stock at each store for 95% level of service (Isafety) = 1.65 x 3464.1 = 5,716.

Order Upto level (OUL) = 30,000 + 5,716 = 35,716.

Average inventory across four stores

= 4 x (Isafety + Q/2) = 4*(5,716+10,000) = 62,864

Annual holding cost for all four stores = $2.5 * 62,864 = $157,160.

Average time unit spends in store (T) = (5,716+10,000)/10,000 = 1.57 weeks.


[b] To compute the optimal order quantity at centralized store observe that this store

faces a cumulative average weekly demand = 4 x 10,000 = 40,000.


Standard deviation of demand during lead time at central store ()

= = 4,000.

Assume review period and lead time remain same as before at 2 weeks and 1 week respectively.

So, average order quantity Q = R×Tr = 10,000×2 = 20,000 units

Standard deviation of demand during review period and lead time at central store.



Safety stock at central store for 95% level of service (Isafety) = 1.65 x 6928.3 = 11,432.

Order Upto level (OUL) = 30,000 + 11432 = 41,432.

Average inventory at central store = (Isafety + Q/2) = 11,432+10,000 = 21,432

Annual holding cost at central stores = $2.5 * 21,432 = $53,580.

Average time unit spends in store (T) = (11,321+10,000)/40,000 = 0.533 week .


7.4 Test Bank

[1] Located in central France, Laurence Garreau’s firm supplies a particular subassembly to the automobile industry. The subassemblies are made in three different shapes (for different car models). These subassemblies need an electric motor, and Laurence is trying to find a supplier for these motors. The specifications of the electric motor required for the three subassemblies are almost identical; however the motors need to be of slightly different shapes to fit into the subassembly. Assume that the firm works for 300 days in a year and the daily requirement of each type of motor is normally distributed with a mean of 200 units and a standard deviation of 50 units. Laurence intends to hold safety stock to meet a 95% service level. The annual holding cost of the motor is 25% of the price of the motor. Also, by convention, the French firm pays for the motors the moment they are shipped.


Laurence has received two quotations. The first is from a firm in Asia. The unit price per motor is 50 Euros. In addition, the transportation cost per unit is 5 Euros. The transit time from Asia is 25 days. The second quotation is from a company in North America. This firm offers to price the motors at 60 Euros/unit, and a transportation cost of 7.5 Euros. It will take only 10 days to deliver the motors.


Evaluate each of the two proposals to determine the most economical alternative. What is your recommendation? (Be specific and show all of your calculations.) (10 points)


Answer:

Daily demand is normally distributed with mean = 200, and standard deviation = 50

For 95% service level, z=1.65

ASIA:

Unit cost, C=50 euros,

Holding cost per unit per year, H = 25% of C = 0.25 × 50 = 12.50 euros

Lead time (LT) = 25 days

Safety stock = z × std dev of demand during lead time = z × sqrt(LT) ×std dev of demand

= 1.65 × sqrt(25) × 50 = 412.5

Since the French firm pays for the motors the moment they are shipped, we also need to consider holding costs for the pipeline stock.

With mean daily demand = 200 units, and lead time = 25 days,

Pipeline stock = 200×25 = 5,000 units

Total inventory = 412.5 + 5,000 = 5,412.5 units

Cost of holding inventory = 12.50 × 5,412.5 = 67,656.25 euros

Unit transportation cost = 5 euros

Total annual transportation cost = annual demand × unit transportation cost

= 200 units/day × 300 days/year × 5 euros = 300,000 euros

Total cost of inventory holding and transportation

= 300,000 + 67,656.25 = 367,656.25 euros

This is the cost for each type of motor. Since there are three slightly different types of motors being shipped, total cost = 3×367,656.25 = 1,102,968.75 euros


NORTH AMERICA

Unit cost, C=60 euros,

Holding cost per unit per year, H = 25% of C = 0.25× 60 = 15 euros

Lead time (LT) = 10 days

Safety stock = z × std dev of demand during lead time = z× sqrt(LT) ×std dev of demand

= 1.65 × sqrt(10) × 50 = 260.88 units

Since the French firm pays for the motors the moment they are shipped, we also need to consider holding costs for the pipeline stock.

With mean daily demand = 200 units, and lead time = 10 days,

Pipeline stock = 200×10 = 2,000 units

Total inventory = 260.88 + 2,000 = 2,260.88 units

Cost of holding inventory = 15 × 2,260.88 = 33,913.20 euros

Unit transportation cost = 7.5 euros

Total annual transportation cost = annual demand × unit transportation cost

= 200 units/day × 300 days/year × 7.5 euros = 450,000 euros

Total cost of inventory holding and transportation

= 450,000 + 33,913.20 = 483,913.20 euros

This is the cost for each type of motor. Since there are three slightly different types of motors being shipped, total cost = 3×483,913.20 = 1,451,739.60 euros

So the North American firm’s quote is more expensive. Go with the Asian firm.


[2] A fashion company decides to market denim jeans in unusual colors: lightning green, electric orange and shocking red. The company buys denim and dyes it in each of the three colors. The next step is producing the jeans - which involves cutting, sewing, etc. - in each of the three colors. In year 1, the company expected demand for each product to be approximately the same, and produced 100,000 units of each product. However, the electric orange jeans were a runaway success (demand was 250,000), while the other two colors sold only 25,000 units each. It is unclear which color will catch on next year, and the company wants to avoid facing the same problem. As a result they decide to reverse their operations, producing basic white denim jeans first, and then dyeing them in different colors. Assume that sufficient capacity exists in both operations, and the dyeing operation is a very simple and quick process. What specific advantages might the company derive from new “operations reversal” process? Does it help the company deal with the problem they faced last year? (5 points)


(Ideal) Answer:

Since the dyeing operation is simple and quick, the company can delay this step till better information is available – and still respond quickly to customer demand. In the current setup, the more complex and long operation (jean production) takes place after jeans have already been dyed in different colors, so it is harder to respond to demand fluctuations quickly. Thus, the company would have had to stock large amounts of inventory of each product (different colors) in order to protect itself from demand uncertainty. The "operations reversal" essentially delays differentiation of the product and improves the company’s ability to respond fast to customer demand, and allows the company to pool their inventory (one basic type of jeans rather than multiple colors).


[3] Daily demand for the ice creams at I-Scream parlor is normally distributed with a mean of 160 quarts and a standard deviation of 100 quarts. The owner has the ice cream supplied by a wholesaler who charges $2 per quart. The wholesaler charges a $400 delivery charge independent of order size. It takes 4 days for an order to be supplied. The opportunity cost of capital to I-Scream is estimated to be 25% per year. Assume 360 days in the year. [Show all work]


(a) The optimal order size of each order is:

  1. 13576 quarts

  2. 9600 quarts

  3. 6788 quarts

  4. 716 quarts

  5. 506 quarts


(b) The owner would like to ensure no stock-outs in 95% of the cycles. The optimal safety stock the store should have is:


  1. 660 quarts

  2. 528 quarts

  3. 330 quarts

  4. 264 quarts

  5. 165 quarts


(c) Currently the owner orders 4000 quarts of ice cream when they have 1680 quarts on hand. The average time spent by a quart of ice cream at the parlor is:


  1. 39.6 days

  2. 20 days

  3. 19 days

  4. 15 days

  5. 12.5 days



[4] Two key components of Benetton’s marketing strategy include (i) Higher variety than competition (ii) Lower price than competition. Clearly these components of their strategy are in conflict. List two operational actions that allow Benetton to support these objectives of the marketing strategy and execute them successfully.


Answer:

  1. extensive use of subcontractors pick up variability at lower cost




  1. fast manufacturing flexibility (postponement) allows Benetton to respond to demand changes so that high variety can be offered at relatively low inventory cost.




  1. regional agents pool demand over the many small stores in a region which results in high service level and high variety at relatively low inventory cost.


[5] Big Old Tires currently owns and operates four warehouses in the Chicago area. Assume that each warehouse serves a geographic region. Weekly demands in any region are normally distributed with a mean of 200 tires and a standard deviation of 100 tires. Orders from suppliers take 4 weeks to be delivered. Big Old Tires is contemplating a consolidation of warehouses from four to one. Upon consolidation, the wholesaler carries the same safety stock in one warehouse as previously in four warehouses combined. The service level (in terms of probability of not stocking out in a cycle) s/he can provide from this single warehouse will be [Give 1-line explanation]:


  1. Higher than before.

  2. Same as before.

  3. Lower than before.


Because: More safety inventory that what is optimal for original service level.


[6] Hewlett-Packard produces deskjet printers for worldwide demand at its Vancouver, WA facility. It is observed that the monthly demand for type AB printers at the European DC averages 15,830 with a standard deviation of 5,624.


  1. The fixed cost of transportation and ordering is $100,000. Each printer costs $100 and HP has a holding cost of 20%. The optimal order quantity for the European DC is




  1. About 43,585

  2. About 12,582

  3. About 125,817

  4. About 435,844

  5. None of the above



EOQ = =43,585


  1. If the delivery lead time from the Vancouver factory is 6 months, how much safety stock of the AB printers should the European DC carry to provide a cycle service level (frequency of no stock-out) of 90%?




  1. About 7,207

  2. About 43,245

  3. About 17,655

  4. About 108,756

  5. None of the above


Safety stock = 1.2816×5624× sqrt(6) = 17,655


(c) HP is considering sending the printers by airfreight instead of sending them by sea. Airfreight will cost an extra $5 per printer but will reduce the delivery lead-time to 1 month. Identify the tradeoff that HP must consider when deciding whether to go with the airfreight option. What would you recommend? Why? Quantify the costs and benefits of the change.


Safety stock decreases by sqrt(6) to 17,655/sqrt(6) = 7,207. This is a reduction of 10,447, which is worth $20/unit,yr x 10,447 = $208,940/yr. Transport costs increases by 15,830units/mo × 12mo/yr × $5/unit = $949,800/yr. Clearly it does not pay to air freight.


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