Solutions for Chapter 6

Problem 6.1 The data in the question is: flow rate R = 50,000 parts/yr, fixed ordering cost S = $800, purchasing cost C = $4/part, and cost of capital r = 20%/yr. Thus, the annual unit holding cost is H = rC = $0.8/yr. The economic order quantity tells us to purchase each time

a) Q = √ 2 RSH

=√ 2×50 ,000×8000 . 8 = 10,000 units.

b) Order R/Q = 5 times per year.

Problem 6.2BIM Computers: Assume 8 working hours per day.

We know Q = 4 wks supply = 1,600 units; R = 400 units/wk = 20,000 units/yr; purchase cost per unit C = $1250*80% = $1,000. Thus, holding cost H = rC = 20%/year × $1,000 = $200/yr. Switch over or setup cost S = $2,000 + (1/2hr×$1,500/day×1day/8hr)= $2,093.75. Thus, # of setups per year = R/Q = 20,000 units/yr / 1600 units/setup = 12.5 setups/yr. Thus,

Annual setup cost = (R/Q) * S = 12.5 setups/yr × $2,093.75/setup = $26,172/yr. Annual Purchasing Cost = R*C = 20,000 units/yr × $1,000/unit = $ 20 M/yr. Annual Holding Cost = (Q/2) ×H = 800 × $200/yr = $160,000/yr. Thus, total annual production and inventory cost = $20,186,172.

EOQ = = 647 units.

number of setups = R/Q = 20,000 /647 = 30.91. Thus, annual setup cost = 30.91setups/yr × $2,093.75/setup = $64,718/yr.

annual holding cost = (Q/2) × H = 323.5 × $200/yr = $64,700/yr (notice that at optimal EOQ annual holding cost equal setup costs)

annual purchasing cost remains $20M/yr The resulting annual savings equals $20,186,172 - $20,129,418 = $56,754.

Problem 6.3Victor's data: flow unit = one dress, flow rate R = 30 units/wk, purchase cost C = $150/unit, order lead time L = 2 weeks, fixed order cost S = $225, cost of capital r = 20%/yr. Victor currently orders ten weeks supply at a time, hence Q = 10wks × 30 units/wk = 300 units.

a. Costs for Victor's current inventory management:

Annual variable ordering (purchasing) cost = RC = $150/unit × 30 units/wk × 52 wks/yr = $234,000/yr.

Annual fixed ordering (setups) cost = (# of orders/yr) S = (R/Q) S = (30×52/yr/300) $225 = $1,170/yr.

Annual holding cost = H (Q/2) = (rC) (Q/2) = $30/yr × 150 = $4,500/yr. Total annual costs = $239,670.

b. To minimize costs, Victor should order in batches of

Q* = EOQ = √ 2 RSH

=√ 2×30×52×22530 = 153 units.

Thus, he should place an order for 153 units two weeks before he expects to run out. That is, whenever current inventory drops to R×L = 30 units/wk × 2 wks = 60 units, which is the re-order point.

His annual cost will be

RC + √2RSH=√2×30×52×225×30 + $234,000 = $4,589 + $234,000 = $238,589.

c. Inventory turns = R/I, where average inventory I = Q/2 with cycle stock only. Current policy: turns = R/(Q/2)=2 × R/Q = 2 × 30units/wk / 300units = 2/10week = 52 × 2/10

per year = 10.4 times per year. Proposed policy: Q is roughly halved, so turns roughly double to 20.4 times per year.

Problem 6.4The retailer: Current fixed costs, S1 = $1000. Current optimal lot size Q1 = 400. New, desired lot size Q2 = 50. We must find the fixed cost S2 at which Q2 is optimal. Since Q1 is optimal for S1, we have

Q1 = 400 = √ 2 RS1

H=√ 2×R×1000

H . So, R/H = 160000/2000 = 80.

Now,

Q2 = 50 = √ 2 RS2

H ,

or S2 = 502 /(2×80) = 15.625. So the retailer should try to reduce her fixed costs to $15.625.

Problem 6.5Major Airlines: This question illustrates the basic tradeoff between fixed and variable costs in a service industry; thus the concepts of EOQ discussed in class in the context of inventory management are much more generic.

The process view here is illuminating and it goes as follows: flow unit = one flight attendant (FA). The process transforms an input (= "un-trained" FA) into an output (= "quitted" FA). The sequence of activities is: undergo training for 6 weeks, go on vacation for one week, wait in a buffer of "trained, but not assigned FA" until being assigned, serve as a FA on flights, and finally quit the job.

UntrainedFA Training Vacation Pool of

trained FAsServe on flights

QuittedFA

T = 6 wks 1 wks ? wks 2 years

R

The question asks for the tradeoff between training costs (higher class size is preferred) versus 'holding costs' in the buffer (smaller class size -> fewer attendants waiting in buffer is preferred).

(a) Flow rate R = 1000 every two years = 500 attendants per year = 10 per week. Fixed costs of training involves hiring ten instructors and support personnel for 6 weeks.

Thus, fixed costs of training S = 10 × ($220+$80) × 6 weeks = $18,000 per training session. Annual holding cost is the cost incurred to hold one flow unit (FA) in the buffer for one year:

H = $500 per month × 12 = $6,000 / person / year. Thus, Economic Class Size (EOQ) = 54.77 or 55 per class. Thus, we should run R/Q = 500 / 55

= 9.09 classes per year Per person variable cost of training is the stipend paid for 6 weeks of training + stipend for a

week of vacation = $500/mo. perperson × 7 wk × 12 mo/yr / 50 wk/yr = $840 per person. Notice that the annual variable cost is constant $840/person × 500 person/yr = $420,000/yr regardless of the class size.

Total Annual Cost = Fixed Costs of Training + Variable Costs of Training + Holding Costs = ($18,000 × 9.09) + ($840 × 500) + (55/2)($6000) = $748,636.36 per year.

Time Between starting consecutive classes (say, T) = Q/R = 5.5 weeks. Thus, we will have two classes overlap for a 1/2 week (and thus we need two sets of trainers and training class rooms). The inventory-time diagram looks as follows (assuming for simplicity that we start the training process at time 0):

t (weeks)0

I (in training)

5.56

55

110

t (weeks)0

I (on vacation)

6 7

55

t (weeks)0

I (in buffer)

6 7

55

Class 1 Class 2 Class 3

Class 1 Class 2 Class 3

Class 1 Class 2 Class 3

(b): This part of the question illustrates the following: Often, in reality, people wish to adopt policies that are simple (e.g., starting training every 6 weeks is simpler than trying to track the exact days to start training when subsequent trainings start every 5.5 weeks. But what is the implication of deviation from the optimal? In this case, quite small. This is because the optimal cost structure near the (optimal) EOQ is quite flat. Thus any solution close to optimality will suffice.

If time between classes (T = Q/R ) has to be 6 weeks, then Q = TR = 6 wks × 10 attendants/wk = 60 attendants.

Total Cost of this policy = ($18,000)(500/60) + ($840)(500) + (60/2)($6000) = $750,000 per year.

Problem 6.6

Fixed cost of filling an ATM m/c, S = $100.

To estimate demand, observe that the average size of each transaction = $80. With 150 transactions per week, annual demand R is estimated to be = 150×52×80 = 624,000.

With cost of money of 10%, unit holding cost, H = $0.10 / year

Then, the economic quantity to place in the ATM machine is given by the EOQ formula:

The number of times the ATM needs to be filled = R/Q = 624000/35327 = 17.66 per year.

Problem 6.7

The annual demand, R = 150,000 lbs/yr. The purchase price per lb is $1.50. However the shipping cost exhibits a quantity discount model. The holding cost per year is then 15% of the sum of the purchase and shipping cost. The administrative costs of placing an order = $50/order.

(a) In addition, rental cost of the forklift truck adds to the fixed cost giving a total fixed cost, S = 50+350 = $400/order. We can use a spreadsheet model as shown in Table TN 6.1. The optimal order quantity = 22,000 lbs with an annual cost of $249,916.77.

(b) If GC buys a forklift and builds a new ramp, then the per-transaction fixed cost will simply be the administrative cost of $50 per order. The economic order quantity and annual operating costs of this option is shown in Table TN 6.2. The economic order quantity is 15000 lbs. with an annual operating cost = $246,833.75. The annual savings = 249,916.77 - 246,833.75 = $3,083.02. The net present value of cost savings (over 5 years) with cost of capital of 15% = $10,334.76. Assuming a useful life of 5 years for the forklift and ramp, an investment of less than $10,334 generates a positive NPV.

Problem 6.8

Changeover time = 4hrs resulting in a fixed cost, S = 4 × 250 = $1,000.

Annual demand, R = 1000/mo × 12 = 12,000 units / yr.

Unit cost, C = 100

Holding cost = $25 / unit / yr.

a) The optimal production batch size is

b) To reduce batch size by a factor of 4, the setup cost needs to be reduced by a factor of (4)2 = 16. That is, S should reduce to 1000/16 = $62.5. This reduction can be achieved by reducing the changeover time or the cost per unit time during changeovers.

Problem 6.9

a) From the EOQ formula, observe that the order quantity is proportional to the square root of annual demand (R). Since cycle inventory is half of the order quantity, it too is proportional to R. Since HP motors has a higher R, the cycle inventory for HP motors is also higher.

b) Average time spent by a motor T = Icycle / R. Since Icycle is proportional to , T is proportional to

1/ . Therefore time spent by a HP motor is less than the time spent by an LP motor.

Problem 6.10

Each retail outlet faces an annual demand, R = 4000/wk × 50 = 200,000 per year. The unit cost of the item, C = $200 / unit. The fixed order cost, S = $900. The unit holding cost per year, H = 20 % × 200 = $40 / unit / year.

a) The optimal order quantity for each outlet

with a cycle inventory of 1500 units. The total cycle inventory across all four outlets equals 6000 units.

b) With centralization of purchasing the fixed order cost, S = $1800. The centralized order quantity is then,

and a cycle inventory of 4242.5 units.

Solutions for Chapter 7

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 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 = 505 560 605 690

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) = 20,000Average inventory (I) = Isafety + Q/2 = 200 + (20,000/2) = 10,200.H = rC = 0.25 x 9.99 = 2.4975 2.50

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 = 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.

Thus, the economic order quantity is

Q = √ 2 SR

H=√ 2×100×50 ,000

2 .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 is

Q¿

=√ 2 RSH =√ 2×1500×52×10000

2 .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 7.4

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) = p – c = $120 – $18 = $102. The marginal cost of stocking an additional unit (MC) = c – v = $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 = $125Cost of understocking by 1 ton = additional cost of using outside trash pick up = MB = $650-$125 =

$525Thus: SL = Prob(RQ) = 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(RQ) = MB/(MB+MC) = 0.7. For uniform distribution between 0 and 20, Prob(RQ)=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/yearR = 10,000 /week = 500,000/yearS = $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 storefaces a cumulative average weekly demand = 4 x 10,000 = 40,000. This gives an annual demand of 2,000,000 units.

Q = EOQ = √ 2 RS

H=√ 2×2 ,000 ,000×1000

2 .5 = 40,000 units.

Standard deviation of demand during lead time at central store (LTD)= √4×2000 = 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

(a) Given that each outlet orders independently and gets its own delivery, the optimal order size at each outlet issqrt(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.

(b) 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

(c) 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.

a) 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.

b) 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 ()

= √4×2000 = 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 .