MBPF Ch7 solutions. Last updated: March 12, 2013
CHAPTER 7: MANAGING FLOW VARIABILITY: SAFETY
INVENTORY
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.
[b]
So the cycle service level is 78.81%.
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
z=
Isafety =
ROP =
80%
0.842
105
505
90%
1.282
160
560
95%
1.645
205
605
99%
2.326
290
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 ) = L R = 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.
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/year
R = 1,000 /week = 50,000/year
S = $100.
61
Thus, the economic order quantity is
2SR
2 ´100 ´ 50, 000
=
= 4, 000, 000 = 2, 000
H
2.50
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.
Q=
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 
(b)
2RS
2  1500  52  10000

 24,980
H
2.5
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 = $15,600
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
MB
30
Prob(R  Q*) =

 0.857 .
 MB  MC  30  5
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
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
Cumulative
Prob.
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 =
2RS
2  500,000  1000 = 20,000 units.

H
2.5
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 =
2RS
2  2,000,000  1000 = 40,000 units.

H
2.5
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 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.
(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 =
sR =
7 ´ 150 = 397
Standard deviation of demand during review period and replenishment leadtime
s R LT D = ( T r + L )s R =
3 ´ 397 = 688
For a 98% service level z = NORMSINV (0.98) = 2.054 and safety stock
I safety = z ´ s RLT D = 2.054 ´ 688 = 1413
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
s R LT D = ( T r + L )s R =
2 ´ 397 = 562
For a 98% service level z = NORMSINV (0.98) = 2.054 and safety stock
I safety = z ´ s RLT D = 2.054 ´ 562 = 1154
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.
s R LT D = ( T r + L )s R =
3 ´ 2000 = 3464.1
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.
s R LT D = ( T r + L )s R =
3 ´ 4000 = 6928.3
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 .