1
Case Questions (30%)
1.What factors influence the decision regarding the opening and closing of stores? Why does 7Eleven choose to have a preponderance of its stores in a particular location?
The decision of opening or closing stores mainly depends on market environment in the region,
such as number of competitors, transportation condition and incomes and population of residents.
7-Eleven applies a market-dominance strategy, which has the following advantages: boosts distribution efficiency, improves brand awareness, increases system efficiency, enhances the efficiency of
franchise support services, improves advertising effectiveness and prevents competitors entrance
into the dominant area.
2.Why has 7-Eleven chosen off-site preparation of fresh foods and subsequent delivery to stores?
7-Eleven centralizes the food production in order to gain economic of scale, guarantee the quality
of fresh foods and reduce transportation costs. Moreover, 7-Eleven can fully utilize the capacity,
appropriately make production plan and aggregate demand for reducing uncertainty.
Another reason is, 7-Eleven has a very effecient distribution system. It has the ability to
distribute the products effeciently to the retailers. For example, every day, it can visit each of the
retailer store several times. By doing this, the service level can be gauranteed while the effeciency
can also be achieved.
3. Why does 7-Eleven discourage direct store delivery from vendors and make an effort to move
all the products through combined DCs?
7-Eleven selects a cross-docking distribution mode. Since all the demands can be aggregated,
therefore, by doing this, the inventory level can be reduced. Also note, in the DC, there is no
inventory kept. The inventory cost is reduced. Since those stores are densely located near the
combined DCs, 7-Eleven can gain the economy of scale for the outbound transportation too. From
the case, we know they achieve a high fill rate with high flexibility and efficiency. This is directly
connected to their location model—to have a preponderance of its stores in a particular location (this
explans why its location choice model can boost distribution efficiency). Because different stores
may had different order sizes, 7-11 use inventory-pooling and combine all the orders together. Just
imangine, without aggregation, if the vednor directly ship to the retailor store, what will happen?
The vendor has to visit many stores. For each store, the order size usually is very small. This
requires high handling and labor cost. By aggraging all the products from all the vendors, the fixed
labor and transporation cost can be shared among all the different products (the total handling
cost and transporation cost with aggrergation can be higher than handling and transporating single
product though). Since all the differnt products are combined in the same truck load, the good
thing is, the order size for each of the product canbe reduced, this is why 7-Eleven can offer short
2
replenishment cycle time for quick response.
4. Why does 7-Eleven combine fresh food shipments by temperature?
The main reason is, to reduce the order size and gain economy of scale in the transporation. This
is one specific example for joint replenishemnt. The good thing is, the fixed transporationcost and
other related cost is reduced. For example, 7-Eleven reduce the number of vehicles required for daily
delivery. The bad thing, there can be some extra handling cost. (Remeber the joint reprenishment
model we present in the class?) Without joint replenishement, for each store, it may visit much
less frequently, like one week per visit. With crossing docking and the joint replenishment, each
store can be visit several times per day. This is the reason that 7-Eleven can make rapid deliver of
a variety of fresh foods.
5. What point of sales data does 7-Eleven gather and what information is made available to store
managers to assist then in their ordering and merchandizing decisions? How did the Information
System at 7-Eleven enable them to become a ‘market differentiator’ ?
POS data includes the information of the item, the time of sales and the age and sex of the
customer. Information analysis of POS data: Sale analysis of product categories over time, SKU
analysis, analysis of waste or disposal, ten-week sales trends by SKU, ten-day sales trends by SKU,
sales trends for new products, sales analysis by day and time, list of slow moving items, sales analysis
by product categories, analysis of sales and number of customers over time, contribution of product
to sections in store display, sales growth by product categories and evaluation of merchandizing. The
information system helps 7-Eleven stores to better match supply with demand, revise merchandize
mix and develop appropriate new products.
By analyzing those data, 7-Eleven can have a better picture of the demand. Therefore, it can
reduce the operating cost such as invnetory ocst. By closely monitoring the fast move item and slow
moving item, it can further explore how to use the retail space which is critical to the convinient
store like 7-Eleven. For example, it can stop displaying and selling the very slow-moving product
(s). At the same time, they can also match the demand with supply better and increases the
customer satisfiction and customer loyalty to 7-Eelven.
From the above discussion, we know that the role of information systems is critical within the
supply chain. In the 7-Eleven case, management of the many Japan stores would be impossible
without the information system. Humans could not physically collect, analyze and incorporate this
data on an hourly basis as many of the programs 7-Eleven do. Immediate feedback of downstream
supply chain data back to the beginning of the supply chain, the moment the transaction occurs is
essential. The real-time feedback loop is the foundation to understanding the marketplace enabling
immediate product consumption analysis and effective change management process. Without such
an information system, a business would have no efficient means to streamline its customers pur-
3
chasing behaviors timely enough to respond to changes in that behavior. Nor would it be able to
predict or help optimize those behaviors to its benefit. When designed effectively, and integrated
with all aspects of the supply chain, information systems allow business’ to adapt quickly and efficiently to the changing needs and behaviors of their customers. Having a sophisticated information
system, 7-Eleven remains competitive and the leader of the market.
6. Does the 7-Eleven have the bullwhip effect? Give a general discussion of reducing the
bullwhip effect for those companies who have the bullwhip effect?
There is no bullwhip effect due to the sophiscated informationsystem. All teh demand inforamtion is available to the centralized system. Each store can palce order, however, the order
doesn’t give the false information to the distribution centor because the sales data is known to the
distributor.
We can reduce the bullwhip effect by following ways: good demand forecasting, reduce the order
size, stabilize he price, eleminating the gaming in shortage situation etc.. For detailed discussion,
please refer tothe lecture note.
2
Forecasting (30%)
For the forecasting, we need to know how to do forecast by various methods and how to evaluate
a method and finally how to choose the best one.
Demand: 8,000; 13,000; 23,000; 10,000; 18,000;24,000. Since the underlying model assumed for
both moving average and exponential smoothing is stationary, the one-step-ahead and multiplestep-ahead forecasts for moving averages and exponential smoothing are the same.
(1) : (7 points) Moving average forecast for the demand in period 7, 8 (N=3);
Month Demand Ft
1
8000
2
13000
3
23000
4
10000
14666.67
5
18000
15333.33
6
24000
17000
7
17333.33
8
17333.33
(2) :(6 points) Exponential method to forecast the demand in period 7,8 (α = 0.1); F1 = 8000.
4
Month
1
2
3
4
5
6
7
8
Demand
8000
13000
23000
10000
18000
24000
Ft
8000
8000
8500
9950
9955
10759.5
12083.55
12083.55
(3) :(8 points) Winter’s method to forecast the demand in period 7,8 (α = β = 0.1, γ =
0.05).Finding initial S0 and G0 by Moving average (N=3) plus linear regression Finding the Seasonal Factor by moving average (c−N+1 , c−N+2 , ..., c0 ). Update All the parameters until you get
c1 ,...,c6 ,S1 ,...,S6 and G1 ,...,G6 .
Using the above parameters to forecast the demand in period 7 and 8.
Month
1
2
3
4
5
6
Demand
8000
13000
23000
10000
18000
24000
MA(3)
Ratio
14666.67
15333.33
17000.00
17333.33
0.89
1.50
0.59
1.04
Sxy = 19333.33, Sxx = 20, S0 = a = 12700, G0 = b = Sxy /Sxx = 966.67
c−2 = 0.59, c−1 = (0.89 + 1.04)/2 = 0.96, c0 = 1.50
Gt
ct
Month
Demand St
Ft
1
8000
13660.00 966.00 0.588
2
13000
14514.17
954.82
0.959
3
23000
15455.42
953.46
1.499
4
10000
16468.37
959.41
0.589
5
18000
17561.81
972.81
0.962
6
24000
18281.79
947.53
1.490
7
11327.26
8
19417.58
(4) :(5 points) Suppose demand satisfies the additive model, that is to say Dt = L+tG+Ct +Rt ,
5
where Dt is the observed demand,Ct is seasonal factor,L + tG is linear trend and Rt is the random
term. Please develop ”Winter’s” method for this given additive model. Basically, for given α, β, γ,
(a) how to get the initial data S0 , G0 , C0 , C−1 , ..., C−N +1 , where N is the number of periods before
the seasonal factor repeats itself; (b) how to get St , Gt , ct and how to update them; (C) how to get
Ft,t+i based on St , Gt , Ct .
(a). We have to do the following steps.
Step 1. Draw a picture and identify the number of periods before the season repeates itself;
Step 2. Indentify the trend using moving average;
Step 3. Calculate {Di − Ti };
Step 4. Group and average.
St = α(Dt − ct−N ) + (1 − α)(St−1 + Gt−1 )
(b).
Gt = β(St − St−1 ) + (1 − β)Gt−1
ct = γ(Dt − St ) + (1 − γ)ct−N
(c).
Ft,t+i = St + i × Gi + Ct−i+N
3
Planning (20%)
A company transforms the raw material AAA into intermediate product BBB by machine EEE.
The machine FFF will transform intermediate product BBB into the final product CCC. Machine
EEE can produce 30 pounds pf product BBB, while machine FFF can produce 20 pounds of CCC
(finished product) per hour. Each pound of product BBB needs one pound raw material. While
each pound of finished product needs one pound of BBB.
While one machine EEE needs 3 operators and one FFF needs 2 operators. Currently, the
company has 14 EEE and 25 FFF. Within the planning horizon, no new machine will be bought.
During any month, the machine EEE or FFF can be idle. If the machine is idle, the associated
worker will be layoff. If the machine is idle, the only savings is the salary of associated workers.
However, layoff each worker costs 500$ (for both operators of machine EEE and FFF). When any
idle machine is brought online, the company needs to spend 400$ as set-up cost for machine EEE
and 300$ set-up cost for machine FFF (set-up cost is the fixed cost). At the same time, for each
new hire people, the company needs to spend 500$ training cost.
For each month, each worker can work 20 days and each day the worker can work 8 hours. Each
worker can work at most 10 hours overtime per month. The salary for all the workers is same. The
regular hour is 10$ per hour, while for the overtime, the salary is 15$ per hour.
There is no subcontracting company available.
The purchasing cost of product AAA is 100$ per 100 pound. There are enough product AAA
in the market so that the company needs not to stock those products. The un-used product BBB
will be stored in the company’s warehouse. For holding BBB, for each month, the holding cost is
6
2$ per 100 pound per month and will be charged by the end of each month. For holding product
CCC, the holding cost is 4$ per 100 pound per month.
All the unsatisfied orders will be backlogged.
The forecast demand for the next 6 months is as following: 30,000; 50,000; 90,000; 25,000;
60,000; 70,000.
Initially the company has 52 operator for EEE and 50 operators for FFF. The initial inventory
is 0 and the the ending inventory is also 0.
We will define the the related variables and then, you are required to get some of the cost
objective functions and fill in some constraints or explain some constraints.
First, let’s define the variables.
met =
number of machines EEE working in month t;,
mft
number of machines FFF working in month t;
=
sbt = back orders of product BBB at the end of month t;
sct = back orders for CCC at the end of month t;
Itb = inventory level of for BBB at the end of month t;
Itc = inventory level of CCC at the end of month t;
Dtc = demand for CCC in month t;
pct = production quantity for CCC in month t;
pbt = production quantity for BBB in month t;
Let = number of machines EEE that working in month t-1
and will be idle in month t;
Lft
= number of machines FFF that working in month t-1
and will be idle in month t
Hte = number of machine EEE that in period t-1 is idle and
will be working in month t;
Htf
= number of machine FFF that in period t-1 is idle
and will be working in month t;
Ote = overtime hours machine EEE in month t;
Otf
= overtime hours for machine FFF in month t;
2 points for each questions.
7
Objective Function : Minimizing the total cost
write to the following cost :
6
X
(1) : regular labor cost
t=1
(2) : layoff cost
6
X
t=1
(3) :
500 × (3 × Let + 2 × Lft )
new-hiring cost
6
X
t=1
or :
(4) :
new-hiring cost
overtime cost
1600 × (3 × met + 2 × mft )
[400 × Hte + 300 × Htf + 500 × (3 × Hte + 2 × Htf )]
6
X
t=1
6
X
t=1
500 × (3 × Hte + 2 × Htf )
15 × (3 × Ote + 2 × Otf )
Answer the following questions :
Part one :
write the constraints
(5) : met−1 − Let + Hte = met evolution of the working machines EEE
(the relationship between the number of working
in month t-1 and month t)
(6) : pct ≤ 20 × 20 × 8 × mft + 20 × Otf constrains on product quantity CCC
based on number of working machine FFF and overtime
Part two :
Explain and make decision
(7) : Hte ≤ 14 − met−1
Is this correct or not? Why?
Yes. The number of new working machine in Month t can not be
l arg er than the number of idle machine in Month t − 1.
c
c
(8) : It−1
+ pct = Dtc + St−1
+ Itc − Stc , Is this correct? Why?
Yes. The left side represnts the total supply available in Month t while
the right side represents the total demand in Month t. Namely, the supply
should be balanced with the demand.
b
(9) : pct ≤ It−1
+ pbt
Is this correct? Why?
Yes. Since each pound of CCC needs one pound of BBB, the production
amount of CCC in Month t can not exceed the amount of BBB available
8
in Month t.
Initial Condition :
(10) : meo = 14, mf0 = 25, I0c = I0b = sb0 = sc0 = 0 Is this correct,why?
Yes. There are 14 EEE and 25 FFF and no initial inventory and backorder.
All decision variables are non-negative.
4
Inventory Control (20%)
(1) (5 points) Estimating Cycle Service Level (Continuous Review Policy): D = 3,500/week(expected
demand); σD = 500 (standard deviation of the one week demand), L = 3 weeks (ordering leadtime);
Q = 10,000 (order quantity each time); ROP = 11,866 (re-order point).
Solution:
RL = 3 ×√
3500 = 10500
σL = 500 3 = 866.025
ROP = 11, 866
CSL = Fs ( ROPσL−RL ) = 0.943
(2) (5 points) Evaluating Safety Inventory Given CSL(Continuous Review Policy) :D = 3,500/week;
σD = 500, L = 3 weeks; Q = 10,000; CSL = 0.90.
Solution:
RL = 3 ×√
3500 = 10500
σL = 500 3 = 866.025
CSL = 0.90
ss = Fs−1 (CSL) × σL = 1.28 × 866.025 = 1109.857
(3) (4 points) (Newsvendor Model) The manager at Sportmart, a sport store, has to decide on
the number of skis to purchase for the winter season. Considering the past demand and weather
forecast, suppose the demand is normally distributed with mean 350 and a standard deviation of
σ = 100. Each item costs c = 100$ and the retail price is p=250,. Any unsold item can be disposed
at the end of the season for 85$. Assume it costs 5$ to hold a pair of skis in inventory for the
season. Evaluate the number of skis that the manager should order to minimize the cost.
Solution:
µ = 350, σ = 100
Cu = 250 − 100 = 150
Co = 100 + 5 − 85 = 20
Cu
Cu +Co = 0.88235
u
) × σ = 350 + 1.1868 × 100 = 468.68
Q∗ = µ + Fs−1 ( CuC+C
o
(4) (6 points) (EOQ for multiple product) Suppose Best Buy sells three models of computers,
the Litepor, Medpor and Heavypor. Annual demands for three products are DL =12,000 for Litepro,
DM =2000 for Medpor and DH =500 for Heavypro. Assume that each model costs best Buy 1000$.
A fixed transportation cost 4000$ is incurred each time an order is delivered. For each model
9
ordered and delivered on the same truck, an additional fixed cost 1000$ is incurred for receiving
and storage. Best Buy incurs a holding cost of 20% per year. Evaluate the lot sizes the Best
Buy should order by considering the following three options: (a) All the products are ordered and
delivered independent. (b) Aggregate all products each time they place an order. (c) Order jointly,
but to be selective about which models they include in each order.
Solution:
(a) Since all the products are ordered and delivered independent, we can just apply EOQ
model to obtain the solution.l
For
q
q
2DL SL
= 2×12000×5000
= 774.597 ' 775
q h
q 1000×0.2
2D
S
M M
Medpor, Q∗M =
= 2×2000×5000
= 316.228 ' 316
q h
q 1000×0.2
2DH SH
2×500×5000
∗
Heavypor, QH =
=
h
1000×0.2 = 158.114 ' 158
For Litepor, Q∗L =
For
(b) Aggregate all products each time they place an order.
The aggregated annual demand D = DL + DM + DH = 14500
The set-up cost per order S = 4000 + 3 × 1000 = 7000
The holding cost per
unit time h = 1000 × 0.2 = 200
q unit per
q
14500×200
n∗ = 2Dh
S =
2×7000 = 14.392
DL
12000
∗
QL = n∗ = 14.392 = 833.77 ' 834
2000
Q∗M = DnM
∗ = 14.392 = 138.96 ' 139
D
500
Q∗H = nH
∗ = 14.392 = 34.74 ' 35
(c) Order jointly, but to be selective about which models they include in each order.
Step 1. Finding the most frequently
q ordered
q product.
hDL
nL = 2SL = 200×12000
2×5000 = 15.49
nM =
q
hDM
q 2SM
hDH
2SH
=
q
200×2000
= 6.32
q 2×5000
= 200×500
2×5000 = 3.16
n = 15.49
Step 2. Identify the ratio of the orders for the other products to the most frequently ordered
nH =
product.
q
q
hDM
= 200×2000
= 14.14
q2sM
q 2×1000
hDH
200×500
nH = 2sH = 2×1000 = 7.07
mM = n/nM = 15.49
14.14 = 1.095, mM = 2
mH = n/nH = 15.49
7.07 = 2.19, mH = 3
nM =
Step 3. Compute the number of orders placed per year for the most frequently ordered
product.
n∗
=
r
200×(DL mL +DM mM +DH mH )
−1
−1
2×[4000+1000×(m−1
L +mM +mH )]
Step 4. Compute the number of orders for all the product.
nL = n∗ /mL = 17.32
nM = n∗ /mM = 8.66
nH = n∗ /mH = 5.77
Step 5. Get the order quantity of each product per order.
10
=
q
200×17500
2×(4000+1000×1.833)
= 17.32
L
Q∗L = D
nL = 692.82 ' 693
D
M
Q∗M = nM
= 230.94 ' 231
DH
∗
QH = nH = 86.60 ' 87
4.1
Optional Question (Bonus 10%)
(EOQ for multiple product) Considering two products A and B. Suppose the annual demand for
A is D1 and annual demand for B is D2 . The cost for A is c1 $ and the cost for B is c2 $. if any
order is placed, a fixed cost S incurs. If product A is ordered, Sa will also incur, if B is ordered,
Sb will incur (that is to say, if A and B is order, the total ordering cost will be S+Sa +Sb , if only A
is order, the ordering cost is S+Sa , if only B is ordered, the ordering cost is S+Sb ). Suppose the
holding cost for holding one year is h$. Suppos if A and B are ordered independently, A will be
ordered more frequently. Considering the following policy, every time when an order is placed, A
will be always ordered, while for product B, every k orders for A, B will be ordered once. That is
to say, suppose na is the order frequency for product A and nb is the order frequency for product
B. Then k=na /nb . Now determine the best k to minimize the total ordering and holding cost.
Solution:
Given k, the total cost is
a )D1
b D2
T C = 12 c1 hQa + (S+S
+ 12 c2 hQb + SQ
Qa
b
D1
2
na = Q
= kD
Qb = nb
a
D2
Qb = kQa D
1
Sb D1
D2
a )D1
1
T C = 12 c1 hQa + (S+S
+
Qa
2 c2 hkQa D1 + kQa
dT C
dQa
2
= 12 h(c1+ c2 k D
D1 ) −
Q∗a =
Q∗b
p
=
r
2
kD
D1
(S+Sa )D1
Q2a
2D1 (S+Sa +Sb /k)
h(c1 +c2 kD2 /D1 )
r
−
Sb D1
kQ2a
2D1 (S+Sa +Sb /k)
h(c1 +c2 kD2 /D1 )
T C(k) = 2h(c1 D1 + c2 kD2 )(S + Sa + Sb /k)
H(k) = (c1 D1 + c2 kD2 )(S + Sa + Sb /k)
dH(k)
c1 D1 Sb
dk = c2 D2 (S + Sa ) −
k2
2
d H(k)
2c1 D1 Sb
=
>
0
dk2
k3
Clearly, T C is convex in k.
k is determined by
k∗ =
q
dT C
dk
= 0.
c1 D1 Sb
c2 D2 (S+Sa )
If k∗ ≥ 1, since k must be an integer, we have two choice.
q
c1 D1 Sb
c2 D2 (S+Sa ) e
q
1 Sb
d c2 Dc12D
(S+Sa ) e
k1 = d
k2 =
−1
k ∗ = arg min{T C(k1 ), T C(k2 )};
If k∗ < 1, since k must be an integer, we have two choice.
11
q
k1 = d
q
k2 = d
c2 D2 (S+Sa )
c1 D1 Sb e
c2 D2 (S+Sa )
c1 D1 Sb e
−1
k ∗ = arg min{T C(k1 ), T C(k2 )}.
5
Appendix
s
hCi Di
2 (S + si )
n = max {ni }
Step 1. ni =
Step 2.
:
s
hCi Di
2si
= n/ni
ni =
mi
For the other products, we calculate the order frequency
mi = dmi e(for the most frequent one, the ratio is one)
Step 3. n∗
vP
u m
(Di mi Ci h)
u
´
= t ³i=1 P
2 S+
Step 4.ni = n∗ /mi
Di
Step 5.Qi =
ni
12
m si
i=1 mi