IE 1079/2079 Logistics and Supply-chain
management
Homework 5: Due Monday July 27
Dr. Louis Luangkesorn
July 23, 2009
1
Problem 1 - Hub location model
In Figure 1, there are three source nodes, A, B, C, four potential hubs 1, 2, 3, 4
and three destination nodes a, b, c. There are deliveries from each of the three
sources to each of the three destinations of Wij ß ∈ A, B, C; j ∈ a, b, c and the
total production at each of the source locations is Oi ß ∈ A, B, C
1. Write out constraint 12.14 (pg. 378) for the following nodes: A, 1, 4, c.
Do not use summations, but list out each term. Define all variables using
complete sentences.
1.1
Solution
Constraint 12.14 is:
X
X
X
Wij Zjk +
Ykli =
Ylki + Oi Zik
j∈V
l∈V
∀i, k ∈ V
l∈V
Where Wij is the flow from i to a destination j. Zjk represents the assignment of the node i to a candidate hub j. Ykli is the flow of items that originate
at i that travel from the hub at k to the hub at l. Oi is the total originating
flow at node i. It could be thought of as the total production at i. V is the set
of all nodes. i, j represent origins and destinations. k, l represent candidate hub
locations.
Constraint 12.14 only refers to hubs, so only 1 and 4 are considered. The
left hand side are flows going to other hubs and destinations while the right
hand side are flows coming from other hubs and origins.
1
a
A
1
2
B
b
4
3
C
c
Figure 1: Figure for problem 1
2
A
A
A
A
WAa Za1 + WAb Zb1 + WAc Zc1 + Y12
+ Y14
= Y21
+ Y41
+ OA ZA1
0+0+0+
A
Y12
WBa Za1 + WBb Zb1 + WBc Zc1 +
B
Y12
B
Y12
+
A
Y14
+
B
Y14
B
Y14
= 0 + 0 + OA ZA1
(1)
(2)
(3)
0+0+0+
+
=
B
Y21
+
B
Y41
+ OB ZB1
= 0 + 0 + OB ZB1
(4)
(5)
(6)
0+0+0+
C
Y12
C
Y12
WAa Za4 + WAb Zb4 + WAc Zc4 +
A
Y41
WCa Za1 + WCb Zb1 + WCc Zc1 +
+
C
Y14
C
Y14
+
A
Y43
+
=
C
Y21
+
C
Y41
+ OC ZC1
= 0 + 0 + OC ZC1
(7)
(8)
(9)
=
WAa Za4 + WAb Zb4 + WAc Zc4 + 0 + 0 =
A
Y34
A
Y34
+
+
A
Y14
A
Y14
+ OA ZA4
(10)
+0
(11)
(12)
WBa Za4 + WBb Zb4 + WBc Zc4 +
B
Y41
+
B
Y43
=
WBa Za4 + WBb Zb4 + WBc Zc4 + 0 + 0 =
B
Y34
B
Y34
+
+
B
Y14
B
Y14
+ OB ZB4
(13)
+0
(14)
(15)
WCa Za4 + WCb Zb4 + WCc Zc4 +
A
Y41
+
A
Y43
=
WCa Za4 + WCb Zb4 + WCc Zc4 + 0 + 0 =
C
Y34
C
Y34
+
+
C
Y14
C
Y14
+ OC ZC4
(16)
+0
(17)
(18)
2
Problem 2 - Stochastic Demands
In Figure 2, two (2) companies at node D, three (3) fire companies are located
at node E, and one (1) fire company is located at node F . The probability that
any given fire company is busy at any point in time is 0.3. Fire companies can
cover up to 12 units of distance.
1. Determine the probability for each node of finding an available fire company that can cover it.
2.1
Solution
First, determine which nodes are covered by which facilities, then add the number of times a node is covered. Next, use 1 − pk , which is equivalent to 1 minus
the probability that all the covering nodes are busy.
So, if you replace the node names with A:1, B:2, C:3, D:4, E:5, F:6, G:7,
H:8, then we get:
> companies = c(0, 0, 0, 2, 3, 1, 0, 0)
> coverage = seq(0, length = 8) * 0
3
A
8
B
8
5
E
13
9
6
7
6
8
11
14
C
12
4
D
G
12
F
9
H
20
Figure 2: Figure for problem 2
>
>
+
>
+
>
>
+
>
+
>
>
>
coverage[1] = companies[1] + companies[2]
coverage[2] = companies[1] + companies[2]
companies[5] + companies[6]
coverage[3] = companies[1] + companies[2]
companies[5] + companies[6]
coverage[4] = companies[1] + companies[2]
coverage[5] = companies[2] + companies[3]
companies[6] + companies[7]
coverage[6] = companies[3] + companies[4]
companies[8]
coverage[7] = companies[5] + companies[7]
coverage[8] = companies[6] + companies[7]
coverage
+ companies[3] + companies[4]
+ companies[3] + companies[4] +
+ companies[3] + companies[4] +
+ companies[3] + companies[4]
+ companies[4] + companies[5] +
+ companies[5] + companies[6] +
+ companies[8]
+ companies[8]
[1] 2 6 6 2 6 6 3 1
> pbusy = 0.3
> pcover = 1 - pbusy^coverage
> pcover
[1] 0.910000 0.999271 0.999271 0.910000 0.999271 0.999271 0.973000 0.700000
where coverage is the number of companies covering a node and pcover is
the probability at least one company that is covering a node is available.
4
3
Problem 3 - Multi-echelon and the Bullwhip
effect
Take the spreadsheet model developed in class (and available on the course
website) and modify it to account for sharing point of sale data with the supplier.
Do this by only having inventory at the retailer and supplier (so the wholesaler
is only a pass through distribution center with no inventory). The retailer will
then order directly from the supplier with a leadtime of L1 + L2 . Set the initial
(s, S) policy Q, r accordingly (i.e. r should be guessed at accounting for a lead
time of L1 + L2 ). One way to do this would be to set this to the value of Q and
r established on day 21 of the model (assuming you are using 20 day moving
average forcasts).
1. Turn in charts and tables corresponding to the lecture. (page 9 of slides
on website)
2. In one paragraph, compare results from the model in the lecture with
results from the model you develop.
4
Problem 4 - Competitive location models
Laura feels like having dinner at an Italian restaurant. She has the following
options:
ˆ The Olive Garden: Less traditional than she would like but still a decent
place to have a meal. The food is average but the prices are fair. The
bathrooms are usually pretty dirty and the tables usually have the film left
over from the dirty rag they were wiped down with, however the service
is excellent in every respect. She lives just west of the city and the Olive
garden is roughly 10 minutes southwest of her apartment.
ˆ Pasta Palace: This epitomizes want-to-be Italian chain restaurants. They
pride themselves on being quick and cheap and succeed in doing this. The
sauce is like your grandmother’s worst nightmare which wouldn’t be so
bad if the meatballs weren’t made of rubber and if the pasta didn’t have
the consistency of oatmeal. In addition to the service being speedy, it is
also punctuated with the smirk and rolling eyes of a sixteen year old kid
working weekends and nights after school. Surprisingly, the place is very
clean. The Pasta Palace is 15 minutes northwest of her apartment.
ˆ Vito’s Kitchen: Walking into this place, you feel like you have just fallen
into a Sicilian dining room. They had it all, from the hanging salami tothe
dust on the imported wine bottles. The sauce has the kind of flavor you
can only get from a fresh herb garden and garlic that was caramelized to
perfection. It’s a small family owned place. The service is with a smile,
but that’s about it. They lack the sense of urgency and attentiveness
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that a corporate trained serve would have. They are more concernedwith
the food and the service suffer accordingly. The bathrooms are always
spotless even though the tables are sticky sometimes. The meals are very
expensive, but it is only 5 minutes west of Laura’s apartment.
1. Derive the utility function based on the following attributes (Note that
to answer this question you have to determine if an attribute is additive
or multiplicative): Quality (Q), Service (S), Atmosphere (A), Cleanliness
(C), Distance (Dij ), Price (P).
2. Assign weights to all of the attributes and solve the utility function to
determine the choice that Laura will most likely make. (Use ranking when
necessary)
4.1
Solution
As discussed in lecture, for restaurants Cleanliness is multiplicative as well as
Distance (according to inverse square law). the other qualities are additive. So
the utility function is
(Q + S + A + P ) ∗ C
d2ij
Using the very opinionated descriptions provided, you can develop a table
for the three restaurant attributes by ranking them. (Note that distance is the
actual distance)
Restaurant
Q S A C dij P Utility
The Olive Garden 2 3 2 1 10 2
0.090
Pasta Palace
1 2 1 3 15 3
0.093
Vito’s Kitchen
3 1 3 2
5
1
0.640
So, given the beliefs about the attributes of quality, service, etc., Laura would
choose the restaurant with the highest utility, or Vito’s Kitchen.
5
Problem 5 - Reverse logistics
Using the Reverse Logistics Network Framework slide as a guide, draw a reverse
flow diagram for the example problem presented in the lecture.
Include all the key facility types
ˆ Collection zones
ˆ Customers outlets
ˆ Facility locations (product storage and remanufacturing are co-located)
Below the diagram, identify the additional costs that are associated with the
different facility types In addition, identify the additional costs associated with
each of the facilities:
6
A
G
I
D
J
F
B
H
E
K
C
L
Figure 3: Figure for problem 5
ˆ Holding costs
ˆ Remanufacturing costs
ˆ Facility location costs
ˆ Path transportation costs
5.1
Solution
ˆ Collection zones - H
ˆ Customers outlets - I, J, K, L
ˆ Facility locations (product storage and remanufacturing are co-located)
H, F, C Note that C would usually be a third party and may not be an
issue.
ˆ Holding costs - There would be reclaimed inventory at the return hub at H
as well as D, E, and F as materials are in production or return processing.
Note that after D the reclaimed parts should be indistinguishable from
original manufacture, so they do not have to be covered separately at G.
ˆ Remanufacturing costs - Remanufacturing costs are incurred at E where
parts are removed/refurbished. At D there would be reintegration into the
supply chain, which may have some costs (inventory) but not necessarily
remanufacturing.
ˆ Facility location costs - C, E, F and H would have facility location costs.
ˆ Path transportation costs - Additional transportation costs exist for J −
H, K − H, L − H, H − F, G − F, F − E, E − D, E − C and potentially
F − A, F − B, F − C.
7
Li
Li (1- ri)
Li ri
Base Repair
Di
Base Stock
Bi
Ai
Depot Repair
Depot Stock
Figure 4: Figure for problem 6
6
Problem 6 - Service Parts Supply Chains METRIC
1. The system in Figure 2 has a single base and a single depot and single part
under consideration. The demand generated at the base is 1.3 parts per day.
Of these, ri , the fraction that can be repaired at the base, is 0.24. The Base
Repair time Bi is 5 days. The Depot repair time for parts from this base, Di
(which includes transportation) is 16 days. The transport time from the depot
to the base, Ai is 4 days.
Assume the base and depot base stock levels are high enough that there are
no backorders, determine the following:
1. Average number of parts in base repair.
2. Average number of parts in depot repair.
3. Average number of parts in transit from depot to base.
6.1
>
>
>
>
>
>
>
Solution
lambda = 1.3
r = 0.24
B = 5
D = 16
A = 4
baserepair = lambda * r * B
baserepair
8
[1] 1.56
> depotrepair = lambda * (1 - r) * D
> depotrepair
[1] 15.808
> enroutetobase = lambda * (1 - r) * A
> enroutetobase
[1] 3.952
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