DECISION MODELING WITH
MICROSOFT EXCEL
Chapter 5
LINEAR OPTIMIZATION:
APPLICATIONS
Part 2
Copyright 2001
Prentice Hall
Dynamic Models
Single time period models (such as those previously
covered) are called static models because time is
not a factor other than to define units of measure.
Dynamic models (or multiperiod models) are defined
for multiple time periods (a more realistic
abstraction of reality).
Many managerial decision models involve decision
making over time in which future decisions are
influenced by decisions made in earlier time periods.
In this model, each time period will have its own
performance measure, unified into a single measure
(e.g., Net Present Value). The timing of events must
be very precisely defined.
Dynamic Models
Dynamic Inventory Models
Dynamic inventory models (multiperiod inventory
models) apply to inventories of materials, cash,
employees, etc. carried from one time period to the
next.
This is a deterministic model because we assume
the demand (i.e., number of orders to be satisfied) in
each future time period to be known at the beginning
of period 1.
Dynamic Inventory Models
For example, a producer of polyurethane has a stock
of orders for the next 6 weeks.
di
denotes the known demand in gallons for
week i
di > 0 assumes that returns of orders are not
allowed for all i
Ci
denotes the cost of producing a gallon
during week i
Ki
denotes the maximum amount that can be
produced in week i
hi
denotes the per unit cost of holding
inventory in stock at the end of week i
I0
initial inventory at the beginning of period
1 (no holding charge is assessed)
The goal is to find a production and inventoryholding plan that satisfies the known delivery
schedule over the next 6 weeks at a min. total cost.
First, develop an expression for the inventory on
hand at the end of each period. Let
Ii
denote the inventory on hand at the end
of week i
xi
gallons of polyurethane produced in week i
So, for any week t, the inventory equation is:
Inventory at
end of week t
It = It-1 + xt - dt
Production in t
Inventory at
beginning of t
Demand in t
Assuming no shrinkage; inventory week t-1 = week t
So, for week 1: I1 = I0 + x1 – d1
For week 2: I2 = I1 + x2 – d2 = I0 + x1 – d1 + x2 – d2
2
2
= I0 + Sxi – Sdi
i=1
I1
i=1
New Inventory = Old Inventory + Production - Demand
For any week t:
t
It = I0 + S(xi – di )
i=1
It is known as a consequential or definitional
variable because it is defined in terms of other
decision variables (the xi values) in the model.
Note that the condition that demand in a time period
t must be satisfied is equivalent to the condition that
inventory It at the end of the time period t must be
nonnegative (as shown below).
Inventory
Ki = constraint on production
di
xi
Ii-1
Ii
Beginning of week i
End of week i
Time
Verbal Model
Minimize production costs + inventory costs
Subject to:
Inventory at the end of week t > 0
t = 1, 2, 3, …, 6
Production in week t < Kt
t = 1, 2, 3, …, 6
Production in week t > 0
t = 1, 2, 3, …, 6
Symbolic Model
xt
production in week t
6
6
t=1
t=1
Min S Ct xt + S ht It
s.t.
It = It-1 + xt – dt
t = 1, 2, 3, …, 6
xt < Kt
xt > 0,
It > 0
The cost in time period t is determined by all
production decisions in time periods 1 through t as
t
shown in the following equation: I = I + S(x – d )
t
0
i=1
i
i
Dynamic Models
A Dynamic Inventory Model
AutoPower generators are assembled in Singapore
from imported parts, tested in Singapore, and
exported to AutoPower’s Asian customers needing
uninterrupted electric power.
The following monthly data are available:
Delivery Requirements
Production Capacity
Jan. Feb. Mar. Apr.
58 36 34 59
60
62
64
66
Unit Production Costs (000s)
$28 $27 $27.8 $29
Unit Inventory Holding Cost
$300 $300 $300 $300
It is the possibility of holding inventory from one
month to the next that makes this model a dynamic
inventory model as opposed to a collection of four
static models.
AutoPower’s goal is to produce and deliver the
required number of generators over the 4 month
interval at the lowest 4 month cost (the January
beginning inventory is 15 generators).
Note in the following Excel spreadsheet model:
one column is devoted to each time period
the ending inventory values are consequential
(definitional) variables
the beginning inventory is equal to the
previous month’s ending inventory
Here is the Excel spreadsheet model after Solving:
=C6+C8-C9
=C4*(C8+C10)/2
=C10
=C3*C6
Note that any shrinkage can be accommodated by
multiplying the ending inventory in a given month by
a fraction (e.g., .99) to produce the next month’s
beginning inventory (showing a 1% shrinkage).
There are two fundamental decisions that must be
addressed in every dynamic model:
1. The planning horizon (the overall time interval
covered by the model; e.g., 4 months).
2. The number of discrete time epochs to
include within that interval (e.g., 1 month
intervals within the 4 month horizon).
The length of the planning horizon means that the
model cannot take into account any production or
delivery requirement beyond that length.
In choosing the time epochs, note that finer time
grids allow more precise measurements of actual
movements of inventory, producing more accurate
inventory cost tacking within the model.
Dynamic Models
Note that you can add realism to the model by using
finer time grids and longer planning horizons. The
temptation to do this is called the “curse of
dimensionality.”
Adding each new column adds a new dimension to
the model in terms of new decision variables,
decision and consequential variable linkages across
time, and parameters to be considered.
Dynamic Models
Every dynamic model must also pay attention to
what are called “edge conditions.”
Edge conditions refer to the set of parameters that
must be specified at the beginning and end of time
in the model (i.e., the initial and ending inventory).
The beginning edge condition is usually not difficult
to find. However, the ending edge condition is more
troublesome because it must stand as a reasonable
starting condition or proxy for all of time beyond the
planning horizon (e.g., AutoPower used 7 as the
ending inventory).
The spreadsheet below shows the Solver results
when each month is treated as a separate static
month.
The resulting production quantity in excess of the
required shipments in a month raises total cost. It is
for this reason that static models are referred to as
myopic because they ignore any consequences of
current decisions upon future payoff.
Here is the Solver Sensitivity Report for the original
dynamic model.
Note that the shadow prices appear in the Reduced
Cost column, hence, no range information is given
for those shadow prices.
This Excel Spreadsheet shows the results of making
each month’s Beginning Inventory a decision
variable instead of a consequential variable.
An added constraint specifies that the Beginning
Inventory be no more than the previous month’s
Ending Inventory.
This way of specifying the constraints changes the
Sensitivity Report as reflected in the shadow prices
and coefficient ranging.
Also, modeling inventory as a decision variable isolates
the unit costs into separate production coefficients and
inventory coefficients, thereby making the objective
function coefficients appear as originally specified in the
spreadsheet model.
Dynamic Models
The Dynamic Inventory Model Recasted as a
Transportation Model
A dynamic model can be modeled as a
transportation model with the sources (From)
representing decisions in a given time period and
the destinations (To) representing one or more time
periods affected, either in the current time period or
a later one.
Unit production cost
per month and
delivery cost per
future month
(including holding
cost for each
additional month).
These cells contain
huge costs, forcing
Solver to avoid
assigning any
positive amounts in
the corresponding
gray cells in the
Shipments decisions
block. They
represent time
reversal decisions
that are impossible
and so must be
avoided by Solver.
Here is the Solver program for optimization:
Here is the Sensitivity Analysis for the model:
A Dynamic Production Scheduling
and Inventory Control Model
Bumles, Inc. uses part of its capacity in its Mexican
plant to make hand-painted teapots. The following
information is available:
One teapot takes 0.5 hours to paint.
There are 30 painters available.
Teapots are made on Thursday, Friday and
Saturday.
Not all 30 painters will be engaged but if they
are, they are available to work any part of an 8hour day, 2 days a week.
A painter can be assigned to any 2 day schedule
and is paid for 16 hours, no matter what.
Slack time is spent on cleaning the plant.
Revenue (ignoring labor costs) for 1 teapot is
equal to $15.
All teapots produced in a week must be shipped
that week.
It costs $0.50 to carry a teapot in ending
inventory from one day to the next.
A unit of lost demand results in a penalty of
$1/unit on Thurs., $3 on Fri. and $5 on Sat.
Painters are paid $8 per hour.
Weekly demand is 100 on Thurs., 300 on Fri. and
600 on Sat.
Here is the resulting spreadsheet model.
A Dynamic Cash Management Model
Winston-Salem Development Corporation (WSDC) is
trying to complete its investment plans for the next
two years.
Currently, WSDC has $2,000,000 on hand and
available for investment. The following table gives
the cash income from previous investments:
Income
6 MONTHS 12 MONTHS 18 MONTHS
$500,000
$400,000
$380,000
There are 2 development projects in which WSDC is
considering participation along with other nonWSDC investors.
1. Foster City Development:
If WSDC participated at a 100% level, the
projected cash flow would be:
Income
INITIAL
6 MONTHS
12 MONTHS
18 MONTHS
24 MONTHS
$- 1,000,000
$- 700,000
$ 1,800,000
$ 400,000
$ 600,000
In order to participate at the 100% level, WSDC
would have to lay out $1,000,000 immediately
and $700,000 again in 6 months.
2. Take over the operation of an old MiddleIncome Housing on the condition that certain
initial repairs be made. At 100%
participation, the cash flow would be:
Income
INITIAL
6 MONTHS
12 MONTHS
18 MONTHS
24 MONTHS
$- 800,000
$ 500,000
$- 200,000
$- 700,000
$ 2,000,000
WSDC can participate in either project at a level less
than 100%. The cash flows would be adjusted
proportionally and outside investors would make up
the difference.
Now WSDC must decide how much of the $2,000,000
on hand should be invested in each of the projects
and how much should simply be invested in a 6month CD for the 7% semiannual return.
The goal (objective function) is to maximize the cash
on hand at the end of 24 months. The decision
variables are:
F = fractional participation in Foster City project
M = fractional participation in Middle-Income
Housing project
S1 = initial surplus funds to be invested in a 7% CD
S2 = 6 mts. surplus funds to be invested in a 7% CD
S3 = 12 mts. surplus funds to be invested in a 7%CD
S4 = 18 mts. surplus funds to be invested in a 7%CD
The constraints in this model must say that at the
beginning of each of the four 6-month periods:
cash invested < cash on hand
The first constraint must say:
Initial investment < initial funds on hand
1,000,000F + 800,000M + S1 < 2,000,000
Because of the interest paid, S1 becomes 1.07S1
after 6 months, and similarly for S2, S3, and S4:
700,000F + S2 < 500,000M + 1.07S1 + 500,000
200,000M + S3 < 1,800,000F + 1.07S2 + 400,000
700,000M + S4 < 400,000F + 1.07S3 + 380,000
These constraints provide material balance of the
cash flows from one time period to another.
Here is the symbolic model for this LP dynamic cash
management model:
Max 600,000F + 2,000,000M + 1.07S4
s.t.
1,000,000F + 800,000M + S1 < 2,000,000
700,000F - 500,000M – 1.07S1 + S2 < 500,000
-1,800,000F + 200,000M – 1.07S2 + S3 < 400,000
- 400,000F + 700,000M – 1.07S3 + S4 < 380,000
F < 1 and M < 1
F > 0, M > 0, Si > 0, i = 1,2,3,4
Here is the
optimal
solution:
Here are the
formulas
and Solver
Parameters:
Here is the partial Sensitivity Analysis:
As can be seen from these results, both investment
projects are attractive with WSDC participating fully,
and from the Sensitivity Report, the marginal return
to WSDCs initial funds is 31% over 24 months.
A Financial and
Production Planning Model
The AutoPower Europe production model for its
diesel powered UPS generators is a product-mix
model for deciding how many of its two models,
BigGen and SmallGen, to produce and sell in each of
the coming months in view of a number of
constraints.
The following information is relevant to this model:
1. The financial data on each BigGen and SmallGen
sold is:
Product Sales
Per Unit Contribution
Price
Material
Margin
$000’s
and Labor
$000’s
$000’s
BigGen
80
75
5
SmallGen
24
20
4
2. Each product is put through machining and
assembly operations in two production
departments, called A and B.
Dept. A has 150 hours available each month.
Dept. B has 160 hours available each month.
3. Each BigGen produced uses 10 hours of
machining in Dept. A and 20 hours of machining
in Dept. B. Each SmallGen uses 15 hours in Dept.
A and 10 hours in Dept. B.
4. Testing is performed in an independent third
department. Each BigGen is given 30 hours of
testing and each SmallGen is given 10.
AutoPower’s labor contract mandates that the
total labor hours devoted to testing cannot fall
below 135 hours each month.
Now, let B be the number of BigGen’s to produce
each month and S be the number of SmallGen’s,
here is the symbolic model:
Max 5B + 4S
(Profit in $000’s)
s.t.
10B + 15S < 150
(hours in department A)
20B + 10S < 160
(hours in department B)
30B + 10S > 135
(testing hours)
B, S > 0
Here are the results of the Solved model.
A Financial and
Production Planning Model
Financial Considerations
The previous AutoPower model ignored important
financial considerations.
Specifically, AutoPower must incur the material and
direct labor costs in the next month, whereas
payments will not be forthcoming for another 3
months.
AutoPower has budgeted $100,000 of cash on hand
to cover the current material and labor costs and
plans to borrow any additional funds needed (at an
annual interest rate of 16%).
In order to hedge against downside risks, the bank
has limited the total due to the bank (principal plus
interest) to be no more than 2/3 of the sum of
AutoPower’s cash on hand and accounts receivable.
If present value of net cash flow is maximized, fewer
BigGen’s and more SamllGen’s should be produced
(because of high material and labor costs).
Also, the committee cannot agree on a discount rate
of 12%, 16% or 20%.
AutoPower needs to formulate a new objective
function, determine how much to borrow (if any),
and devise a production plan incorporating this new
information.
Now, let D = debt (i.e., total borrowed in $000s).
The net cash flow next month in $ will be:
D – 75B – 20S
The net cash flow three months later (after payments
are received) will be:
80B + 24S – 1.040D
Discount factor a (based on the
annual discount rate R) is equal
to:
a=
1
1+R/4
AutoPower can
borrow at 16% per
annum (4% for 3
months.
The objective is to Max. present value of net cash
flow, so the objective function (in $000s) becomes:
Max 1D – 75B – 20S + a (80B + 24S – 1.040D)
Additional constraints required are:
1. Since the total uses of funds cannot exceed the
total sources of funds, the inequality is:
Material and labor costs < debt + cash on hand
75E + 20F < D + 100
2. The bank requires that the total amount due the
bank must be no greater than 2/3 of the cash on
hand plus the accounts receivable:
debt + interest < 2/3(cash on hand + accts. rec.)
D + .04D < 2/3(100 + 80B + 24S)
1.5(D + .04D) < 100 + 80B + 24S
Here is the production and financial model Excel
spreadsheet with optimal solution for the case
R=20%.
The surplus funds
can be invested to
earn 20% interest
while the cost of the
funds is only 16%.
AutoPower Europe should borrow $279.49 thousand
and reduce production of BigGens.
Here are the Solver parameters:
A Financial and
Production Planning Model
Effect of Financial Considerations
Inserting the financial considerations in the original
production mix model leads to a different plan and a
smaller profit.
The objective function in the production and finance
model is less because future cash flows are
discounted and interest costs of borrowing are
included.
SolverTable1 can be used to show the profits based
on varying values of R (discount rate).
If R is < 16%, borrow $192.50
If R is > 16%, borrow $279.49
If R is = 16%, then there are alternative
optimal solutions where D lies between
$192.50 and $279.49.
SolverTable1 can be used to show the payment
delays between zero and 4 months to illustrate the
benefit to AutoPower Europe of improving its
accounts receivables by encouraging its customers
to pay invoices promptly.
% Improvement =
(Profit – 30.162)/30.162
If AutoPower’s customers paid for their generators
in one month instead of delaying 3 months, their
profit would increase by more than 40% per month.
Network Models
Transportation and assignment models are
members of a more general class of models called
network models.
Network models involve from-to sources and
destinations.
Applied to management logistics and distribution,
network models are important because:
They can be applied to a wide variety of real
world models.
Flows may represent physical quantities,
Internet data packets, cash, airplanes, cars,
ships, products, …
Network Models
A Capacitated Transshipment Model
Zigwell Inc. is AutoPower’s largest US distributor of
UPS generators in five Midwestern states.
Zigwell has 10 BigGen’s at site 1
These generators must be delivered to construction
sites in two cities denoted 3 and 4
3 BigGen’s are required at site 3 and 7 are required
at site 4
Because of prearranged schedules concerning
drivers available, these generators may be
distributed only according to any of the following
alternative routes.
-3
Supply
3
+10
1
2
This is a network diagram or
network flow diagram.
Each arrow is called an arc or branch.
Each site is termed a node.
4
5
-7
There are many allowable routes. Which of these
allowable routes is selected will be determined by:
cij
uij
the costs (per unit) of traversing the
routes
the capacities along the routes
Costs are primarily due to fuel, tolls, and the cost of
the driver for the average time it takes to traverse
the arc.
Because of pre-established agreements with the
teamsters, Zigwell must change drivers at each site
it encounters on a route.
Because of limitations on the current availability of
drivers, there is an upper bound, uij, on the number
of generators that may traverse an arc.
The arc capacities (uij) and costs (cij) have been
added to the network model.
-3
3
c23
u23
+10
1
c12
u12
2
c24
u24
c34
u34
c43
u43
4
c25
u25
5
-7
u53 c53
Network Models
A Capacitated Transshipment Model
LP Formulation of the Model
The goal is to find a shipment plan that satisfies the
demands at minimum cost, subject to the capacity
constraints.
The capacitated transshipment model is basically
identical to the transportation model except that:
1. Any plant or warehouse can ship to any other
plant or warehouse
2. There can be upper and/or lower bounds
(capacities) on each shipment (branch)
The decision variables are:
xij = total number of BigGen’s sent on arc (i, j)
= flow from node i to node j
The model becomes:
Min c12x12 + c23x23 + c24x24 + c25x25 + c34x34 +
c43x43 + c53x53 + c54x54
s.t.
+x12 = 10
-x12 + x23 + x24 + x25 = 0
-x23 – x43 – x53 + x34 = -3
-x24 + x43 – x34 – x54 = -7
-x25 + x53 + x54 = 0
0 < xij < uij all arcs (i, j) in the network
Properties of the Model
1. xij is associated with each of the 8 arcs in the
network. Therefore, there are 8 corresponding
variables: x12, x23, x24, x25, x34, x43, x53, and x54
The objective is to minimize total cost.
2. There is one material flow balance equation
associated with each node in the network. For
example:
Total flow out of node 1 is 10 units
Total flow out of node 2 minus the total flow
into node 2 is zero (i.e., total flow out must
equal total flow into node 2 ).
Total flow out of node 3 must be 3 units less
than the total flow into node 3 .
Intermediate nodes that are neither supply points
nor demand points are often termed
transshipment nodes.
3. The positive right-hand sides correspond to
nodes that are net suppliers (origins).
The negative right-hand sides correspond to
nodes that are net destinations.
The zero right-hand sides correspond to nodes
that have neither supply nor demand.
The sum of all right-hand-side terms is zero (i.e.,
total supply in the network equals total demand).
In general, flow balance for a given node, j, is:
Total flow out of node j – total flow into node j = supply at node j
Negative supply is a requirement. Nodes with
negative supply are called destinations, sinks, or
demand points.
Nodes with positive supply are called origins,
sources, or supply points.
Nodes with zero supply are called transshipment
points.
4. A small model can be optimized with Solver.
Here is the Excel formulation and optimal solution
for the transshipment model.
=C10*J3
=SUM(C10:C14)
=Sum(C10:G10)
=H10-C15; Note that you can use the TRANSPOSE
function. Highlight cells C16:G16 and enter
=TRANSPOSE(H10:H14)-C15:G15
Here are the Solver Parameters:
Network Models
A Capacitated Transshipment Model
Integer Optimal Solutions
The integer property of the network model can be
stated thus:
If all the RHS terms and arc capacities, uij, are
integers in the capacitated transshipment
model, there will always be an integer-valued
optimal solution to this model.
Network Models
A Capacitated Transshipment Model
Efficient Solution Procedures
The structure of this model makes it possible to
apply special solution methods and software that
optimize the model much more quickly than the
more general simplex method used by Solver.
This makes it possible to optimize very large scale
network models quickly and cheaply.
Network Models
A Shortest-Route Model
The shortest-route model refers to a network for
which each arc (i,j) has an associated number, cij,
which is interpreted as the distance (or cost, or time)
from node i to node j.
A route or path between two nodes is any sequence
of arcs connecting the two nodes.
The objective is to find the shortest (or least-cost or
least-time) routes from a specific node to each of the
other nodes in the network.
In this example, Aaron Drunner makes frequent wine
deliveries to 7 different sites:
Note that the arcs are undirected (flow is permitted
in either direction).
7
Distance between
1
nodes.
2
8
6
3
7
4
H
Home Base
1
3
3
4
1
1
1
6
2
3
5
2
The goal is to minimize overall costs by making sure
that any future delivery to any given site is made
along the shortest route to that site.
The goal is to minimize overall costs by finding the
shortest route from node H to any of the other 7
nodes. Note that in this model, the task is to find an
optimal route, not optimal xij’s.
The following Excel model finds the shortest path
between any two nodes. The starting and ending
nodes are specified by the values 1 and –1,
respectively.
Indicator variables (cells C3:J10) are used to signify
the node-arc connectivity.
The decision variables are constrained to be
between 0 and 1 and the objective function (U21)
minimizes total distance.
Here is the Excel formulation and optimal solution
for the shortest route model.
Here are the Solver Parameters:
Network Models
An Equipment Replacement Model
In this example, Michael Carr is responsible for
obtaining a high speed printing press for his
newspaper company.
In a given year he must choose between purchasing:
New Printing Press
high annual
acquisition cost
Old Printing Press
no annual
acquisition cost
low initial
maintenance cost
high initial
maintenance cost
Assume a 4-year time horizon. Let:
cij
denote the cost of buying new equipment
at the beginning of year i, i = 1, 2, 3, 4
and maintaining it to the beginning of year
j, j = 2, 3, 4, 5
Three alternative feasible policies are:
1. Buying new equipment at the beginning of
each year.
Total (acquisition + maintenance) cost =
c12 + c23 + c34 + c45
2. Buy new equipment only at the beginning of
year 1 and maintain it through all successive
years.
Total (buying + maintenance) cost = c15
3. Buy new equipment at the beginning of years
1 and 4.
Total cost = c14 + c45
The solution to this model is obtained by finding the
shortest (i.e., minimum cost) route from node 1 to
node 5 of the network.
Each node on the shortest route denotes a
replacement, that is, a year at which new equipment
should be bought.
Here is the network model for this problem.
c15
c14
c35
c13
1
c12
2
c23
3
c24
c34
4
c45
5
c25
Assume the following costs:
$1,600,000 purchase cost
$500,000 maintenance cost in purchase year
$1,000,000, $1,500,000, and $2,200,000 for
each additional year the equipment is kept
Here is the Excel spreadsheet and solution.
Indicator
variables
used
Cost Parameters
to signify the node-arc
connectivity
Cost consequences of
Time reversals
buying and
that have no
maintaining a printing
meaning
press beginning in one
year and continuing to
Costs
a terminal
year.
associated
Represents beginning year
with decisions
(1) and ending year (-1)
Here are the Solver Parameters:
The optimal strategy is to buy a printing press at
the beginning of year 1 and keep it for 2 years,
replacing it at the beginning of year 3 with a new
printing press which he keeps until the
beginning of year 5 for a total 4-year cost of $6.2
million.
Network Models
A Maximal Flow Model
In the maximal-flow model, there is a single source
node (the input node) and a single sink node (the
output node).
The goal is to find the maximum amount of total flow
that can be routed through a physical network (from
source to sink) in a unit of time.
The amount of flow per unit time on each arc is
limited by capacity restrictions.
The only requirement is that for each node (other
than the source or the sink):
flow out of the node = flow into the node
Network Models
A Maximal Flow Model
An Application of Maximal-Flow:
The Urban Development Planning Commission
The UDPC (Urban Development Planning
Commission) is an ad hoc special interest study
group.
The group’s current responsibility is to coordinate
the construction of the new subway system with the
state’s highway maintenance department.
Because the new subway system is being built near
the city’s beltway, the eastbound traffic on the
beltway must be detoured.
The proposed network of alternative routes and the
different speed limits and traffic patterns (producing
different flow capacities) are given below:
6
0
Detour
2
4 6
Detour
0
2
Begins
Ends
0
0
4
1
Indicates 0 capacity in the
direction of the arrow.
6
Indicates a capacity of
0 1
6000 vehicles3per hour
in 0
4
2
5
the direction0 of 3the arrow.
0
6
0
Here is the Solved Excel model:
The optimal solution is a maximal flow of 8000
vehicles per hour.
Translating the Excel solution to the original
network diagram gives the following traffic pattern:
6
2
4
6
4
8
1
6
2
4
2
5
3
2
8
As is common in network models, there are
alternative optimal solutions to the model: