From Hillier, F. S., & Lieberman, G. J. (2001). Introduction to Operations Research (7th ed.).
McGraw-Hill Higher Education.
PB3 – Transport, Assignment and Network Problems
Exercise 1
Suppose that England, France, and Spain produce all the wheat, barley, and oats in the world.
The world demand for wheat requires 125 million acres of land devoted to wheat production.
Similarly, 60 million acres of land are required for barley and 75 million acres of land for oats.
The total amount of land available for these purposes in England, France, and Spain is 70
million acres, 110 million acres, and 80 million acres, respectively. The number of hours of
labor needed in England, France and Spain to produce an acre of wheat is 18, 13, and 16,
respectively. The number of hours of labor needed in England, France, and Spain to produce
an acre of barley is 15, 12, and 12, respectively. The number of hours of labor needed in
England, France, and Spain to produce an acre of oats is 12, 10, and 16, respectively. The labor
cost per hour in producing wheat is $9.00, $7.20, and $9.90 in England, France, and Spain,
respectively. The labor cost per hour in producing barley is $8.10, $9.00, and $8.40 in England,
France, and Spain respectively. The labor cost per hour in producing oats is $6.90, $7.50, and
$6.30 in England, France, and Spain, respectively. The problem is to allocate land use in each
country so as to meet the world food requirement and minimize the total labor cost.
a) Formulate this problem as a transportation problem by constructing the appropriate
parameter table.
b) Draw the network representation of this problem.
c) Starting with the northwest corner rule, interactively apply the transportation simplex
method to obtain an optimal solution for this problem.
Exercise 2
Sarah has just graduated from high school. As a graduation present, her parents have given
her a car fund of $21,000 to help purchase and maintain a certain three-year-old used car for
college. Since operating and maintenance costs go up rapidly as the car ages, Sarah's parents
tell her that she will be welcome to trade in her car on another three-year-old car one or more
times during the next three summers if she determines that this would minimize her total net
cost. They also inform her that they will give her a new car in four years as a college graduation
present, so she should definitely plan to trade in her car then.
The table gives the relevant data for each time Sarah purchases a three-year-old car. For
example, if she trades in her car after two years, the next car will be in ownership year 1 during
her junior year, etc.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Sarah's Data Each Time She Purchases a Three-Year Old Car
Purchase
Price
Operating and Maintenance Costs
for Ownership Year
1
2
3
4
$12,000
$2,000
$3,000
$4,500
$6,500
1
$8,500
Trade-in Value at End
of Ownership Year
2
3
$6,500
$4,500
4
$3,000
When should Sarah trade in her car (if at all) during the next three summers to minimize her
total net cost of purchasing, operating, and maintaining the cars over her four years of
college?
1. Formulate this problem as a shortest-path problem.
2. Use the corresponding algorithm to solve this shortest-path problem.
Exercise 3
Consider the following network.
1. Using the corresponding algorithm, find the minimum spanning tree for this
network.
Exercise 4
For the network shown below, use the augmenting path algorithm to find the flow pattern
giving the maximum flow from the source to the sink, given that the arc capacity from node
i to node j is the number nearest node i along the arc between these nodes.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Exercise 5
A company will be producing the same new product at two different factories, and then the
product must be shipped to two warehouses. Factory 1 can send an unlimited amount by rail
to warehouse 1 only, whereas factory 2 can send an unlimited amount by rail to warehouse 2
only. However, independent truckers can be used to ship up to 50 units from each factory to
a distribution center, from which up to 50 units can be shipped to each warehouse. The
shipping cost per unit for each alternative is shown in the following table, along with the
amounts to be produced at the factories and the amounts needed at the warehouses.
1. Formulate the network representation of this problem as a minimum cost
flow problem.
2. Formulate the linear programming model for this problem.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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PB3 – Transport, Assignment and Network Problems - Implementation
Exercise 6
A contractor, Susan Meyer, has to haul gravel to three building sites. She can purchase as
much as 18 tons at a gravel pit in the north of the city and 14 tons at one in the south. She
needs 10, 5, and 10 tons at sites 1, 2, and 3, respectively. The purchase price per ton at each
gravel pit and the hauling cost per ton are given in the table below. Susan wishes to determine
how much to haul from each pit to each site to minimize the total cost for purchasing and
hauling gravel.
Pit
North
South
Hauling Cost per Ton at Site
1
2
3
$30
$60
$50
$60
$30
$40
Price per Ton
$100
$120
Now suppose that trucks (and their drivers) need to be hired to do the hauling, where each truck can
only be used to haul gravel from a single pit to a single site. Each truck can haul 5 tons, and the cost per
truck is five times the hauling cost per ton given above. Only full trucks would be used to supply each
site.
a) Formulate this problem as an assignment problem by constructing the appropriate cost
table, including identifying the assignees and tasks.
b) Obtain an optimal solution using Excel Solver, Python and/or MATLAB.
c) Reformulate this assignment problem as an equivalent transportation problem with two
sources and three destinations by constructing the appropriate parameter table.
d) Obtain an optimal solution for the problem as formulated in part c) using Excel Solver,
Python and/or MATLAB.
Exercise 7
Formulate and solve a spreadsheet model for Exercise 2.
Exercise 8
Consider the network from Exercise 3.
1. Using the corresponding algorithm, find the shortest path through the
network, where the numbers represent actual distances between the
corresponding nodes.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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PB3 – Problem solutions
Exercise 1 solution
a) Formulate this problem as a transportation problem by constructing the appropriate
parameter table.
Let England, France, and Spain be the three sources, where their supplies are the millions
of acres of land that are available for growing these crops. Let Wheat, Barley, and Oats be
the three destinations, where their demands are the millions of acres of land that are
needed to fulfill the world demand for these respective crops. The unit cost (in millions of
dollars) is the labor cost per million acres, so the number of hours of labor needed is
multiplied by the cost per hour. The parameter table is as follows.
Source
Wheat
162
93.6
158.4
125
England
France
Spain
Demand
Unit Cost ($ million)
Destination
Barley
121.5
108
100.8
60
Oats
82.8
75
100.8
75
Supply
70
110
80
b) Draw the network representation of this problem.
The network presentation of this problem is given below.
[70]
162
E
W [-125]
93.6
158.4
121.5
[110]
108
F
B
[-60]
O
[-75]
100.8
75
82.8
[80]
S
100.8
c) Starting with the northwest corner rule, interactively apply the transportation simplex
method to obtain an optimal solution for this problem.
For this problem, the initial BF solution obtained by the northwest corner rule is shown
below.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Des tination
Wheat
162
England
Barley
121.5
Oats
82.8
70
70
93.6
108
Source France
100.8
Spain
125
Demand
75
110
55
55
158.4
Supply
100.8
5
75
60
75
80
The Transportation Simplex Method Optimality Test:
1) Derive ui and vj by selecting the row having the largest number of allocations, setting
its ui=0, and then solve the set of equations cij = ui+ vj for each (i,j) such that xij is basic.
2) If cij − ui − vj  0 for every (i,j) such that xij is nonbasic, then the current solution is
optimal, so stop. Otherwise, go to an iteration.
Optimality Test:
Since cij - ui - vj = 0 if xij is a basic variable,
cij = ui + vj
for each (i, j) such that xij is basic.
Because the number of unknowns (the ui and vj) exceed the number of these equations
by one, we can set one unknown equal to an arbitrary value, say 0. These equations can
then be solved as outlined below.
x21:
93.6 = u2 + v1.
Set u2 = 0, so v1 = 93.6,
x22:
108 = u2 + v2.
v2 = 108.
x11:
162 = u1 + v1.
Know v1 = 93.6, so u1 = 68.4.
x32:
100.8 = u3 + v2.
Know v2 = 108, so u3 = -7.2.
x33:
100.8 = u3 + v3.
Know u3 = -7.2, so v3 = 108.
Since cij - ui - vj represents the rate at which the objective function will change as a
nonbasic variable xij is increased, we now can check whether increasing any nonbasic
variable will decrease the total cost Z.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Nonbasic
variable
x12
x13
x23
x31
cij - ui - vj
121.5 - 68.4 - 108
82.8 - 68.4 - 108
75 - 0 - 108
158.4 -(-7.2) - 93.6
= -54.9
= -93.6
= -33
= 72
Because some of these (cij - uij - vj) values are negative, the initial BF solution is not
optimal.
Iteration 1:

𝑐𝑖𝑗 − 𝑢𝑖 − 𝑣𝑗 represents the rate at which the objective function will change as the
nonbasic variable 𝑥𝑖𝑗 is increased.
We select the nonbasic variable x13 to be the entering basic variable because it has the
largest negative value of (cij - ui - vj).
When x13 is increased from 0 by any particular amount, a chain reaction is set off that
requires alternately decreasing and increasing current basic variables by the same
amount in order to continue satisfying the supply and demand constraints. This chain
reaction is depicted in the next figure, where the + sign inside a box in cell (1, 3) indicates
that the entering basic variable is being increased there and the + or - sign next to other
circles indicate that a basic variable is being increased or decreased there.
Des tination
1
1
70
2
55
2
+
Source
3
Demand
55
5
125
3
Supply
+
70
-
+
60
110
75
-
80
75
Each donor cell (indicated by a minus sign) decreases its allocation by exactly the same
amount as the entering basic variable and each recipient cell (indicated by a plus sign) is
increased. The entering basic variable will be increased as far as possible until the
allocation for one of the donor cells drops all the way down to 0. Since the original
allocations for the donor cells are
x11 = 70,
x22 = 55,
x33 = 75,
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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x22 will be the one that drops to 0 as x13 is increased (by 55). Therefore, x22 is the leaving
basic variable.
Since each of the basic variables is being increased or decreased by 55, the values of the
basic variables in the new BF solution are
x11 = 15,
x13 = 55,
x21 = 110,
x32 = 60,
x33 = 20.
Optimality Test After Iteration 1:
Since Source 1 now has two basic variables (tied for the maximum number), let us set u1
= 0 this time. The cij = ui + vj equations then would be solved as follows.
x11:
162 = u1 + v1.
Set u1 = 0, so v1 = 162,
x13:
82.8 = u1 + v3.
v3 = 82.8.
x21:
93.6 = u2 + v1.
Know v1 = 162, so u2 = -68.4.
x33:
100.8 = u3 + v3. Know v3 = 82.8, so u3 = 18.
x32:
100.8 = u3 + v2. Know u3 = 18, so v2 = 82.8.
We next calculate (cij - ui - vj) for the nonbasic variables.
Nonbasic
variable
x12
x22
x23
x31
cij - ui - vj
121.5 - 0 - 82.8
= 38.7
108 - (-68.4) - 82.8 = 93.6
75 - (-68.4) - 82.8 = 60.6
158.4 - 18 - 162
= -21.6
We still have one negative value of (cij - ui - vj), so the current BF solution is not optimal.
Iteration 2:
Since x31 is the one nonbasic variable with a negative value of (cij - ui - vj), x31 becomes the
entering basic variable. The resulting chain reaction is depicted next.
Des tination
1
1
Source
15
2
-
Supply
3
55
+
70
110
2
3
+
Demand
125
20
60
-
80
75
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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The donor cells have allocations of x11 = 15 and x33 = 20. Because 15 < 20, the leaving
basic variable is x11.
Since the basic variables x21 and x32 were not part of this chain reaction, their values do
not change. However, x31 and x13 increase by 15 while x11 and x33 decrease by 15. Therefore,
the values of the basic variables in the new BF solution are
x13 = 70,
x21 = 110,
x31 = 15,
x32 = 60,
x33 = 5
Optimality Test After Iteration 2:
Because Source 3 now has the largest number of basic variables, we set u3 = 0 this time.
The resulting calculations are shown below.
x31:
158.4 = u3 + v1.
Set u3 = 0, so v1 = 158.4,
x32:
100.8 = u3 + v2.
v2 = 100.8.
x33:
100.8 = u3 + v3.
v3 = 100.8.
x13:
82.8 = u1 + v3.
Know v3 = 100.8, so u1 = -18.
x21:
93.6 = u2 + v1.
Know v1 = 158.4, so u2 = -64.8.
Nonbasic
variable
x11
x12
x22
x23
cij - ui - vj
162 - (-18) - 158.4 = 21.6
121.5 - (-18) - 100.8 = 38.7
108 - (-68.4) - 100.8 = 72
75 - (-64.8) - 100.8 = 39
Since all of these values of (cij - ui - vj) are nonnegative, the current BF solution is optimal.
Thus, the optimal allocation of land to crops is
70
110
15
60
5
million acres in England for oats,
million acres in France for wheat,
million acres in Spain for wheat,
million acres in Spain for barley,
million acres in Spain for oats.
The total cost of this grand enterprise would be Z = $25.02 billion.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Exercise 2 solution
1.
Formulate this problem as a shortest-path problem.
The following figure shows the network formulation of this problem as a shortest path problem.
Nodes 1, 2, 3, and 4 are the end of Sarah's year 1, 2, 3, and 4 of college, respectively. Node 0 is now,
before starting college. Each arc from one node to a second node corresponds to the activity of
purchasing a car at the time indicated by the first of these two nodes and then trading it in at the
time indicated by the second node. Sarah begins by purchasing a car now, and she ends by trading
in a car at the end of year 4, so node 0 is the origin and node 4 is the destination.
The number of arcs on the path chosen from the origin to the destination indicates how many times
Sarah will purchase and trade in a car. For example, consider the path:
This corresponds to purchasing a car now, then trading it in at the end of year 1 to purchase a second
car, then trading in the second car at the end of year 3 to purchase a third car, and then trading in
this third car at the end of year 4.
Since Sarah wants to minimize her total net cost from now (node 0) to the end of year 4 (node 4),
each arc length needs to measure the net cost of that arc's cycle of purchasing, maintaining, and
trading in a car. Therefore,
Arc length = purchase price + operating and maintenance costs - trade-in value.
For example, consider the arc from node 1 to node 3. This arc corresponds to
purchasing a car at the end of year 1, operating and maintaining it during ownership years 1 and 2,
and then trading it in at the end of ownership year 2. Consequently,
Length of arc from to = 12,000 + 2,000 + 3,000 - 6,500 = 10,500 (in dollars).
The arc lengths calculated in this way are shown next to the arcs in the figure. Adding up the lengths
of the arcs on any path from node 0 to node 4 then gives the total net cost for that particular plan
for trading in cars over the next four years. Therefore, finding the shortest path from the origin to
the destination identifies the plan that will minimize Sarah's total net cost.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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2.
Use the shortest-path algorithm to solve this shortest-path problem.
Shortest-path algorithm:
Objective of nth iteration: Find the nth nearest node to the origin (to be repeated for n=1,2,… until
the nth nearest node is the destination).
Input for nth iteration: n-1 nearest nodes to the origin (solved for at the previous iterations),
including their shortest path and distance from the origin. (These nodes, plus
the origin, will be called solved nodes; the others are unsolved nodes.)
Candidates for nth nearest node: Each solved node that is directly connected by a link to one or more
unsolved nodes provides one candidate – the unsolved node with
the shortest connecting link. (Ties provide additional candidates.)
Calculation of the nth nearest node: For each such solved node and its candidate, add the distance
between them and the distance of the shortest path from the
origin to this solved node. The candidate with the smallest
such distance is the nth nearest node (ties provide additional
solved nodes), and its shortest path is the one generating this
distance.
n
1
2
Solved nodes
connected to
unsolved nodes
0
Its closest
connected
unsolved node
1
0
2
10,500
1
2
5,500+5,500
0
3
= 11,000
17,000
1
3
5,500+10,500
3
Total cost
involved
5,500
nth
nearest
node
1
2
Its minimum
Its last
cost
connection
5,500
10,500
0 →
0→2
3
16,000
1→3
3
16,000
2→3
4
21,000
2→4
= 16,000
2
3
10,500+5,500
0
4
= 16,000
25,000
1
4
5,500+17,000
= 22,500
4
2
4
10,500+10,500
= 21,000
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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3
4
16,000+5,500
= 21,500
Thus, the shortest path turns out to be
Trade in the first car at the end of Year 2.
Trade in the second car at the end of Year 4.
The length of this path is 10,500 + 10,500 = 21,000, so Sarah's total net cost is $21,000. Recall that
this is exactly the amount in Sarah's car fund provided by her parents (I guess they did the math
first…).
Exercise 3 Solution
1. Using the corresponding algorithm, find the minimum spanning tree for this
network.
Exercise 4 Solution
Iteration 0: The initial residual network is
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
12
Iteration 1: One of the several augmenting paths is 1→ 3 → 8 → 9, which has a residual capacity of
min{9, 6, 7} = 6. Any of the augmenting paths could be chosen, but suppose we select this one. By
assigning a flow of 6 to this path, the resulting residual network is
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Iteration 2: Assign a flow of 4 to the augmenting path 1→ 2 → 4 → 7 → 9. The resulting residual
network is
Iteration 3: Assign a flow of 3 to the augmenting path 1→ 3 → 5 → 7 → 9. The resulting residual
network is
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Iteration 4: Assign a flow of 2 to the augmenting path 1→ 2 → 5 → 9. The resulting residual network
is
Iteration 5: Assign a flow of 3 to the augmenting path 1→ 5 → 9. The resulting residual network is
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Iteration 6: Assign a flow of 2 to the augmenting path 1→ 2 → 7 → 5 → 9. (Although flow between
nodes 5 and 7 can only go in the direction from node 5 to node 7, this assignment of a flow of 2 to 7
→is, in reality, simply reducing the previously assigned flow from node 5 to node 7 by 2 units.) The
resulting residual network is
There are no more augmenting paths, so the current flow (given by the number at the end of the
respective arcs) in the following network is optimal. The maximum flow is 20.
Exercise 5 Solution
1.
Formulate the network representation of this problem as a minimum cost flow
problem.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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2.
Formulate the linear programming model for this problem.
PB2 – Implementation solutions
Exercise 6 solution
a) Formulate this problem as an assignment problem by constructing the appropriate cost table,
including identifying the assignees and tasks.
The tasks are the loads needed at sites 1, 2, and 3. The assignees are the three trucks from the
North pit and the two trucks from the South pit. Considering the purchase price for the gravel and
the hauling cost per truck, the cost table is constructed as follows.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
17
Assignee
1a
650
650
650
900
900
North 1
North 2
North 3
South 1
South 2
1b
650
650
650
900
900
Task (Site)
2
800
800
800
750
750
3a
750
750
750
800
800
3b
750
750
750
800
800
3a
3b
b) Obtain an optimal solution.
Obtain the following optimal solution with a minimum cost of $3600.
Assignee
North 1
North 2
North 3
South 1
South 2
1.
Excel Solver
2.
Python (+ Pyomo)
1a
X
1b
Task (Site)
2
X
X
X
X
import pyomo.environ as pyo
model = pyo.ConcreteModel()
model.x = pyo.Var(range(25), domain=pyo.NonNegativeIntegers)
costs = [
650, 650, 800, 750, 750,
650, 650, 800, 750, 750,
650, 650, 800, 750, 750,
900, 900, 750, 800, 800,
900, 900, 750, 800, 800,
]
supply = [1, 1, 1, 1, 1]
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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demand = [1, 1, 1, 1, 1]
model.OBJ = pyo.Objective(expr=sum([costs[i]*model.x[i] for i in
range(25)]), sense=pyo.minimize)
model.supply_constraint = pyo.ConstraintList()
model.demand_constraint = pyo.ConstraintList()
for i in range(5):
expr = [model.x[i] for i in range(5*i, 5*i + 5)] # constrain for
each supply row
expr2 = [model.x[i] for i in range(i, 25, 5)] # constrain for each
demand column
model.supply_constraint.add(expr=sum(expr) == supply[i])
model.demand_constraint.add(expr=sum(expr2) == demand[i])
Running pyomo solve pb3-ex1-pyomo.py --solver=glpk generates results.yml, which shows
Objective:
OBJ:
Value: 3600
Variable:
x[10]:
Value: 1
x[17]:
Value: 1
x[24]:
Value: 1
x[3]:
Value: 1
x[6]:
Value: 1
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
19
3.
MATLAB
% Performance measure coefficients
costs = [
650, 650, 800, 750, 750
650, 650, 800, 750, 750
650, 650, 800, 750, 750
900, 900, 750, 800, 800
900, 900, 750, 800, 800
];
% Demand constraints
a = [ones(1, 5) zeros(1, 20)];
demand_constraints = zeros(5, 25);
for i = 1:5
demand_constraints(i,:) = circshift(a, i*5);
end
% Supply constraints
supply_constraints = repmat(eye(5), 1, 5);
% Merge equality constraints
Aeq = [demand_constraints
supply_constraints];
lb = zeros(25,1); % non-negativity constraint
[x, zeval] = linprog(costs, [], [], Aeq, ones(10, 1), lb)
c)
Reformulate this assignment problem as an equivalent transportation problem with two
sources and three destinations by constructing the appropriate parameter table.
The parameter table for the formulation as an equivalent transportation problem is given below.
Source
North
South
Demand
1
650
900
2
Destination
2
800
750
1
3
750
800
2
Supply
3
2
d) Obtain an optimal solution for the problem as formulated in part (c).
We use the Excel Solver to obtain the following optimal solution with a minimum cost of $3600.
Source
North
South
Demand
1
2
2
Destination
2
1
1
3
1
1
2
Supply
3
2
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
20
Exercise 7 solution
Formulate and solve a spreadsheet model for Exercise 2.
1.
Excel
The following figure shows a spreadsheet model for this problem. The next figure shows the
equations entered in the target cell TotalCost (D23) and the other output cells Cost (E12:E21) and
NetFlow (H12:H16). After applying the Solver, the values of 1 in the changing cells OnRoute
(D12:D21) identify the shortest (least expensive) path for scheduling trade-ins.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
21
2. Python (+CPLEX)
from docplex.mp.model import Model
''' Parameters '''
N = [n for n in range(10)] # 10 variables
C = [5500, 10500, 17000, 25000, 5500, 10500, 17000, 5500, 10500, 5500]
''' Model creation '''
mdl = Model('LP')
x = mdl.continuous_var_dict(N, lb=0, name='x')
''' Constraints '''
mdl.add_constraint(x[0]+x[1]+x[2]+x[3] == 1)
mdl.add_constraint(x[4]+x[5]+x[6]-x[0] == 0)
mdl.add_constraint(x[7]+x[8]-x[1]-x[4] == 0)
mdl.add_constraint(x[9]-x[2]-x[7] == 0)
mdl.add_constraint(x[3]-x[6]-x[8]-x[9] == -1)
''' Performance measure '''
Z = mdl.continuous_var(lb=0, name='Z')
mdl.add_constraint(Z == sum(x[i] * C[i] for i in N))
mdl.minimize(Z)
mdl.print_information()
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
22
solution = mdl.solve()
print(solution)
print(solution._solve_details)
Which outputs
x_1=1.000
x_8=1.000
Z=21000.000
3. MATLAB
% Equality constraints
Aeq = [
1 1 1 1 0 0 0 0 0 0
-1 0 0 0 1 1 1 0 0 0
0 -1 0 0 -1 0 0 1 1 0
0 0 -1 0 0 -1 0 -1 0 1
0 0 0 -1 0 0 -1 0 -1 -1];
% Right side
beq = [1 0 0 0 -1];
% Performance measure parameters
f = [5500 10500 17000 25000 5500 10500 17000 5500 10500 5500];
lb = zeros(10,1); % non-negativity constraint
[x, zeval] = linprog(f, [], [], Aeq, beq, lb)
Exercise 8 Solution
1.
Using the corresponding algorithm, find the shortest path through the
network, where the numbers represent actual distances between the
corresponding nodes.
Optimization and Decision - PB #3 Transport, Assignment and Network Problems
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Optimization and Decision - PB #3 Transport, Assignment and Network Problems
24