1. Transportation Problem

2. Terminology Supply Point Demand Point Supply Constraints
Demand Constraints

3. Formulating Transportation Problems

5. Supply Constraints
Demand Constraints
All Xij ≥ 0

6. LP formulation of Powerco’s problem

7. LP Solution
Z = 1020

8. General Description of a Transportation Problem

9. Constraints

10. Balanced Transportation Problem
If total supply equals total demand, and the problem is said to be a balanced transportation problem

11. Balanced transportation problem

12. Solving Transportation Problems on the Computer
You know how to do it in Lindo
You may also solve it in Lingo
Note: Using statement, LINGO can read from a spreadsheet the values of data that are defined in the Sets portion of a program (see pg 370 of Winston Book).

13. defined by two sets of symbols: nodes and arcs
Network Analysis
defined by two sets of symbols: nodes and arcs

14. Network Problem SEERVADA PARK: sightseeing and backpack hiking
Park road system is shown in figure where nodes show locations of ranger stations
The numbers give the distances of these roads in miles
Scenic Wonder
A limited amount of sightseeing and backpack hiking. Cars are not allowed into the park, but there is a narrow, winding road system for trams and for jeeps driven by the park rangers. This road system is shown (without the curves) in Fig. 9.1, where location O is the entrance into the park; other letters designate the locations of ranger stations (and other limited facilities). The numbers give the distances of these winding roads in miles. The park contains a scenic wonder at station T. A small number of trams are used to transport sightseers from the park entrance to station T and back.
Entrance

15. The park management currently faces three problems
To determine which route from the park entrance to station T has the smallest total distance for the operation of the trams
This is shortest-path problem
Where the telephone lines should be laid to provide some connection between every pair of stations with a minimum total number of miles of line installed
This is an example of the minimum spanning tree problem
How to route the various trips to maximize the number of trips that can be made per day without violating the limits on any individual road
This is an example of the maximum flow problem

16. Terminology of Networks
Node
Arcs
Path
Directed and undirected arcs
A connected network is a network where every pair of nodes is connected
A network with n nodes requires only (n - 1) links to provide a path between each pair of nodes
A path between two nodes is a sequence of distinct arcs connecting these nodes.

18. Shortest Path Problem
The objective is to find the shortest path (the path with the minimum total distance) from the origin to the destination
Problem: The Seervada Park management needs to find the shortest path from the park entrance (node O) to the scenic wonder (node T) through the road system

19. The minimum spanning tree problem
Chose links that provide a path between each pair of nodes
To find the spanning tree with a minimum total length of the links

20. Maximum Flow Problem
To transport the maximum amount of flow from a starting point (called the source) to a terminal point (called the sink).
How to route the various tram trips from the park entrance (station O in Fig. 9.1) to the scenic wonder (station T) to maximize the number of trips per day

21. Minimum Cost Flow Problem
All previously studied problems are special cases of the minimum cost flow problem
Can be solved so efficiently is that it can be formulated as a linear programming problem

22. Network Model

23. Network Methods
Used for the planning, management and control of projects
Project activities are described by a network
Any project has a beginning and an end
A project can be broken down into:
Activities (tasks/jobs) with their associated completion times
Precedence relationships (certain activities must be finished before other can start)
Two widely used project management techniques
PERT (Program Evaluation and Review Technique)
CPM (Critical Path Method)

24. Project Management Techniques
PERT
Ability to deal with uncertain activity completion times (allows randomness)
CPM
Deterministic method
Time vs. cost trade off
Combined PERT and CPM
Uncertain activity completion times
Project completion time/project cost trade-offs analysis

25. How long would it take to complete a project?
Need to know
The completion time of each activity (CT)
What activities must be finished before an activity can start (precedence relationship )
Combine them into a network (the reason it is called a network analysis)
The collection of series and parallel tasks can be modeled as a network

26. C1 = review of A (4 days) C2= review of B (10 days)
Example : Writing of a Research paper by 3 authors (2 students and their advisor)
A- write the background and the literature review (CT=10 days) by author 1
B- write the remaining paper (CT=30 days) by author 2
C- review by the advisor
C1 = review of A (4 days)
C2= review of B (10 days)
D- submit the paper
Option 1: = 54
Option 2: start A and B simultaneously, when A is finished start C1. A and C1 would take 10+4=14 days while B would take 30 days. After the 30th day start C2. Total time to complete the paper = =40 days

27. Network Planning Describe the project network Activity-On-Node (AON)
Activity = Task that must be performed
Event = Milestone marking the completion of one or more activities
Activity-On-Node (AON)
Nodes are activities
Arrows indicate the precedence relationship
Used frequently in practical non-optimization situations
Activity-On-Arc (AOA)
Arcs represent activities
Nodes are milestones (starting and ending points)
Used in optimization settings

28. Network Scheduling Determine the Critical Path (CP)
The expected length of each path is equal to the sum of the expected durations E(T) of all the activities on each path
E(T) = (a+4m+b)/6
Where;
a = Optimistic time
b = Pessimistic time
m = Most likely time
The critical path is the path with the longest expected length
E(T) = (a+4m+b)/6 (follow beta distribution)
The Beta distribution models events which are constrained to take place within an interval defined by a minimum and maximum value.  For this reason, the Beta distribution is used extensively in PERT, CPM and other project planning/control systems to describe the time to completion of a task.

29. Network Control
Monitor the project progress based on network scheduling
Take correction actions if required

30. AON Forward Pass (FP) Backward Pass (BP) Total Slack (TS)
During FP it is assumed that each activity will begin at its earliest starting time (ES)
An activity can begin as soon as its predecessors have finished
Completion of FP determines the earliest completion time (EC) of the project
Backward Pass (BP)
Determines the latest completion time (LC) for each activity
The latest start time (LS) for an activity is its LC minus the activity duration
Total Slack (TS)
Amount of time an activity may be delayed from its ES without delaying the LC of the project
TS = LS(i) –ES(i) OR LC(i) – EC(i)
Any activity with a total slack of zero is a critical activity
A path from start to end nodes consisting entirely of critical activities is a critical path

31. Immediate Predecessor
Example: Determine the CP and the project duration for the given problem
Activity
Completion Time
Immediate Predecessor A 3 -- B 4 C 1 D 6
A, B E 7
D, C F 10 G 5
E, F

32. Critical Path AON Example (3,4) (0,3) (4,14) (0,0) (6,7) (1,4) (7,17)
F, 10
(0,0)
(6,7)
(1,4)
(7,17)
Start
(22,22)
(17,22)
(10,17)
(0,0)
(0,4)
(4,10)
G, 5
End
E, 7
B, 4
D, 6
(10,17)
(17,22)
(22,.22)
(0,4)
(4,10)
Critical Path

33. AOA Representation Arc (i, j) = Activity (i, j)
Node
Arc
Arc (i, j) = Activity (i, j)
Node i = Immediate predecessor node of arc (i, j) or start node for the activity
Node j = Immediate successor node of arc (i, j) or end node for the activity
Note: Nodes are numbered sequentially such that the ending node of an activity has a higher number than the starting node

34. AOA Notation
ET(i) = Earliest time at which the event corresponding to node ‘i’ can occur
LT(i) = Late event time at which the event corresponding to node ‘i’ can occur without delaying the completion of the project
tij = Duration of activity (i, j)
TF(i, j) = Total float (slack). Amount by which the starting time of activity (i, j) could be delayed beyond its earliest possible starting time without delaying the project (An activity with a TF=0 is a critical activity)
TF(i, j) = LT(j) - ET(i) - tij
FF(i,j)= Free Float. Amount by which the starting time of activity (i,j) can be delayed w/o delaying the start of any later activity beyond its earliest possible starting time
FF(i, j) = ET(j) - ET(i) - tij
Dummy Arc: An activity network allows only one arc between any two nodes
If there is a same predecessor and same immediate successor for many activities use dummy arc

35. Rules in AOA Network
Node 1 represents the start of the project. An arc should lead from node 1 to represent each activity that has no predecessors.
A node (called the finish or end node) representing completion of the project should be included in the network.
Number the nodes in the network so that the node representing the completion of an activity always has a larger number that the node for the start of an activity
An activity should not be represented by more than one arc in the network
Two nodes can be connected by at most one arc
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

36. AOA: Optimization
LP formulation
Let
Xj = time that the event corresponding node j occurs
Tij = time to complete activity (i, j)
Constraints
Xj ≥ Xi + tij for all (i, j)
LP Objective function
MINIMIZE XF-XI
I = index of start node
F = index of finish node

37. AOA Example 4 7 2 6 1 3 5 ET(j) ET(i) i j LT(i) LT(j) (4) (17) (3)
(22)
C (1)
F (10)
G (5) 4 7 2 6
A (3)
(0)
(4)
(7)
(17)
(22) 1
E (7)
B (4)
(4)
(10)
(0)
D (6) 3 5
(4)
(10)
ET(j)
ET(i) i j
LT(i)
LT(j)

38. AOA Optimization X2 ≥ X1 + 3 X6 ≥ X5 + 7 X3 ≥ X1 + 4 X6 ≥ X4 + 10
Min X7 - X1
S.T.
X2 ≥ X1 + 3
X3 ≥ X1 + 4
X4 ≥ X2 + 1
X5 ≥ X3 + 6
X3 ≥ X2 + 0
X5 ≥ X4 + 0
X6 ≥ X5 + 7
X6 ≥ X4 + 10
X7 ≥ X6 + 5

39. Examples 1- Shortest Path
Xij = millions of barrels of oil per hour that will pass through arc (i,j) of pipeline
source node
Sink Node
Artificial Arc

40. Feasible flow? Two conditions to be satisfied
0 ≤ flow through each arc ≤ arc capacity
Flow into node i = flow out of node I
let x0 be the flow through the artificial arc
conservation of flow implies that x0 = total amount of oil entering the sink
Sunco’s goal is to maximize x0 subject to (1) and (2):
We assume that no oil gets lost while being pumped through the network, so at each node, a feasible flow must satisfy (2), the conservation-of-flow constraint.
The introduction of the artificial arc a0 allows us to write the conservation-of-flow constraint for the source and sink.

41. LP

42. Using LP to find a CP in an AOA Network:
Activity
Description
Immediate Predecessors
Duration (days) a
Train workers -- 6 b
Purchase new materials
--- 9 c
Produce product 1
a, b 8 d
Produce product 2 7 e
Test Product 2 10 f
Assemble products 1 & 2 into new product 3
c, e 12
Example: Introducing new product f c 5 4 6 3
Source ENCE 667 (UMD) Professor Gabriel’s class notes. b 1 d e
Problem???? a 2 4 5

43. LP Min X6 – X1 S.t. X2 ≥ X1 + 6 (arc 1-2) X3 ≥ X1 + 9 X3 ≥ X2 + 0
Variables unrestricted in signs
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

44. Lindo Input File

45. Lindo Output File Project completed in 38 days
In general, the value of Xi may assume any value between ET(i) and LT(i)
Critical path goes from start to finish node in which each arc corresponds to a constraint with dual price =-1

46. Critical Path f c 4 5 6 3 b 1 d e a 2 4 5 CP

47. Time-cost Tradeoff in Scheduling
Project Crashing: to expedite the project schedule
Project crashing is usually done at a cost
Tradeoff between more cost and a shorter schedule
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

48. Project crashing and time-cost Analysis
Time-cost tradeoff can be analyzed using the following LP
Min 𝑖, 𝑗 ( bij – aij yij)
S.t.
xi + tij ≤ xj (for all i & j)
xn = T
dij ≤ yij ≤ uij (for all i & j)
aij = unit cost of crashing activity
bij =fixed cost of crashing activity
tij =duration of activity
dij =crash duration of activity
uij =normal duration of activity
xi =realization time for node i
xn =realization time for last node n
T = target time (deadline)
yij = original duration – crash varaible
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

49. Redo Example: Introducing new product
Crashing detail
No fixed cost for crashing (bij = 0)
Activities can be crashed at most 5 days beyond normal crashing time (yij =5)
Target completion time T = 25 days
Costs of crashing individual activities are as follows
A = # of days by which activity a is reduced (unit cost =$10)
B = # of days by which activity b is reduced (unit cost =$20)
C = # of days by which activity c is reduced (unit cost =$3)
D = # of days by which activity d is reduced (unit cost =$30)
E = # of days by which activity e is reduced (unit cost =$40)
F = # of days by which activity f is reduced (unit cost =$50)
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

50. LP: Project Crashing and Time-cost Analysis
Min 10A + 20B + 3C + 30D + 40E + 50F
S.t.
A ≤ 5
B ≤ 5
C ≤ 5
D ≤ 5
E ≤ 5
F ≤ 5
x2 ≥ x1 + 6 – A (arc 1-2 constraint)
x3 ≥ x B
x3 ≥ x2 + 0
x4 ≥ x D
x5 ≥ x C
x5 ≥ x E
x6 ≥ x – F
x6 – x1 ≤ 25
A, B,…..F ≥ 0
Xj urs
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

51. Lindo Input file

52. Lindo Output file

53. f c 5 4 6 3 b 1
Normal AOA network
Total Project time = 38 days d e a 2 4 5 f c 5 4 6 3 b
Crashed AOA network
Total Project time = 25 days 1 d e a 2 4 5 CP

54. Scheduling under limited resources
Assignment Problem
Min 𝑖=1 𝑛 𝑗=1 𝑛 𝐶𝑖𝑗 𝑥𝑖𝑗 Min Total cost
s.t.
𝑗=1 𝑛 𝑥𝑖𝑗 = 1, i = 1,2, ……n ith worker must get assigned to exactly 1 task
𝑖=1 𝑛 𝑥𝑖𝑗 = 1, j = 1,2, ……n jth task must get assigned to exactly 1 worker
Xij ≥ 0 , i ,j = 1,2,…….n
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

55. Example
Consider the following cost ($ matrix for an assignment problem with n=5
Task 1
Task 2
Task 3
Task 4
Task 5
Worker 1
200
400
500
100
Worker 2
700
800
1,100
Worker 3
300
900
1,000
Worker 4
Worker 5
Source ENCE 667 (UMD) Professor Gabriel’s class notes.

56. LP Min 200x11 + 400x12 + 500x13…………………………..200x55 S.t
x11+x12 +…..x15 = !worker 1 …. …
x51+x52……..x55 = ! Worker 5
x11+ x21 + x31 + x41+ x51 = 1 ! Task 1
…..
x15 + x25 + x35 + x45 + x55 = 1 ! Task 5
All variables non-negative

57. Solution Optimal solution at a cost of $1,500 is as follows Task 1
Worker 1
Worker 2
Worker 3
Worker 4
Worker 5
Source ENCE 667 (UMD) Professor Gabriel’s class notes.