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The Supply Chain

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Market research data

scheduling information

Engineering and design data

Order flow and cash flow

Supplier

Customer

Ideas and design to satisfy end customer

Material flow

Credit flow

Inventory

Supplier

Customer

Manufacturer

Inventory

Inventory

Supplier

Customer

Distributor

Inventory

- Facilities, functions, activities for producing & delivering product or service from supplier to customer
- Production planning
- Selecting suppliers
- Purchasing materials
- Identifying facility locations
- Managing inventories
- Distributing product

Given a set of facilities/locations, identify the shipping strategy that will minimize the total costs of distributing the product from the supply nodes to the demand nodes.

Supply nodes = sources where units are being sent from

Demand nodes = destinations where units are being received

Harbors (Sources)

Plants (Destinations)

$120

Amsterdam (500)

Leipzig (400)

$130

$62

$41

Nancy(900)

$61

$100

Antwerp (700)

$40

$110

Liege (200)

$90

$102.5

$122

Le Havre (800)

Tilburg (500)

$42

To

Supply

Leipzig

Nancy

Liege

From

Tilburg

120

130

41

62

500

Amsterdam

61

100

40

110

700

Antwerp

102.5

90

122

42

800

Le Havre

Demand

400

900

200

500

- Create a table for the unit shipping costs
- Create a table for the shipping quantity from each source to each destination (sink)
- Calculate the total shipped from each source
- Calculate the total received at each destination
- Calculate the total shipping cost for a shipping strategy
- Create the spreadsheet model for the problem using the template outlined in TPmodels.xls

- Total shipped from each source <= Supply
- Total received at each destination = Demand
- Amount shipped from each source to each destination >= 0
- Note: if you are minimizing transportation costs and you do not require the model to satisfy demand, Solver will choose to not ship anything and incur a cost of $0.

- A transportation problem is balanced if
Total supply at all of the sources =

Total demand at all of the destinations

- The KPilller transportation problem is currently balanced with Total Supply = Total Demand = 2000 engines
- In this case, all of the units are shipped from the sources (harbors) and all of the destinations (plants) receive their demand

- If Total supply at all of the sources >
Total demand at all of the destinations,

the problem is feasible. There will be unshipped units at some of the source locations though.

- (Resolve model with Nancy’s plant demand set equal to 700 engines)

Total demand at all of the destinations,

the problem will be infeasible.

- (Resolve model with Nancy’s plant demand set equal to 1000 engines)

- If the objective is changed to maximize profit or revenue, the problem can then be solved by changing the demand constraint:
- Total received at each destination <= Demand

- Note: with the new objective, Solver will choose to ship as much as possible. Not all demand will be satisfied even though the problem is now feasible. To ensure that all destinations receive a reasonable amount of product ,a minimum demand constraint can be added to the current demand constraint for each destination:
- Total received at each destination >= Minimum Acceptable Demand

- The model needs to be re-balanced in order to identify an optimal shipping strategy when minimizing costs remains the objective
- Include an extra source into the model to supply the current shortage or
- Set the minimal percent of demand at each destination that must be met to keep customer happy and yet does not result in an overall shortage (e.g. 95% of each plant’s demand) .

- Include an Extra Source to Supply the Current Shortage
- Extra capacity needed = Total demand at all destinations – Total supply at all current sources
- To create this additional source of supply/capacity, either
- Acquire a new facility/harbor and include it in the network design and spreadsheet model’s table structure or
- add a Dummy source into the model’s table structure

- In this problem, the total demand exceeds the total supply by 2100 – 2000 = 100 engines
- Insert a dummy harbor with a capacity of 100 engines and a unit shipping cost of $0 to each plant. Edit the spreadsheet model and Solver dialog box to include this new imaginary source.
- The identified optimal solution will identify how many engines to ship from each harbor to each of the plants. The engines shipped from the dummy harbor are units that will not actually be distributed; these are the amounts that the receiving plants will be short in the eventual distribution.

- In this problem, the total demand still exceeds the total supply by 2100 – 2000 = 100 engines
- Insert a possible location for a harbor with a capacity of at least 100 engines along with the identified unit shipping costs from this location to each plant. Edit the spreadsheet model and Solver dialog box to include the new harbor warehouse at this location.
- The identified optimal solution will identify how many engines to ship from each harbor, including the additional harbor at the new location, to each of the plants so as to minimize total costs

- How would you use the transportation model to identify whether Hamburg or Gdansk might be a better location for an additional harbor?
- What happens when you do not add a new “real location” to the network but use a dummy source which ends up shipping primarily to one destination? What can you do to resolve this problem?

- “Putting the Link in the Supply Chain”
- What type of models have we studied in this class to help you analyze this case? Sketch out the layout of the different models that you would need to integrate on a piece of paper.
- How would you link the models together?

- There are three types of nodes in network flow models:
- Supply
- Demand
- Transshipment

- Transshipment nodes can both send to and receive from other nodes in the network
- In transshipment models, negative numbers represent supplies at a node and positive numbers represent demand.

1

$120

7

Amsterdam (-500)

Leipzig (400)

$62

$90

6

$61

Nancy(900)

2

Antwerp (-700)

$40

$55

$110

5

Liege (200)

$90

$122

3

$60

Le Havre (-800)

4

Tilburg (500)

$42

For each arc in a network flow model

we define a decision variable as:

Xij = the amount being shipped (or flowing) from node ito node j

For example…

X14 = the # of engines shipped from node 1 (Amsterdam) to node 4 (Tilburg)

X56 = the # of engines shipped from node 5 (Liege) to node 6 (Nancy)

Note: The number of arcs determines the number of variables!

Minimize total shipping costs.

MIN: 62X14 + 120X17 + 110X24 + 40X25

+61X27 + 42X34 + 122X35 + 90X36

+ 60X45 + 55X56 + 90X67

For Minimum Cost Network Apply This Balance-of-Flow

Flow Problems Where:Rule At Each Node:

Total Supply > Total DemandInflow-Outflow >= Supply or Demand

Total Supply < Total DemandInflow-Outflow <=Supply or Demand

Total Supply = Total DemandInflow-Outflow = Supply or Demand

- In this illustration:
Total Supply = 2000

Total Demand = 2000

- For each node we need a constraint:
Inflow - Outflow = Supply or Demand

- Constraint for node 1:
–X14 – X17 = – 500

This is equivalent to: +X14 + X17 = 500

- Flow constraints
–X14 – X17 = –500} node 1

-X24 – X25 – X27 = -700} node 2

-X34 – X35 – X36 = -800} node 3

+ X14 + X24 + X34 – X45 = +500 } node 4

+ X25 + X35 + X45 –X56 = +200} node 5

+ X36 + X56 – X67 = +900} node 6

+ X17 + X27 + X67 = +400} node 7

- Nonnegativity conditions
Xij >= 0 for allij

- The sumif function adds the cells specified by a given condition or criteria
- =sumif(range, criteria, sum range)
- range are the cells which will be looked at to see if they meet a specified criteria
- Criteria specifies the conditions that will allow a row or column to be included in a sum calculation (i.e. = # or label, >0)
- Sum range are the cells that are to summed assuming the row or column meets the criteria

- Use cautiously as is not always linear

1

$120

7

Amsterdam (-500)

Leipzig (400)

$62

500

$90

400

6

$61

Nancy(900)

2

800

Antwerp (-700)

$40

$55

100

$110

300

5

Liege (200)

$90

$122

3

$60

Le Havre (-800)

4

Tilburg (500)

$42