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Supply chain and logistic optimization

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Supply chain and logistic optimization

Road Map

- Definition and concept of supply chain.
- Primary tool box at strategic level (software).
- Models at strategic level.
- Models at tactical level.

- A box of cereal spends, on average, more than 100 days from factory to sale.
- A car spend around 2 weeks from factory to dealer.

- National Semiconductor (USA)used air transportation and closed 6 warehouses, 34% increase in sales and 47% decrease in delivery lead time.

- Compaqestimates it lost $0.5 billion to $1 billion in sales in 1995 because laptops were not available when and where needed.
- When the 1 gig processor was introduced by AMD (Advanced Micro Devices), the price of the 800 meg processor dropped by 30%.
- P&Gestimates it saved retail customers $65 million (in 18 months) by collaboration resulting in a better match of supply and demand.

- IBM claims that it lost a major market share for desktops in 93, for not been able to purchase enough of a display chip.
- US companies spent $898 B for SC activities in 98. Out of the above 58% of SC costs were incurred for transportation and 38% for inventory.

- Less time in storage– less deteriorate-High quality.
- Effective distribution process (Fast delivery to customers)
- Switching from old technology to new technology without scraping lot of products.
- Of course less storage cost.

Gartner Group:

“By 2004 90% of enterprises that fail to apply supply-chain management technology and processes to increase their agility will lose their status as preferred suppliers”.

AMR Research: “The biggest issue enterprises face today is intelligent visibility of their supply chains – both upstream and down”

Supplier

Retailer

Demand

Retailer

Demand (Q)= A - B*Z

Assume A = 120

B = 2

Supplier buys a goods at price X

Sells to retailer at a price Y

Retailer sells to customer at a price Z

Retailers profit (R) = (Z-Y) * (A-B*Z)

Retailers profit (R) = (Z-Y) * (A-B*Z)

Y=40

Suppliers profit = Q * (Y-X)

Combined profit

CP=0.25 (A2 / B -2 . A .X + 2 . B. X. Y – B. Y2)

Traditional scenario

Demand 20 units

Retail price $ 50.0

Retailer’s profit $ 200.0

Supplier’s profit $ 200.0

Collaborative scenario

Demand 30 units

Retail price $ 45.0

Retailer’s profit $ 225.0

Supplier’s profit $ 225.0

10000

Demand

P=2000-0.22Q

2000

Price

- A Retailer and a manufacturer.
- Retailer faces customer demand.
- Retailer orders from manufacturer.

Demand Curve

Variable Production Cost=$200

Selling Price=?

Manufacturer

Retailer

Wholesale Price=$900

- Retailer profit=(PR-PM)(1/0.22)(2,000 - PR)
- Manufacturer profit=(PM-CM) (1/0.22)(2,000 - PR)
Retailer takes PM=$900

Sets PR=$1450 to maximize (PR -900) (1/0.22)(2,000 - PR)

Q = (1/0.22)(2,000 – 1,450) = 2,500 units

Retailer Profit = (1,450-900)∙2,500 = $1,375,000

- Manufacturer takes CM=variable cost
Manufacturer profit=(900-200)∙2,500 = $1,750,000

- Case with $100 discount
New demand

Q = (1/0.22) [2,000 – (PR-Discount)] = (1/0.22) [2,000 – (1450-100)] = 2954

Retailer Profit = (1,450-900)∙2,955 = $1,625,250

Manufacturer profit=(900-200-100)∙2,955 = $1,773,000

- $100 wholesale discount to retailer
- Retailer takes PM=$800
Sets PR=$1400 to maximize (PR -800) (1/0.22)(2,000 - PR)

Q = (1/0.22)(2,000 – 1,400) = 2,727 units

Retailer Profit = (1,400-800)∙2,727 = $1,499,850

- Manufacturer takes PR=$800 and CM=variable cost
- Manufacturer profit=(800-200)∙2,727 = $1,499,850

- What happens if both collaborate (SCM)?
Manufacturer sets PR=$1,100 to maximize (PM -200) (1/0.22)(2,000 - PM)

Q = (1/0.22)(2,000 – 1,100) = 4,091 units

Net profit =(1100-200)∙ 4,091 = $3,681,900

- Low price.
- High quality.
- Product customization.
- Fast delivery.
- Fast technology induction.

Strategy: Right product + Right quantity + Right customer + Right time

Cost reduction and value addition at each stage

Sharing information will lead to reducing uncertainty for all the partners and hence:

- reducing safety stocks,

- reducing lead times,

- improving brand name.

Result

- Value added product/services

- large market segment (Leading brand),

- long lasting supply chain,

- consistent and growing profits,

- Satisfied customer.

It is better to collaborate and co-ordinate to achieve a win-win situation.

Definition

Supply Chain refers to the distribution channel of a product, from its sourcing, to its delivery to the end consumer.

(some time referred as the value chain)

A supply chain is a global network of organization that cooperate to improve the flows of material and information between suppliers and customers at the lowest cost and the highest speed.

Objective is customer satisfaction and to get competency over others

Information flow

Material Flow

Cooperation

Value addition

Customer satisfaction

Cost

Value added product

Operational performance

Coordination

Information sharing

Assets Management (Inventory)

- Increasing product variety.
- Shrinking product life cycle.
- Fragmentation of supply chain.
- Soaring randomness because of new comers.

Responsiveness

(High cost)

Zone of strategic fit

Low cost

DeterministicUncertainty

FMCG (Tooth paste, soaps etc )

Price sensitive, Low uncertainty

Fashionable products (Garments), customize products etc

High responsiveness, high uncertainty

Ineffective marketing

Wrong material

High Inventories

Low order fill rates

Supply shortages

Inefficient logistics

High stock outs

Local objectives vs Global objective

- Production
- - Low production cost.
- High utilization.
- High quality raw material.
- Big order size.
- Stable production.
- Short delivery lead times (Raw materials)

- Marketing
- - High inventories levels.
- Low prices.
- Ability to accept every customer order.
- Short product delivery lead times.
- Various order sizes and product mixes.

SCM is a tool which integrate necessary activities and decomposing irrelevant activities.

Management by departments

Management by projects

Finished products

Orders

Primary decisions

Secondary decisions

Buy

Sell

Make

Move

Store

Recruitment of employees

Salary, TA, DA

Training, relocation, etc

Selection of providers

- Provider provides the quantity between two limits.
- Cost is a concave function of the shipped quantity.
- Objective is to select one or more provider to satisfy the demand.

- Case of single manufacturing unit.
- Case of several manufacturing unit.

Cost function

Problem to be solved

- Algorithm
- Select the cheapest provider if the quantity to be supplied is maximum.
- Adjust the remaining quantity among rest of the providers so that the solution remain feasible.

- Compute the cost for the first provider for q=1,2,…A.
- For provider=2,…N
- Compute the Q=1,2,…,A

F(Pro,Q)=Min0<=x<=Q (F(Pro-1, Q-x), fpro(x)

Case of several manufacturing units:

- Heuristic approach.
- Piece wise linearization for integer programming.
- Bender decomposition.

Providers

Retailers

Manufacturers

- Idle processing capacity with providers in different periods.
- Idle transportation capacity with providers in different periods.
- Idle manufacturing capacity with manufacturers.
- Idle transportation capacity with manufacturers.
- Demand at various retailers and the demand of one period may differ from another period.

Objective is to decide how much and where to invest.

Supply <= Available Capacity + Added Capacity

Added capacity <= BigNumber* Binary Variable {0,1}

Investment Cost

BinaryVariable*FixedCost + Slope*Added capacity

- Results
- There exist at least one optimal solution in which all the binary constraints are saturated.
- Replacing big number by corresponding capacity, if greater than zero, and denote new problem by P2m, then m is definite.

- Construct the several instances of the problem from the relaxed solution of the problem . These instance converge towards the solution of the problem P. Unfortunately, we do not know the conversion time.
- For this reason, we derive sub-optimal solution from the instances and select the best solution.

- Langrangean heuristic approach to find good lower bound.
- Branch and price approach.
Presented approach was similar to the langrangean approach.

- Multi-echelon system.
- Selection of a partner from each echelon.
- Expected demand is known for a given horizon.
- Objective is not to invest at any location.
- Utilization of idle capacity (Production, transportation, storage etc).

- Solution should be feasible for entire horizon.
- Decision: Whether the new chain exists and profitable?

Demand

- Storage cost at entry and at exit (running).
- Production cost (running).
- Connection cost (Fixed)

Note: It is possible that solution may not exist

Minimisation

- Resolve two echelon problem for each pair of nodes using final demand.
- Consider the cost corresponding to these arcs as surrogate length.
- Solve the k-shortest path problem and compute the k-shortest path.
- For each shortest path, compute the real cost.
- If the relax path length is bigger than best real cost, stop.

D1

D3

D2

D1

D3

D2

Demand

1

2

3

1

2

3

1

5

Problem data

1

(2, 1)

2

(5,1)

3

(2,0)

Solution

1

3

Formulation

Min

s.t.

An optimal algorithm is known for the above case.

Simple assembly

Algorithm

Computing two times: Early start time – Latest start time

Select the common interval.

3

4

1

7

5

6

14

2

8

9

12

13

10

11

- Simple- easy to program but time consuming.
- Little tidy – difficult to program but on average performance is better.
- Both gives the optimal schedule.
- Worst case complexity is also same.

S1

1

3

4

7

14

S2

2

3

4

7

14

S3

8

9

12

13

14

S4

5

6

7

14

S5

10

11

12

13

14

- For each assembly operation
- If the idle windows for two different lines are not the same then select the window which has higher lower limit (beginning time) and restart the calculation.
- If the windows are the same but starting time are different, then set the lower limit of this window as the greatest starting time of the two and restart the calculation.

If neither of the above case is present, then the solution is optimal.

The algorithm converge towards an optimal solution.

- Decompose the assembly into sub-assemblies.
- Solve the sub-assemblies.
- Coordinate the timing of sub-assemblies.

- Recursive approach.
- Each time early start time (EST) and latest start time (LST) of assembly has to be computed.
- Advantage:
- If the time lies between EST and LST then re-computation is not required.

3

4

1

7

5

6

14

2

8

9

12

13

10

11

3

4

7

1

5

6

5

2

7

8

9

14

12

13

10

11

13

WIP Control (Extension)

First case

Number of finished jobs at the exit of each machine is not limited in quantity but limited by time.

Approach

Introduce one virtual machine, following the real machine with operation time 0 and flexibility [0, T]

- Case 2
Number of finished jobs are limited in time and in quantity too.

Approach

Introduce as many virtual machines as the number of finished jobs permitted.

In both the cases, the algorithm presented before are applicable.

Delivery date

Backlogging cost

Inventory holding cost

Instant of ordering

Delivery instant

Three cots are to be considered

I1, Inventory cost between the arrival of first component and last component.

I2, Inventory cost between the arrival of last component and the delivery date.

B, Backlogging cost if the last component arrives after the delivery date.

Next

is continuous and differentiable in [R, +∞]

Basic property

1

1

Propriété fondamentale:

- Algorithme général
- Partir d’une solution admissible.
- Chercher une solution admissible R1 du voisinage de R qui vérifié (1)
- Conserver ou rejeter R1 (recuit simulé)
- Retour à 2.

- Algorithm
- Start with one feasible solution.
- (First feasible solution can easily be generated considering the same ordering time for all components.)
- 2. Define new solution R1in the neighborhood of R that satisfies relation (1) (Use gradient method)
- 3. Conserve or reject R1 (Simulated annealing)
- 4. Go to 2.

The behavior of cost I1 depends upon the density function. The I1 may be convex or concave based on the nature of density functions. Hence, with an exact information of density functions an specialize algorithm could have been devised.

With uniform densities, the problem can be solved using gradient method only.

For general problem we proposed an approach based on simulated annealing which looks, in each iteration, for a closer solution which satisfies the relation 1.

Assumptions

- Customer is price sensitive.
- Average customer demand depleted as price increases.
- Customer demand is stochastic.
- Backordering cost at supplier is higher than at retailer.
- Inventory cost at supplier is cheaper than retailer.
- Inventory can be transferred between supplier and retailer in negligible time. In other words customer is ready to wait during the lead time.

Example

Model

- Each maintain a stock of Ip and Ir.
- Retailer pays for the holding and also pays for stock out.
- Supplier pays for holding and also pays for stock out. This stock out penalty goes to retailer (Compensation).
- Profit is shared according to their relative risk i.e. investment.

Objective is to maximize total benefit.

Fractional demand is given by f(wr), a concave function of selling price.

Fractional demand

Lost sell due to high price

Price wr

- Take any starting price wr
- 2. Optimize Ip and Ir => Ip and Ir , keepingWr fix.
- Total profit function is convex w.r.t stocks and wr constant.
- 3. Optimize Wr, Ip and Ir are fixed.
- Combined profit function is concave w.r.t Wr.
- 4. Go to 2 until profit increases.

- Dynamic pricing
- Online bidding, (Ebay.com)

- Price setting and discount for perishable goods in supermarket.

- Customize pricing: different prices for different customer segment.

- Inventory management in advance demand information sharing.