Supply chain and logistic Supply chain and logistic optimizationoptimization

Definition and concept of supply chain.

Primary tool box at strategic level (software).

Models at strategic level.

Models at tactical level.

Road Map

Potential of SCM

• 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.

• Compaq estimates 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&G estimates 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.

Advantage of low inventories• 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”

Why supply chain: A tutorial

Supplier RetailerDemand

Retailer

Traditional scenario

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=40Price Demand Profit

40 40 041 38 3842 36 7243 34 10244 32 12845 30 15046 28 16847 26 18248 24 19249 22 19850 20 20051 18 19852 16 19253 14 18254 12 16855 10 15056 8 12857 6 10258 4 7259 2 3860 0 0

Profit

0

50

100

150

200

250

Z

Pro

fit

Profit

Suppliers profit = Q * (Y-X)

selling cost supplier profit30 031 2932 5633 8134 10435 12536 14437 16138 17639 18940 20041 20942 21643 22144 22445 22546 22447 22148 21649 20950 20051 18952 17653 161

Suppliers profit

0

50

100

150

200

250

Selling price

pro

fit

Suppliers profit

Retailers profit w.r.t. Y

selling price (Y) Retailer profit36 28837 264.538 24239 220.540 20041 180.542 16243 144.544 12845 112.546 9847 84.548 7249 60.550 5051 40.552 3253 24.555 12.556 857 4.558 259 0.560 0

Retailer prof it

0

50

100

150

200

250

300

350

400

450

500

Y

Pro

fit

Retailer prof it

Combined profit

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

Y SC profit38 41839 409.540 40041 389.542 37843 365.544 35245 337.546 32247 305.548 28849 269.550 25051 229.552 20853 185.554 16255 137.556 11257 85.558 5859 29.560 0

Supply chain

0

50

100

150

200

250

300

350

400

450

500

30 33 36 39 42 45 48 51 54 57 60

Y

Co

mb

ined

pro

fit

Supply chain

Retailers profit w.r.t. Y

Profit

0

100

200

300

400

500

30 32 34 36 38 40 42 44 46 48

Z (Retail price)

Selling priceZ Profit30 031 5832 11233 16234 20835 25036 28837 32238 35239 37840 40041 41842 43243 44244 44845 45046 44847 44248 43249 418

Summery

Demand 20 units

Retail price $ 50.0

Retailer’s profit $ 200.0

Supplier’s profit $ 200.0

Demand 30 units

Retail price $ 45.0

Retailer’s profit $ 225.0

Supplier’s profit $ 225.0

Collaborative scenario

Traditional scenario

Example

• A Retailer and a manufacturer.– Retailer faces

customer demand.– Retailer orders from

manufacturer.

Selling Price=?

Wholesale Price=$900

Retailer Manufacturer

Variable Production Cost=$200

Demand

Price

10000

2000

P=2000-0.22Q

Demand Curve

Example• 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 costManufacturer profit=(900-200)∙2,500 = $1,750,000

Example: Discount

• Case with $100 discountNew 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

Wholesale discount

Global Optimization

• 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

Strategy Comparison

Strategy Retailer Manufacturer TotalNo discount 1,370,096 1,750,000 3,120,096 Wth ($100) discount 1,625,250 1,773,000 3,398,250 Discount to retailer only 1,499,850 1,499,850 2,999,700 SCM scenario - - 3,681,900

Market Requirement

• 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.

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)

Definition

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

Supply chain

Material Flow

Cooperation

Value addition

Information flow

Assets Management (Inventory)

Information sharing

Coordination

Co

st

Operational perform

ance Val

ue

add

ed p

rod

uct

Customer satisfaction

Difficulties

• Increasing product variety.

• Shrinking product life cycle.

• Fragmentation of supply chain.

• Soaring randomness because of new comers.

Deterministic Uncertainty

Responsiveness

(High cost)

Zone of stra

tegic fit

Low cost

FMCG (Tooth paste, soaps etc )

Price sensitive, Low uncertainty

Fashionable products (Garments), customize products etc

High responsiveness, high uncertainty

Low order fill rates

Supply shortages

Wrong material

High Inventories

Inefficient logistics High stock outs

Ineffective marketing

Local objectives vs Global objective

Marketing

- High inventories levels.

- Low prices.

- Ability to accept every customer order.

- Short product delivery lead times.

-Various order sizes and product mixes.

Production

- Low production cost.

- High utilization.

- High quality raw material.

- Big order size.

- Stable production.

-Short delivery lead times (Raw materials)

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

Management by departmentsM

anag

emen

t by

proj

ects

Orders Finished products

Primary decisions

Secondary decisions

Store

Buy

Move Make

Sell

Recruitment of employees

Training, relocation, etc Salary, TA, DA

Strategic models

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.

Exact algorithm for the integer demand

1. Compute the cost for the first provider for q=1,2,…A.

2. 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:

Approches

• Heuristic approach.

• Piece wise linearization for integer programming.

• Bender decomposition.

Capacity Planning

RetailersProviders

Manufacturers

1. Idle processing capacity with providers in different periods.

2. Idle transportation capacity with providers in different periods.

3. Idle manufacturing capacity with manufacturers.

4. Idle transportation capacity with manufacturers.

5. 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

Formulation

Results

1. There exist at least one optimal solution in which all the binary constraints are saturated.

2. 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.

Approach

...,P,P 12

02

1P

Algorithm Optimal solution Error

1. 276518 275781 0.267

2. 257359 257359 0.0

3. 324934 322973 0.607

4. 292007 292007 0.0

5. 354365 354116 0.070

6. 322126 325881 1.910

7. 334013 333499 0.154

8. 321299 319871 0.446

9. 291253 290190 0.366

10. 310869 310869 0.0

Other approaches

- Langrangean heuristic approach to find good lower bound.

- Branch and price approach.

Presented approach was similar to the langrangean approach.

Short term supply chain formation

• 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

Costs

• Storage cost at entry and at exit (running).

• Production cost (running).

• Connection cost (Fixed)

Note: It is possible that solution may not exist

Minimisation

Path relaxation approach

1.Resolve two echelon problem for each pair of nodes using final demand.

2.Consider the cost corresponding to these arcs as surrogate length.

3.Solve the k-shortest path problem and compute the k-shortest path.

4.For each shortest path, compute the real cost.

5. If the relax path length is bigger than best real cost, stop.

D1 D2 D3

D1 D2 D3

Demand

Insertion of new project

1 2 3

1

2

3

Problem data

Solution

1

3

1(2, 1)

2(5,1)

3(2,0)

5 1

Formulation

An optimal algorithm is known for the above case.

Min 1mx

s.t.

Simple assembly

1. Computing two times: Early start time – Latest start time

2. Select the common interval.

Algorithm

12

5

37

6

4

14

13

1

10 11

9

2

8

Complex schedule

Two approaches

• 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.

Simple approach

1 14743S1

10 14131211S5

14765S4

8 1413129S3

2 14743S2

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.

Second approach

1. Decompose the assembly into sub-assemblies.

2. Solve the sub-assemblies.

3. 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.

Second approach

12

5

37

6

4

14

13

1

10 11

9

2

8

51

2

5

37

6

4

12 13

10 11

98

7

14

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.

Inventory holding cost

Backlogging

cost

Instant of ordering

Delivery instant

Delivery date

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

BI)Z(W 2 is continuous and differentiable in [R, +∞]

n

kn

kk

kksh

h)rZ(F1

1

1

Basic property

Propriété fondamentale:

n

kn

kk

kksh

h)rZ(F1

1

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.

1

n, r...,,rrR 21

Algorithm

1. Start with one feasible solution.

(First feasible solution can easily be generated considering the same ordering time for all components.)

2. Define new solution R1 in 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.

Partnership formation: A model

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.

Fra

ctio

nal d

eman

d

Price wr

Lost sell due to high price

Algorithm

1.Take any starting price wr

2. Optimize Ip and Ir => Ip and Ir , keeping Wr 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.

Numerical illustration

Continue

Hot areas in SCM

• 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.