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GILI TEST July 15th, 2014

EOQ and Savings Matrix

 INTRODUCTION TO  THE TEST  OF JULY 15th, 2014

In the following the solution of the test is provided. The subject concern the Economic Order Quantity, the reorder point s and Savings Matrix Methos. 

Please refer to Silver, Pyke, Peterson, Inventory Management and Production Planning and Scheduling, 1998 for any further information or to Nahmias, Production and Operations Analysis, 2013

 EXERCISE:

A Distribution Center supplies products (in packages) to a set of 20 clients.

Package weekly demand is normal distributed for each client with average and standard deviations shown in the following table. Weekly days in a week are 5 days/week. The price v of each package is 20 €. A, the order setup cost is 10 €/order. Order lead time is 1 day. r, the carrying cost interest rate on daily basis is 10% €/€-day.

The total amount of money allocated for the safety stocks is 230 €.

 

Determine:

A.      EOQ for each client (approximate EOQ to the nearest integer)

B.The reorder point s of each client, by imposing the same Client service Level P1 (approximate s to the nearest integer)

At the end of each day all the clients with inventories below or equal to the reorder point orders and EOQ to the Central Depot.

C.      By considering the Present values of inventory in the table, find the clients that need to launch and order to the Distribution Center. (hint: the total of the packages to be order is less than 50).

D.      Determine the routing by using the saving matrix methods, considering that the Distribution Center can use 2 trucks of 25 packages of capacity each one. In the following table the matrix of distances between the 20 clients and the Distribution Center is shown. 

 

QUESTIONS: 

a)Write a Linear programming model for the optimization of the Aggregate Planning, taking into account workers, people hiring, layout, material costs, inventory carrying costs, stock out costs (considering a backorder model), overtime costs, production level, outsourcing costs. 

b)Show the rule for determine the k coefficient of safety stocks in the case a model cost B2 is considered.

c)What is it the Mix Production capacity?


 

a)

 ANSWER TO POINT a

  A.      Calculate the EOQ for each client:

 

A = 10 €

D [day] = D/5

r = 0.1 €/€/day

 

v = 20 €

 

Repeat the same for all the other 19 clients.


 

 ANSWER TO POINT b 

A.     B. Calculate the Standar Deviations of Lead Time for all the 20 clients:

By knowing that the total amount of safety stocks is 230€, we can find k:

 

It is the possible to calculate the safety stocks for each client:

You can calculate the reorder point for each client:


 

 ANSWER TO POINT c

C.      Routing and sequencing will  include only the clients that launch an order, the clients with the present Inventory level below or equal the reorder point. Those are clients number:: 4, 5, 10, 13, 17, 18 e 19. The DC deliveres to those clients the order quantities (equal to EOQs). Total packages to be delivered are 48, less than 50.


 

 ANSWER TO POINT d

D. Determine the routing by using the saving matrix methods, considering that the Distribution Center can use 2 trucks of 25 packages of capacity each one. In the following table the matrix of distances between the 20 clients and the Distribution Center is shown.

 Truck capacity is  t1 = t2 = 25 packages.

The  reduced matrix of distance to be considered is shown:

This matrix includes only the 7 clients and the DC.

Next the Saving Matrix is computed:

 

sij = dDCi + dDCj – dij

Routing and Sequencing of First Truck

Greatest saving is between client 4 and 5: 13017.

C4 = 8 packages

C5 = 8 packages

C4+ C5 = 16 < t1 = 25

 

Next saving is 11742 between clients  5 and 18. Client 18 is the inserted in the routing.

C18 = 8

Ctot= 24.

 

In truck 1 no other clients can be added.

 

 

The route of Truck 1is:  DCà18à4à5àDC

The total number of packages delivered is 24.

 

The lenghs of the path is:

dDC,18 + d18,4+ d 4,5 + d 5,DC = 6461+1952+2596+8106 = 19115.

 

 

TRUCK 2

Greates saving is (neglecting clients included in TRUCK 1 route) is  7647, between clients 13 and 19.

C13 = 6 packages

C19 = 8 packages

C4+ C5 = 14 < t1 = 25

Next savings is between clients 17 and 19 (7207).

C17 = 2 packages

Ctot= 16.

 

Next one between 10 and 13 (1152).

C10 = 8 packages

Ctot= 24

 

Truck 2 route is  DCà 10à13à19à17àDC.

The total number of packages delivered by TRUCK 2 is 24.

Truck 2 path lenght is:

dDC,10 + d10,13 + d 13,19 + d 19,17 + d 17,DC = 628+ 5785+5811+6728+6786= 25738.