GILI TEST Sept 4th, 2014
Bakery Case: Inventory Aggregation and Saving Matrix
(Re-adapted from Nahmias, Production and Operations Analysis, 2013, pp 334)
INTRODUCTION TO THE TEST OF September 4th, 2014
In the following the solution of the test is provided. The subject concern the Inventory Aggregation, Saving Matrix Methods.
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:
Grain Mill is a small bakery that supply 5 major customers with bread each morning. Daily Demand in each customer follows a normal distribution and it is independent one customer to the others. Considering the Location Bakery in (0,0) and the following data for the customers:
Customer | Coordinates |
Average Daily Demand (units/day) |
Daily Standard Deviations (units/day) |
1 | (15,30) | 85 | 25 |
2 | (5,30) | 162 | 35 |
3 | (-10,20) | 26 | 13 |
4 | (-5,5) | 140 | 90 |
5 | (-20,10) | 110 | 25 |
a) What are the quantities to be supplied to each customer to guarantee a P1, the cycle service level of 80%? And 99%?
b) By the Saving Matrix Method, considering trucks of capacity of 300 units, determine the number of trucks and the routes for the distribution of the bread in the 3 cases:
b1. Quantities equal to the average daily value
b2. Quantities that guarantee P1=80%
b3. Quantities that guarantee P1=99%
Now consider a new scenario when bread is made and stocked in the Bakery each morning. At beginning of the day clients book the bread and then bread is distributed between customers in the exact quantities booked.
c) What is the reduction of the Total inventory that we can reach, keeping the same P1=80% and P1=99%?
d) What is the probability than surely more than 3 trucks are necessary for the distribution in the case c)?
e) What are the Fill Rates in cases a and c considering a complete lost sales scenario?
QUESTIONS:
a)Explain the Wagner Within method.
b)Show the rule for determine the k coefficient of safety stocks in the case a model cost B1 is considered.
c)What is it the Availability?
a)
ANSWER TO POINT a
We have to calculate k from the p(k) =80% k=0,841. Quantity to order are kσ_{i }for each customers. The same for p(k)=99%. Results are the following.
Customer | 80% | 90% |
1 | 106 | 143 |
2 | 191 | 243 |
3 | 37 | 56 |
4 | 216 | 349 |
5 | 131 | 168 |
Total | 682 | 961 |
ANSWER TO POINT b
By Using the Saving Matrix method we can determine the routes of distribution. First of all we evaluate the distance matrix, then the saving matrix.
distance matrix | 0 | 1 | 2 | 3 | 4 | 5 |
0 | 0 | 33,54 | 30,41 | 22,36 | 7,07 | 22,36 |
1 | 0 | 10,00 | 26,93 | 32,02 | 40,31 | |
2 | 0 | 18,03 | 26,93 | 32,02 | ||
3 | 0 | 15,81 | 14,14 | |||
4 | 0 | 15,81 | ||||
5 | 0 |
savings | |||||
1 | 2 | 3 | 4 | 5 | |
1 | |||||
2 | 53,95 | ||||
3 | 28,98 | 34,75 | |||
4 | 8,60 | 10,56 | 13,62 | ||
5 | 15,59 | 20,76 | 30,58 | 13,62 |
by ranking the savings it is possible to give priorities to the bounds to be connected frist
Bound | Savings |
1,2 | 53,95 |
2,3 | 34,75 |
3,5 | 30,58 |
1,3 | 28,98 |
2,5 | 20,76 |
1,5 | 15,59 |
3,4 | 13,62 |
4,5 | 13,62 |
2,4 | 10,56 |
1,4 | 8,6 |
B1.Now considering the case where average value are considered those are the routes:
First route
Bound (1,2) c_{1}=85 , c_{2}=162 c_{tot}=247
(2,3) c_{3}=26 c_{tot}=247+26=273<300
(3,5) c_{5}=110 c_{tot}>300
First route 0,1,2,3,0
Second Route
(4,5) c_{4}=140, c_{5}=110, c_{tot}=250 <300
Second route 0,4,5,0
In the same manner
B2.with Cycle Service Level =80%
First route 0,1,2,0
Second route 0,3,5,0
Third route 0,4,0
B3.with Cycle Service Level =99%
Considering that c_{4}=349>300, this load is split in 2 routes
c_{41}=300 and c_{42}=49:
First route 0,4,0
Second route 0,2,3,0
Third route 0,4,5,0
Forth route 0,1,0
ANSWER TO POINT c
The global demand distribution GD is a gaussian with average equal to 523 and standard deviation=104.
Then when P_{1}=Cycle Service Level=80% the Global quantity is 610 with a reduction respect to point a) equal to 71 units or 10,42%.
When P_{1}=Cycle Service Level=99% the Global quantity is 764 with a reduction respect to point a) equal to 196 units or 20,43%.
ANSWER TO POINT d
Considering the Global demand GD of point c, and the capacity of 3 trucks=900 units. The probability to be sure to use at least 4 trucs is P(GD>900). Considering the distribution of GD as in point c, to the point 900 corresponds in the Gaussian Standard Distribution a value of k=3,64 and then a probability P(GD>900)=0,01%.
ANSWER TO POINT e
Considering the 5 customers in point a) and the special Global Customer in point c), for each one of them we know:
Di=the average value of the demand
σ_{di}=the standard deviation of the demand
where i=1,2,3,4,5 in the case of single customer or GD in the case of global customer.
We have to use the formula for the P_{2}=Fill Rate in the case of Lost Sales P_{2}=1- ESPRC_{i}/(D_{i}+ESPRC_{i})
Where ESPRC means Expected Shortages for Replenishment Cycle ESPRC=σ_{di }G_{u}(k).
In the case of P_{1}=80% k=0.8416 and Gu(0,8416)=0.112
in the Case of P_{1}=99% k=2.33 and Gu(0.00352).
the calcule of P_{2} gives:
P_{2} | ||
Customers | P_{1}=80% | P_{2}=99% |
1 | 0.97% | 0.999% |
2 | 0.98% | 0.999% |
3 | 0.95% | 0.998% |
4 | 0.93% | 0.997% |
5 | 0.98% | 0.999% |
Global Demand | 0.98% | 0.999% |