Integrated Master Program
Industrial and Financial Economy
Gothenburg University
2003/11/20
Risk Management and Economics of Information
Case 2 : Arla
Supervisor: Prof. Göran Bergendahl
Authors:
Aijun Hou
Aránzazu Muñoz-Luengo
Emma Aer
GuangBin Zhao
Muna Girgis
TABLE OF CONTENTS
1.0 Introduction 1
2.0 Objectives 1
3.0 Problem Situation Description 1
3.1 Minimize Cost of Cream / Butter Production (in Götene) 2
3.2 Minimize Cost for Liquid Milk Production 2
3.3 Minimize Cost for Cheese Production 3
3.4 Minimize Total Cost 3
4.0 Proposed method 5
4.1 Linear Programming 5
4.2 Simple Multi-Attribute Rating Technique (SMART) 8
4.3 Heuristic Method 9
5.0 SWOT Analysis 10
5.1 Strengths and opportunities 10
5.2 Threats and weaknesses 11
6.0 Conclusion 13
7.0 Reference 15
1.0 Introduction
Arla is one of the largest Swedish milk producers, and it was built up on 26th April, 1915 in Stockholm. In the beginning of 1970’s , Arla got a new name-Mjölkcentral Arla, which is a name from the merger between Mjölkcentralen, lantbrukamas Mjölkcentral, Sydöstmejerier, and örebro-Ortens Mejerförening. Nowadays, bedsides milk, Arla also produces other products like butter, cheese, yoghurt, however, liquid milk, cheese, and butter are there main product.
Now, Arla is planning to expend market to Skaraborg, there are 27 farms to produce milk and 10 potential locations for dairies to produce Cheese, Cream/butter, and Liquid milk. While butter/cream only could be produced in Götene, Liquid milk is to be produced at one single site only, cream and skim milk will be the by- products during production process. Cheese could be produced at any of the 10 dairies. Skim milk and whey will be the by-product. The cream produced at cheese and liquid dairies will be transported to Götene as the input for butter production.
2.0 Objectives
Under this situation, we have there main tasks:
1. Propose one or several objectives to determine :
· The best location for the cheese production
· The best single location for the production of liquid milk.
2. Propose a method to find out the best size, location and time-phasing for the
Investment in dairies
3. Analyze which effect that would come from :
· A substantial annual growth (2%) of the supply of the raw milk
· A substantial expansion (3%) in the sales of milk products like certain high quality cheeses or an assortment of yogurt products (“high quality liquid milk products)
3.0 Problem Situation Description
Our objective is to minimize total cost to determine the best location for cheese production, the best single location for the production of liquid milk. In order to minimize total cost, we have to minimize the cost from cheese, liquid milk, and butter/cream dairies.
3.1 Minimize Cost of Cream / Butter Production (in Götene)
Our objective here is to minimize total cost of cream/butter production. As the butter production is produced only in Götene, we only need to consider the transportation cost of raw milk.
Cost (butter/cream) = Transportation cost of raw milk ----(A)
As we know, from the liquid milk dairy and cheese dairy would also ship cream to Götene. The amount of cream from cheese making and liquid milk dairies is equal to the demand for raw milk at each dairies multiply the input-output coefficient for the cream (Table 7.3).The demand raw milk we can divide the demand for each production by coefficient factor from table 7.3
FROM / SUMMER / WINTER / TOTALCheese production / 3,492 / 4056.94 / 7549
Liquid milk production / 325.19 / 612.8 / 938
TOTAL / 3817 / 4670 / 8487
Table1. Transported cream to Götene
Transported amount of cream summer time
= (60,000/0,756)*0,044+(11,500/0,7780)*0,022
=79,365*0,044+14,781*0,022
=3,492+325
=3817
Transported amount of cream winter time
=(60,000/0,7025) *0,0475+ 16,100/0,7225)*0,027
=85,409*0,0475+22283,74*0,027
=4056,94+612,80
=4670
This amount of cream will lower then the need for the total amount of raw milk for cream/ butter production, by getting cream from cheese and liquid dairies, will reduced the transport cost of raw milk from farmers to dairy.
3.2 Minimize Cost for Liquid Milk Production
In order to minimize cost for liquid milk production, we have to minimize:
Total cost for liquid milk production = Transportation cost of the raw milk +
Delivery cost of cream to Götene +
Distribution cost of liquid milk. ------(B)
As we know from case, the liquid milk production will generate two by product: cream and feed. Cream will be delivered to Götene. The amount of cream to be delivered is equal to demand for raw milk at this dairy times coefficient factor of, which equal to 11500/0,778 *0,022 + 16100/0,7225 *0,0275 = 938 (see table 1) . The total cost will also depends upon where the dairy is located and the distribution cost from dairy to market, which is also depend on where the dairy is located.
3.3 Minimize Cost for Cheese Production
We would minimize the cost for cheese production, hence we will minimize:
Total cost for cheese production = Transportation cost of the raw milk +
Delivery cost of cream to Götene +
Annual net investment cost. ----- (C)
As for the liquid milk production, the cheese production also generates cream during the process, which will be transported to Götene. The transportation cost will depend on where the dairy is located and the demand for raw milk, which equals 60,000/0,756) *0,044 +(60,000/0,7025) *0,0475=7549 (see table 1) besides this, total cost also depends on the distribution cost to market.
The important thing here is that from table 7.6, we know that economic of scale could be done by producing more and hence reducing cost.
Figure 1. Milk for cheese tons/year
3.4 Minimize Total Cost
Our main task now is to minimize the total cost for the company. In order to minimize total cost we have to minimize total cost of A+B+C.
It is really crucial to know the demand of raw milk at each dairy. From case, we know the average annual (in tons) demand for different product during summer time and winter time, and then we can calculate the total demand for raw milk at each dairy.
PRODUCT / SUMMER / WINTER / TOTALCheese / 79365 / 85409 / 164774.3
Cream / 34530 / 20900 / 55430
Liquid milk / 14781.491 / 22283.737 / 37065.23
TOTAL / 128676.574 / 128592.9897 / 257269.6
Table 2. Demand for raw milk
We now calculate the demand for raw milk by dividing the demand of each product by input and output coefficient in table 7.3. :
Demand for raw milk at cheese production
= 60000/0,7560=79 365 tons
Demand for raw milk at winter
=60000/0,7025=85 409tons
Total demand for raw milk at cheese production= 79365+85409=164 774
However, for the demand of raw milk at cream dairy, we deduct the amount transported from cheese and liquid dairies.
Summer time demand=(7270-3817)/0.10=34530 tons
Winter time demand =(6760-4670)/0,10=20900tons
Total demand for raw milk at Götene=34530+20900=55430 tons
Demand for raw milk at liquid milk production
Summer time= 11500/0,7780=14,781 tons
Winter time = 16,100/0,7225=22,283tons
Total demand for raw milk at liquid milk production= 14781+22283=37065 tons
Total demand for raw milk at the dairies =257,269 ton / year
When we calculate total cost, we didn’t take distribution cost of butter and cheese into account, the reason behind this, is that we didn’t get any information about this, hence we assume that Arla has different marketing strategy regarding about this. The marketing strategy might be is that cheese and butter will be sold to other area or focus on in Sweden besides Skaraborg because of that cheese butter could be stored for long, while liquid milk could not be fresh for long, hence Arla’s distribution channel must be very short. Now we know the total demand for raw milk at each dairy during summer and winter time. Hence, we have to decide the best location for each production.
4.0 Proposed method
In order to solve this transportation problem, we could choose from different methods which range from more mathematical application based to more descriptive foundation. We will present and discuss two methods of great convenience for these kinds of problems, which are Linear Programming and SMART.
4.1 Linear Programming
In the previous section we explained the basic problem to be solved which will help us understanding the concepts to take into account to implement this method. We considered that this specific case is of great complexity and therefore we decided to deeply describe the steps of its implementation. This sort of problem could be solved by different software such as Solver, SAS or SPSS.
We first start specifying the objective function in words. The objective is to minimize the total transportation costs compounded by the costs of the raw milk to the available dairies, the delivery costs of cream to the butter factory in Götene from the liquid milk factory and the cheese factory, and finally the distribution costs of the liquid milk to the markets.
Another optimization problem is the cheese production based on the decision regarding the number and size of cheese dairies, transportation costs and delivery costs.
In the mathematical formulation, as we specified in the previous section we do not contemplate the delivery costs for cheese and butter as we are not provided with these costs in the case.
Now we formulate the mathematical problem for solving the optimal dairies and the amount of flow from the farms to the optimal dairies:
m = number of lorries (m = 27)
n = number of potential dairies (n = 10)
= amount of raw milk collected in region i to dairy j
= cost of transporting one unit from i to j
for,
Total cost of all transportations is:
The objective function is results:
Min. 3.96x0101+9.57x0102 +9.08x0103 +10.4x0104 +16.83 x0105…9.572708 +13.04x2709 +5.78x2710
At this point we should define the constraints, the purpose of the constraints are to guarantee that the total amount of raw material, in this case the milk that is shipped to each factory equals the transportation capacity of each lorry. Also the constraints will define that the number of products shipped to a dairy coincides with the number of the required products.
We will find a number of constraints, n + m, besides the nonnegative constraints.
The variables are:
ai = total amount to be transported from lorry i
bj = total amount to be transported to dairy j.
The constraints are as follows:
, for
, for
, for and.
If we apply the constraints to our case: (We are supposed to obtain some of the x, decision variables equal to zero since not all the routes are optimal).
Transportation from lorries
x0101 + x0102 + x0103 + …+ x0110 = 9200 (Transportation out of Hova)
x0201 + x0202 + x0203 + … + x0210 = 8900 (Transportation out of Toreboda)
………..
x2701 + x2702 + x2703 + … + x2710 = 9000 (Transportation out of Sandhem)
Where Xn is the total amount to be shipped from each farm. We can calculate Xn from table 7.1 by adding up the forecasted production of milk at winter and summer time for each farm.
Transportation to dairies
x0101 + x0201 + x0301 + …+ x2701 = M1 (Transportation into Toreboda)
x0102 + x0202 + x0302 + …+ x2702 = M2 (Transportation into Tibro)
………..
x0110 + x0210 + x0310 + …+ x2710 = M10 (Transportation into Stenstorp)
Where Mn is the total amount to be shipped to each dairy or the input capacity for each facility. Although we calculated the demand for raw milk for cheese, butter and liquid milk we still do not know the allocation of raw milk for the different dairies.
Next, we continue by formulating the problem for the butter production in the site of Götene, following the same reasoning, the constraint function is, (from table 7.2, the third column which corresponds to the specified dairy number for Götene that is number 3):
x0103 + x0203 + x0303 + …+ x2703 0 (Transportation into Götene)
We also should minimize the delivery costs for cream to Götene in the linear programming method, and then the constraint would be (data from table 7.4):
Min. 12.5Q1 + 13Q2 + 10.5Q4 +…4.75Q8 + 5.5Q9 + 11.5Q10
Where Qn is the cream deliveries in dairy n.
We now proceed to analyze the best location for the liquid milk production, to do so we gather the information of forecasted cost for liquid milk distribution (data from table 7.5) and should also be taken into account the lowest distribution costs for the optimal liquid milk dairy.
Then, the objective function for this constraint is:
Min. 608299M1, 484589M2 .… 430939M10,
Subsequently, we should try to decide the number and capacity of cheese factories that could be constructed.
Using the available information in the case (table 7.6), we can calculate the annual net investment per ton of cheese making, and by looking at the results we observe that the smallest annual costs is for the facility that produces more tons of milk for cheese, so Arla can take advantage of so called economies of scale, which are defined as primary advantage of expanding or installing a big facility, that is to say the marginal cost decreases by producing one more unit.
Total investment costs (SEK)(1) / Maximum quantity of milk for cheese (ton)
(2) / Annual net investment costs per ton (SEK/ton)
(3)
580 000 / 10 / 58 000,00
580 000 / 13 400 / 43,28
1 123 000 / 40 000 / 28,08
1 754 000 / 80 000 / 21,93
2 202 000 / 120 000 / 18,35
Table 3. Annual net investment costs per ton
Next we calculated the raw milk requirements for every kind of available facility to facilitate the understanding of our proposal, and then we will compare these figures with the previously calculated amounts of raw milk needed in order to fulfill the annual demand of cheese.
Total investment costs (SEK)(1) / Maximum quantity of milk for cheese (ton)
(2) / Demand of cheese in summer (ton)
(3) / Demand of cheese in winter (ton)
(4) / Quantity of raw milk for cheese (ton)
(3)/0,756+(4)/0,7025
580000 / 10 / 5 / 5 / 13,73
580000 / 13400 / 6700 / 6700 / 18399,80
1123000 / 40000 / 20000 / 20000 / 54924,78
1754000 / 80000 / 40000 / 40000 / 109849,55
2202000 / 120000 / 60000 / 60000 / 164774,33
Table 4. Quantity of raw milk for cheese for each facility size