Capital Budgeting Models
Involvement of the model where limited investment funds are assigned amongst competing investment alternatives, it is stated to be the work of capital budgeting models. During a certain period, the available alternatives are identified with the help of investment cost and its associated advantages. It is an extremely simple process to determine an investment cost. Although benefit estimation can be a little tough, when the projects are distinguished by lower tangible returns.
The only issue related to this model is the selection of alternative sets. Any hindrance in choosing a correct alternative can influence the choice of a project. It is a difficult task to choose a set of alternatives which will increase a subject’s overall benefit even if there is a budget restriction.
It can be clearly stated with the help of an example.
A central government has invested 950 crores in a project of non-conventional sources. the amount invested is in the form of grants. After a thorough evaluation by experts, state governments have decided on 6 projects and submitted them.
For the next 10 years, a roughly calculate benefit of every project is stated in the table below.The government’s funding request is also highlighted in the table.For example, in project A, the given value (net benefit) is 5.5. an estimated return benefit over the next 10 years is counted to be Rs. 5.50.
Upto the limit of the maximum indicated funding aggregate, government permits any amount. You can see in the table that project E is considered as an extremely important project as half of the funds can be easily asked for. Due to this advantage, this project is sought by thecentral government.
Project A and B in combination are considered as natural priority projects as when its amount estimation is put together, at an average it can give at least 250 crores.
Problem formulation as alinear programming model.
Project Net benefit / Rupee invested Request for funds (crores of rupees)
A 5.5 150
B 3.5 250
C 2.4 120
D 2.0 100
E 6.0 500
F 3.1 160
For each destination and source of origin there is a certain cost of shipping unit. The cost usually encompasses the time and distance between both the points where the commodities are transported and from where it is being sent. The common problem in this aspect is determining the total number of units that needs to be supplied to each destination from every source point.
The aim of this model is to reduce the total transportation cost and also ensure few factors in this regard.
There are 2 locations in a town which are located at a higher altitude. The army stored petrolandkerosene there where these items would be used in 4 different zones during winter. The storage is based on the condition when highways close in this season, and the supply and availability of petrol and keroseneare not possible in these regions.
Supply cost of 1 kilolitres of petroland kerosene is provided in this table, that too from each stock location to every zone. Also, normal demand level as well as storing location capacity for every given zone are specified in this table.
Zone Maximum Supply
1 2 3 4
Storage Location 1 4 6 2.50 3.00 1000
Storage Location 2 5 2 3.50 4.50 800
Demand (Kilolitres) 300 500 400 350
To find out the total quantity of petrol and kerosene container (in kilolitres) that requires being delivered from the storage location of each zona, there are 8 decisions that are needed to be taken. But it is not mandatory that the decision is only about transportation of commodity units.
To resolve this problem, few equations are needed to be solved.
To indicate the total kilolitres delivered from 1 location to another in different zones (location 1 to zone 2, location 1 to zone 2, location 1 to zone 3 and so on), let us assume its equation to be x11, x12, x13,x14…
Similarly for location 2, the equation is indicated with the help of x21,x22, x23, x24…
Its total cost is given as:
4 x11 + 6x12+ 2· 50 x13 + 3· 50x14+ 5 x21 + 2 x22+ 300 x23+ 4· 50 x24
The reduction in function has to be made.
The constraints are:
x11+ x12+ x13+ x14 ≤100 (For location 1)
x21 + x22+ x23+ x24 ≤800 (For location 2)
These constraints make sure that the demanded quantities are met in every zone.
After the calculation, it is found that the total transportation for zone 1 comes to be at 300 Kilolitres.
x11 + x21= 300
To increase the net benefit, theamount of money is required to be found out which can be assigned to each project. The project is divided into 6 parts which are measured in the form of x1 ,x2, ….x6. The amount value is designated in crores of rupees.
Increased objective function is denote by:
Z = 5· 5 x1+ 3· 5 x2+ 2· 4 x3+ 2· 0 x4+ 6· 0 x5+ 3.1 x6
There are 4 types of constraints used in this problem.
x1+ x2+ x3 + x4+ x5+.x6 = 950
X5 ≥50% of 500≥ 250
Z = 5 . 5 x1+ 3.5 x2+ 2.4 x3+ 2.0 x4+ 6.0 x5+ 3.1x6
It has been subjected to:
x1+ x2+ x3+ x4+ x5+ x6 950
x1+x2 ≥ 250
x1,x2, x3, x4,x5,x6, ≥ 0