Lecture6

37 capital rationing chapter 139 of swk in practice

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Unformatted text preview: he market value at the end of study period. If no market value information available, use the imputed market value technique . 37 Capital Rationing (Chapter 13.9 of SWK) In practice, projects do not have to be mutually exclusive. Some of them are independent, and some of them are mutually exclusive. Example: A company has different departments, sales, research, manufacturing etc. Within each department the projects may be mutually exclusive. But on the firm level, a project of the sales department may be independent of the project from the research. There is often a capital constraint (limited amount of money for investment). Question: How do we allocate our capital to each project in an optimal way? This is called capital rationing. Objective: Maximizing the total PW. Caveat: Maximizing the IRR is not the right criterion! SEEM5740/ECLT5930 38 Example 6.8 4 independent projects (numbers are in millions of dollars). Project CF0 CF1 CF2 PW(10%) A B C D -10 -5 -5 0 30 5 5 -40 5 20 15 60 21 16 12 13 Budget constraint: capital investment cannot exceed 10 million at time 0. SEEM5740/ECLT5930 39 Example 6.8 xA : either 0 or 1. xA = 0 means we do not invest in A. xA = 1 means we invest in A. xB , xC and xD are defined similarly. The present worth of all selected projects is 21xA + 16xB + 12xC + 13xD The total capital investment is 10xA + 5xB + 5xC SEEM5740/ECLT5930 40 Example 6.8 To solve this problem, we can formulate it mathematically as follows: maximize 21xA + 16xB + 12xC + 13xD subject to 10xA + 5xB + 5xC ≤ 10 xA , xB , xC , xD is either 0 or 1 This is a constrained optimization problem (called binary integer programming). It can be easily solved by an optimization solver (Excel has one). ∗ ∗ ∗ ∗ Optimal solution: xA = 0, xB = xC = xD = 1, PW ∗ = 41. SEEM5740/ECLT5930 41 Example 6.8 Now suppose we have additional constraint: the capital investment at the end of year 1 cannot exceed 10 million. So the problem becomes maximize 21xA + 16xB + 12xC + 13xD subject to 10xA + 5xB + 5xC ≤ 10 − 30xA − 5xB − 5xC + 40xD ≤ 10 xA , xB , xC , xD is either 0 or 1 ∗...
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