Conclusion

The practical problem of banking can thus be solved by using a rather simple technique of optimization used so often in Physics.
The result is perfect in the sense that no approximation is involved in its derivation.
Also, the result is simple enough to be implemented by any practical banker.
The outcome is thus interesting not only for its mathematical exactitude but also for its easy applicability.

The Final Prescription

If the demand loan component were set at 44 units, the overall burden of interest would have been (44 ´ 10% + 1 ´ 12% ´ 6/12 + 6 ´ 12% ´ 6/12) = 4.82 units.
On the other hand, if the demand loan component were set at 46 units, the overall interest burden would have been (46 ´ 10% + 4 ´ 12% ´ 6/12) = 4.84 units.
Obviously, the overall burden due to interest is minimized if the demand loan is set at 45 units.

The Final Prescription

The readymade prescription for minimizing I now becomes:
(i)                  Select x at such a level that n/12 = a/(a+b) or, if a/(a+b) is not available, the next lower available value.
(ii)                If a certain range of values of x satisfies (i), choose the lowest value of x from the range. Once we set the optimal of x i.e. x0 at 45 units, the interest burden for the borrowing company over the next one year comes to (45 ´ 10% + 5 ´ 12% ´ 6/12) = 4.8 units.

The Implementation

B. If the levels of borrowing be 45 units for 6 months and 50 units for the other 6 months, n/12 = 1 if the demand loan component is set below 45 units; n/12 = ½ if the demand loan component is set at 45 units or above, but below 50 units.
Let a = 10% p.a. and a+b = 12% p.a. Then n/12 = 10/12 as per our formula. But, n/12 can be equal to only 1, ½, or 0.
We shall choose the next lower permissible value for n/12, viz. ½. Further, as n/12 = ½ for all values of the demand loan from 45 units and above but less than 50 units, we shall select the lowest permissible value, viz. 45 units as the desired value of x0.

The Implementation

A. Let the levels of borrowing projected by a company for the next twelve months be 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51 and 52 units (the levels of borrowing need not occur in this chronological order). Also, let a = 10% p.a. and b=2% p.a.
In such a situation, n/12 = 10/(10+2), or n=10. In other words, the demand loan component should be set at such a level that the level of borrowing would exceed the demand loan for 10 months. Thus, x0 = 42 units or more but less than 43 units.
The actual solution will be corresponding to the lowest value of x0, i.e. x0 = 42 units

The Solution

In other words, for I to be minimum, x is to be so chosen that for n months Wi>x, where n/12 = a/(a+b)

The Solution

The Solution
Or,  
(5) n/12 = a/(a+b) for = 0

The Solution

The Solution
Or, 12
(4) = a -[(a+b)/12] Σ[ θ(Wi – x) + (Wi – x)δ(Wi – x)] i=1
 
12
= a – [(a+b)/12] n - [(a+b)/12] Σ (Wi – x)δ(Wi – x)
i=1
 
= a – [(a+b)/12] n
where n = number of months for which Wi >x

The Solution

If a company works out its requirement of working capital finance for the next twelve months i.e. Wi for i=1,2, …., 12, then the total interest burden for the company during the next twelve months works out to:
12 12
(3) I(x) = Σ x.(a/12) + Σ (Wi – x). θ(Wi – x). [(a+b)/12].
i=1 i=1
 
12
= a.x +[(a+b)/12] Σ (Wi – x). θ(Wi – x),
i=1