# Manage your Finances with Maxima

This twenty-second article of the mathematical journey through open source, show cases managing our usual financial computations using the financial package of Maxima.

<< Twenty-first Article

In today’s credit-oriented world, loan, EMIs, the principal, interest rates, savings, etc are the common lingo. How well do we really understand these, or for that matter are able to compute the actual finances related to such things? Or, do we just accept what the other party has to provide us. One may say, either get into the gory details to understand and verify, or forget about understanding and just accept it if the offering suits you – you cannot not understand and also verify at the same time. However, with the finance package of Maxima, you can actually verify without getting much into the computational details. This article is going to exactly walk you through that.

## Basic operations

To be able to use the finance package of Maxima, the first thing to do, after getting the Maxima prompt is to load the finance package, using load(). And, then go ahead to do the various computations. days360() is the simplest function to give the number of days between two dates, assuming 360 days per year, 30 days per month – one of the common interest calculation norm.

\$ maxima -q
(%o1)     /usr/share/maxima/5.24.0/share/contrib/finance/finance.mac
(%i2) days360(2014, 1, 1, 2014, 10, 1);
(%o2)                                 270
(%i3) days360(2014, 1, 1, 2014, 12, 31);
(%o3)                                 360
(%i4) days360(2014, 1, 1, 2014, 3, 1);
(%o4)                                 60
(%i5) days360(2014, 1, 1, 2015, 1, 1);
(%o5)                                 360
(%i6) quit();

Note that days360() takes from and to dates, each as a triplet of year, month, date.

One common computation, we often deal with, is the final amount we would get after applying a particular rate of compound interest on a particular principal amount – just use fv(<rate>, <principal>, <period>) for computing the future value of the <principal>, at the compound interest rate of <rate>, after <period> periods of the rate. As an example, what would be the future value of investing ₹10,000 for 10 years at the compound interest rate of 15% per year – just call fv(0.15, 10000, 10);

\$ maxima -q
(%i1) load(finance)\$ /* \$ to suppress its output */
(%i2) fv(0.15, 10000, 10);
(%o2)                          40455.57735707907

If you are interested in, how exactly did it calculate, or what is the formula, that is where Maxima is fun to play with symbols. Just use some symbols, instead of actual numbers:

(%i3) string(fv(r, p, n));
(%o3)                              p*(r+1)^n
(%i4) quit();

And, there you go. p*(r+1)^n is the future value for investing p amount for n periods at the compounded interest rate of r per period.

How about doing an inverse computation? Suppose, I want to get ₹10,00,000 after 5 years from my investment today at the interest rate of 10.75%. Now for that, how much should I invest today? Don’t scratch your head, just call pv(<rate>, <future_val>, <period>);

(%i2) pv(0.1075, 1000000, 5);
(%o2)                          600179.7323625274
(%i3) fv(0.1075, 600180, 5);
(%o3)                          1000000.445928875
(%i4) quit();

That requires an investment of ₹6,00,180. Yes, so if you invest that amount, the future value would be ₹10,00,000 – that’s the check done above using fv().

## Loans and EMIs

Fundamental thought behind today’s culture of buying a house, a car, or even smaller items is “let’s buy it on loan and pay it off in EMIs (equated monthly installments)”. These terms would be familiar to most of you, so no need to explain them. But, how do you compute these? One would say, there are some complicated formulas to do so. Yes, you are right. But, you don’t need to worry about any of them. Just tell Maxima to give you the complete schedule for your loan, at a given rate of interest, for a given period of time, using amortization(). The first example below provides the schedule for a home loan of ₹20 lakhs at an interest rate of 9.25% per annum (p.a.) for 5 years. The various columns in the output schedule provide the following information:

• n: installment payment time – year for our example
• Balance: principal + interest left over to be paid out, after the current installment is paid out
• Interest: interest part being paid out in the current installment
• Amortization: principal part being paid out in the current installment
• Payment: current installment to be paid out, i.e. the EYI (equated yearly installment)

What is this EYI? We were supposed to talk only about EMI, right? Okay, in that case, we need to convert the rate of interest and the period, both in terms of months. So, we need to divide the per annum rate of interest by 12 to get it per month, and multiply the number of years by 12 to get that in number of months. That is exactly what the second example below shows. Note the 60 EMIs to be paid out over the period of 5 years.

\$ maxima -q
(%i2) amortization(0.0925, 2000000, 5)\$
"n"      "Balance"    "Interest"  "Amortization"  "Payment"
0.000   2000000.000         0.000         0.000         0.000
1.000   1667475.420    185000.000    332524.580    517524.580
2.000   1304192.317    154241.476    363283.103    517524.580
3.000    907305.527    120637.789    396886.790    517524.580
4.000    473706.709     83925.761    433598.818    517524.580
5.000         0.000     43817.871    473706.709    517524.580

(%i3) amortization(0.0925/12, 2000000, 5*12)\$
"n"      "Balance"    "Interest"  "Amortization"  "Payment"
0.000   2000000.000         0.000         0.000         0.000
1.000   1973656.870     15416.667     26343.130     41759.797
2.000   1947110.679     15213.605     26546.192     41759.797
3.000   1920359.860     15008.978     26750.818     41759.797
4.000   1893402.837     14802.774     26957.023     41759.797
5.000   1866238.021     14594.980     27164.816     41759.797
6.000   1838863.809     14385.585     27374.212     41759.797
7.000   1811278.588     14174.575     27585.221     41759.797
8.000   1783480.730     13961.939     27797.857     41759.797
9.000   1755468.598     13747.664     28012.133     41759.797
10.000   1727240.538     13531.737     28228.059     41759.797
11.000   1698794.887     13314.146     28445.651     41759.797
12.000   1670129.968     13094.877     28664.919     41759.797
13.000   1641244.090     12873.919     28885.878     41759.797
14.000   1612135.550     12651.257     29108.540     41759.797
15.000   1582802.632     12426.878     29332.918     41759.797
16.000   1553243.605     12200.770     29559.026     41759.797
17.000   1523456.728     11972.919     29786.877     41759.797
18.000   1493440.244     11743.312     30016.484     41759.797
19.000   1463192.382     11511.935     30247.861     41759.797
20.000   1432711.360     11278.775     30481.022     41759.797
21.000   1401995.381     11043.817     30715.980     41759.797
22.000   1371042.632     10807.048     30952.749     41759.797
23.000   1339851.289     10568.454     31191.343     41759.797
24.000   1308419.513     10328.020     31431.776     41759.797
25.000   1276745.450     10085.734     31674.063     41759.797
26.000   1244827.233      9841.580     31918.217     41759.797
27.000   1212662.979      9595.543     32164.253     41759.797
28.000   1180250.793      9347.610     32412.186     41759.797
29.000   1147588.763      9097.767     32662.030     41759.797
30.000   1114674.963      8845.997     32913.800     41759.797
31.000   1081507.453      8592.286     33167.510     41759.797
32.000   1048084.276      8336.620     33423.177     41759.797
33.000   1014403.463      8078.983     33680.814     41759.797
34.000    980463.026      7819.360     33940.437     41759.797
35.000    946260.965      7557.736     34202.061     41759.797
36.000    911795.264      7294.095     34465.702     41759.797
37.000    877063.889      7028.422     34731.375     41759.797
38.000    842064.793      6760.701     34999.096     41759.797
39.000    806795.913      6490.916     35268.880     41759.797
40.000    771255.168      6219.052     35540.745     41759.797
41.000    735440.463      5945.092     35814.705     41759.797
42.000    699349.687      5669.020     36090.776     41759.797
43.000    662980.711      5390.821     36368.976     41759.797
44.000    626331.390      5110.476     36649.320     41759.797
45.000    589399.565      4827.971     36931.825     41759.797
46.000    552183.057      4543.288     37216.508     41759.797
47.000    514679.671      4256.411     37503.386     41759.797
48.000    476887.197      3967.322     37792.474     41759.797
49.000    438803.406      3676.005     38083.791     41759.797
50.000    400426.052      3382.443     38377.354     41759.797
51.000    361752.873      3086.617     38673.179     41759.797
52.000    322781.588      2788.512     38971.285     41759.797
53.000    283509.900      2488.108     39271.689     41759.797
54.000    243935.492      2185.389     39574.408     41759.797
55.000    204056.031      1880.336     39879.461     41759.797
56.000    163869.167      1572.932     40186.865     41759.797
57.000    123372.528      1263.158     40496.638     41759.797
58.000     82563.728       950.997     40808.800     41759.797
59.000     41440.360       636.429     41123.368     41759.797
60.000         0.000       319.436     41440.360     41759.797

(%i4) quit();

If you want to be a little stylish in your monthly payments, that is, instead of equal payments, you want to do increasing payments, Maxima could help you with that as well. arit_amortization() and geo_amortization() are two such functions, which provides the schedule for increasing payments with fixed amount increments and fixed rate increments, respectively. Here’s a small demo of the same:

\$ maxima -q
(%i2) amortization(0.10, 100, 5)\$
"n"       "Balance"     "Interest"   "Amortization"  "Payment"
0.000       100.000         0.000         0.000         0.000
1.000        83.620        10.000        16.380        26.380
2.000        65.603         8.362        18.018        26.380
3.000        45.783         6.560        19.819        26.380
4.000        23.982         4.578        21.801        26.380
5.000         0.000         2.398        23.982        26.380

(%i3) arit_amortization(0.10, 10, 100, 5)\$
"n"       "Balance"     "Interest"   "Amortization"  "Payment"
0.000       100.000         0.000         0.000         0.000
1.000       101.722        10.000        -1.722         8.278
2.000        93.615        10.172         8.106        18.278
3.000        74.698         9.362        18.917        28.278
4.000        43.890         7.470        30.809        38.278
5.000         0.000         4.389        43.890        48.278

(%i4) geo_amortization(0.10, 0.05, 100, 5)\$
"n"       "Balance"     "Interest"   "Amortization"  "Payment"
0.000       100.000         0.000         0.000         0.000
1.000        85.907        10.000        14.093        24.093
2.000        69.200         8.591        16.707        25.298
3.000        49.558         6.920        19.642        26.562
4.000        26.623         4.956        22.935        27.891
5.000         0.000         2.662        26.623        29.285

(%i5) quit();

%i2 has been provided for comparative analysis. %i3 shows the incremental payout with increments of ₹10 (the second parameter of arit_amortization()). %i4 shows the incremental payout with increments at the rate of 5% (the second parameter of geo_amortization()). Both these computations could be done in decrements as well – just pass the second parameter negative.

## Plan your savings

Say, you have a savings account like PPF (public provident fund), giving you interest at a rate of 8% p.a., and at the end of 5 years, you want to have save ₹1,00,000. So, what should be your minimum yearly deposit into your account. It is not just divide by 5, as interest would be also added to your savings. saving() shows you the complete schedule as follows:

\$ maxima -q
(%i2) saving(0.08 /* interest rate */, 100000 /* final savings */, 5 /* years */)\$
"n"       "Balance"     "Interest"    "Payment"
0.000          0.000         0.000         0.000
1.000      17045.645         0.000     17045.645
2.000      35454.943      1363.652     17045.645
3.000      55336.983      2836.395     17045.645
4.000      76809.588      4426.959     17045.645
5.000     100000.000      6144.767     17045.645

And, the minimum yearly deposit to be done is ₹17,046. The “Balance” and “Interest” columns, respectively, tell you about the balance and the interest accumulated in the corresponding year. If you are only interested in knowing the minimum amount to be deposited, you may just use the annuity_fv() function – basically computing the annuity of ₹17,046 every year for 5 years to have a future saving of ₹1,00,000 after 5 years.

(%i3) annuity_fv(0.08, 100000, 5);
(%o3)                         17045.64545668365
(%i4) quit();

## Project planning

Finance management is a key ingredient of any project planning, whether it be a professional or personal project. Assume a project would take n years, with given yearly expenses, and say the available interest rate is r p.a. Then, a common question, every project manager needs to answer is what is the net present value (NPV) of the project, which needs to be invested for the project. The answer to this basic question is more often than not, one of the important factors for deciding whether to take up this project or not. Maxima provides npv() to compute the same. As an example, if a project needs ₹100, ₹200, ₹150, ₹600 over 4 years, respectively, and the current interest rate is 7%, what is the NPV? It would be ₹848 as shown below:

\$ maxima -q
(%i2) npv(0.07, [100, 200, 150, 600]);
(%o2)                         848.3274983420189
(%i3) quit();

Another common practice to select between various projects is to compute the benefit to cost ratio of the various projects, and select the one with best ratio. The benefit to cost ratio for a given interest rate r could be computed using benefit_cost(). Here’s an example to demonstrate the same, assuming 18% as the rate of interest:

• Project P1 (2 years): Yearly Benefits (100, 200), Yearly Costs (150, 50)
• Project P2 (3 years): Yearly Benefits (0, 200, 100), Yearly Costs (100, 100, 0)
• Project P3 (4 years): Yearly Benefits (0, 200, 200, 100), Yearly Costs (100, 100, 50, 50)
\$ maxima -q