Swift 中 EXCEL 的 RATE 函数提供不同的结果

2024-04-27

func RATE(nper: Double, pmt: Double, pv: Double, fv: Double, type: Double, guess: Double) -> Double{
    var rate = guess
    var y: Double = 0
    var f: Double = 0
    var FINANCIAL_MAX_ITERATIONS: Double = 128
    var FINANCIAL_PRECISION = 1.0e-08
    if (abs(rate) < FINANCIAL_PRECISION) {
        y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv
    } else {
        f = exp(nper * log(1 + rate))
        y = pv * f + pmt * (1 / rate + type) * (f - 1) + fv
    }
    var y0 = pv + pmt * nper + fv
    var y1 = pv * f + pmt * (1 / rate + type) * (f - 1) + fv

    // find root by secant method
    var i: Double  = 0
    var x0: Double = 0
    var x1 = rate
    while ((abs(y0 - y1) > FINANCIAL_PRECISION) && (i < FINANCIAL_MAX_ITERATIONS)) {
        rate = (y1 * x0 - y0 * x1) / (y1 - y0)
        x0 = x1
        x1 = rate

        if (abs(rate) < FINANCIAL_PRECISION) {
            y = pv * (1 + nper * rate) + pmt * (1 + rate * type) * nper + fv
        } else {
            f = exp(nper * log(1 + rate))
            y = pv * f + pmt * (1 / rate + type) * (f - 1) + fv
        }

        y0 = y1
        y1 = y
        i += 1
    }

    return rate
}

RATE(nper: 252, pmt: -29002.85, pv: 2500000, fv: 0, type: 0,猜测: 0.1) // -1.347153369879729 -- 错误(正确的是0.010833331)

RATE(nper:24,pmt:-46.14,pv:1000,fv:0,类型:0,猜测:0.1)//0.008324438477500274 --- 正确

在上面,第一个值给出的结果比我在 Excel 中得到的结果错误,但在第二个值上它给出了正确的结果。


以下是用于 LibreOffice 中 RATE 函数的 C++ 代码的 Swift 移植。原始源代码可以在以下位置找到https://cgit.freedesktop.org/libreoffice/core/tree/sc/source/core/tool/interpr2.cxx https://cgit.freedesktop.org/libreoffice/core/tree/sc/source/core/tool/interpr2.cxx

这些设置为返回nil如果尝试计算费率时出现任何错误。

func rateIteration(nper: Double, pmt: Double, pval: Double, fval: Double, type: Bool, guess: Double) -> Double? {
    // See also #i15090#
    // Newton-Raphson method: x(i+1) = x(i) - f(x(i)) / f'(x(i))
    // This solution handles integer and non-integer values of Nper different.
    // If ODFF will constraint Nper to integer, the distinction of cases can be
    // removed; only the integer-part is needed then.
    var valid = true
    var found = false
    var x = 0.0
    var xNew = 0.0
    var term = 0.0
    var termDerivation = 0.0
    var geoSeries = 0.0
    var geoSeriesDerivation = 0.0
    let iterationsMax = 150
    var count = 0
    let epsilonSmall = 1.0E-14
    let SCdEpsilon = 1.0E-7

    var pv = pval
    var fv = fval
    if type {
        // payment at beginning of each period
        fv = fv - pmt
        pv = pv + pmt
    }

    if nper == nper.rounded() {
        // Integer nper
        x = guess
        while !found && count < iterationsMax {
            let powNminues1 = pow(1 + x, nper - 1)
            let powN = powNminues1 * (1 + x)
            if x == 0.0 {
                geoSeries = nper
                geoSeriesDerivation = nper * (nper - 1) / 2
            } else {
                geoSeries = (powN - 1) / x
                geoSeriesDerivation = nper * powNminues1 / x - geoSeries / x
            }

            term = fv + pv * powN + pmt * geoSeries
            termDerivation = pv * nper * powNminues1 + pmt * geoSeriesDerivation
            if abs(term) < epsilonSmall {
                found = true // will catch root which is at an extreme
            } else {
                if termDerivation == 0.0 {
                    xNew = x + 1.1 * SCdEpsilon // move away from zero slope
                } else {
                    xNew = x - term / termDerivation
                }

                count += 1
                // more accuracy not possible in oscillating cases
                found = abs(xNew - x) < SCdEpsilon
                x = xNew
            }
        }

        valid = x > -1.0
    } else {
        // nper is not an integer value
        x = (guess < -1.0) ? -1.0 : guess
        while valid && !found && count < iterationsMax {
            if x == 0.0 {
                geoSeries = nper
                geoSeriesDerivation = nper * (nper - 1) / 2
            } else {
                geoSeries = (pow(1 + x, nper) - 1) / x
                geoSeriesDerivation = nper * pow(1 + x, nper - 1) / x - geoSeries / x
            }

            term = fv + pv * pow(1 + x, nper) + pmt * geoSeries
            termDerivation = pv * nper * pow(1 + x, nper - 1) + pmt * geoSeriesDerivation
            if abs(term) < epsilonSmall {
                found = true // will catch root which is at an extreme
            } else {
                if termDerivation == 0.0 {
                    xNew = x + 1.1 * SCdEpsilon
                } else {
                    xNew = x - term / termDerivation
                }

                count += 1
                // more accuracy not possible in oscillating cases
                found = abs(xNew - x) < SCdEpsilon
                x = xNew
                valid = x >= -1.0 // otherwise pow(1 + x, nper) will fail
            }
        }
    }

    if valid && found {
        return x
    } else {
        return nil
    }
}

func RATE(nper: Double, pmt: Double, pv: Double, fv: Double, type: Double = 0, guess: Double = 0.1) -> Double? {
    let payType = type != 0.0

    if nper <= 0.0 { // constraint from ODFF spec
        return nil
    }

    if let res = rateIteration(nper: nper, pmt: pmt, pval: pv, fval: fv, type: payType, guess: guess) {
        return res
    } else {
        if guess == 0.1 {
            /* TODO: this is rather ugly, instead of looping over different
             * guess values and doing a Newton goal seek for each we could
             * first insert the values into the RATE equation to obtain a set
             * of y values and then do a bisecting goal seek, possibly using
             * different algorithms. */
            var x = guess
            for step in 2...10 {
                var g = x * Double(step)
                if let res = rateIteration(nper: nper, pmt: pmt, pval: pv, fval: fv, type: payType, guess: g) {
                    return res
                } else {
                    g = x / Double(step)
                    if let res = rateIteration(nper: nper, pmt: pmt, pval: pv, fval: fv, type: payType, guess: g) {
                        return res
                    }
                }
            }
        }
    }

    return nil
}

这是您的两个测试用例:

RATE(nper: 252, pmt: -29002.85, pv: 2500000, fv: 0)
RATE(nper: 24, pmt: -46.14, pv: 1000, fv: 0)

结果是:

0.010833331202068584
0.0083244384774994358

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