IFEJM: New Intuitionistic Fuzzy Expert Judgment Method for Effort Estimation in Agile Software Development

Abstract

Many agile projects use expert judgment-based methods for estimating effort. Commonly, the judgments made during estimating project features are consensual. However, this will hardly be achieved when a conflict arises between estimators. Besides, estimate depends on the experience and skills of the estimator and could be threatened by his uncertainty to make reliable and accurate assessments. To fill these gaps, an intuitionistic fuzzy expert judgment method is proposed. The latter allows making fuzzified assessments and integrates estimators’ priorities according to a set of human factors. As well, it provides consensual estimates either by the end of the estimation rounds or automatically using an iterative algorithm. On the other hand, an initial empirical study has been conducted on an agile project in which user stories have been estimated by students and experts. The first findings have revealed that the proposal is more suitable for inexperienced estimators or in the first sprints of the project where disagreement is still significant. Nevertheless, when group agreement is increased during the estimating process, the proposal maintains a null bias toward the overestimation or the underestimation of the user stories.

Appendices

Appendix A: All Individual Estimates Are Identical

Proof

Replacing all $\mu _i^{(l)}$ by $\mu _i^{(1)}$ and all $\nu _i^{(l)}$ by $\nu _i^{(1)}$ in Eq. (13), we have:

$$\begin{aligned} (\mu _i^*,\nu _i^*)= & {} IFWA_\lambda \left( {\tilde{e}}_i^{(1)},{\tilde{e}}_i^{(2)},\cdots ,{\tilde{e}}_i^{(m)}\right) \\= & {} \left( 1-(1-\mu _i^{(1)})^{\sum _{l=1}^{m} \lambda _l},(\nu _i^{(1)})^{\sum _{l=1}^{m} \lambda _l}\right) \end{aligned}$$

Since $\sum _{l=1}^{m} \lambda _l =1$, consequently

$$\begin{aligned}(\mu _i^*,\nu _i^*)\!=\!(\mu _i^{(1)},\nu _i^{(1)})\!=\!(\mu _i^{(2)},\nu _i^{(2)}) \!=\!\cdots \!=\!(\mu _i^{(m)},\nu _i^{(m)})\end{aligned}$$

$\square$

Appendix B: Proof of Equation (20)

Proof

By differentiating the Lagrangian function in Eq. 19, we can get the derivatives:

$$\begin{aligned}{\left\{ \begin{array}{ll} \frac{\partial L(c_i,\xi )}{\partial c_i}=2c_i \displaystyle \sum _{l=1}^{m}d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)} )-2\xi \\ \frac{\partial L(c_i,\xi )}{\partial \xi }=\displaystyle \sum _{u_i\in D}c_i -1 \end{array}\right. }\end{aligned}$$

According to the Lagrange multiplier method [74], the stationary points of $L(c_i,\xi )$ could be found by setting the above derivatives to zero: $\frac{\partial L(c_i,\xi )}{\partial c_i}=\frac{\partial L(c_i,\xi )}{\partial \xi }=0$. We obtain afterward the following equations:

$$\begin{aligned}{} & {} c_i \sum _{l=1}^{m}d_l({\tilde{e}}_i^{*},{\tilde{e}}_i^{(l)} )-\xi =0 \end{aligned}$$

(B1)

$$\begin{aligned}{} & {} \sum _{u_i\in D}c_i -1 =0 \end{aligned}$$

(B2)

Using Eq. (B1), we have:

$$\begin{aligned} c_i=\frac{\xi }{\displaystyle \sum _{l=1}^{m}d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)})} \end{aligned}$$

By replacing $c_i$ in Eq. (B2), we have:

$$\begin{aligned} \displaystyle \sum _{u_i \in D}\frac{\xi }{\sum _{l=1}^{m}d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)})} =1 \end{aligned}$$

so that

$$\begin{aligned} \xi = \frac{1}{\displaystyle \sum _{u_i \in D}\frac{1}{\sum _{l=1}^{m} d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)})}} \end{aligned}$$

Therefore, the coefficients of adjustment are obtained as follows:

$$\begin{aligned} c_i=\frac{\frac{\displaystyle 1}{\displaystyle \sum _{l=1}^{m}d_l({\tilde{e}}_i^{*},{\tilde{e}}_i^{(l)} )}}{\displaystyle \sum _{u_i\in D}\frac{1}{\displaystyle \sum _{l=1}^{m}d_l({\tilde{e}}_i^{*},{\tilde{e}}_i^{(l)} )}} \end{aligned}$$

(B3)

On the other hand, the Hessian matrix of Lagrangian function:

$$\begin{aligned} H=\frac{\partial ^2 L(c_i )}{\partial c_i\partial c_j} \end{aligned}$$

where

$$\begin{aligned} \frac{\partial ^2 L(c_i )}{\partial c_i\partial c_j}=0 \, for \,i\ne j\end{aligned}$$

and

$$\begin{aligned}\frac{\partial ^2 L(c_i )}{\partial c_i\partial c_j}=2 \sum _{l=1}^{m}d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)}) \end{aligned}$$

The eigenvalues of H are $2 \sum _{l=1}^{m}d_l({\tilde{e}}_i^*,{\tilde{e}}_i^{(l)})$ greater than 0. According to [74], H is positive definite. Thereby, $c_i$ is a strict local minimum. $\square$

Appendix C: Proof of Theorem 1

Proof

Using Eq. (17), we consider $\mu _i^{(l,j)}$ as a sequence:

$$\begin{aligned} \mu _i^{(l,j+1)} = c_i\mu _i^{(l,j)}+(1-c_i)\mu _i^{*(j)} \end{aligned}$$

(C4)

Based on the monotonic sequence theorem, every bounded, monotonic sequence converges. First, we demonstrate that $\mu _i^{(l,j)}$ is a bounded sequence, i.e., $0 \le \mu _i^{(l,j)}\le 1$ for all $j\in {\mathbb {N}}$. By definition $\mu _i^{(l,1)}$ is a membership function, and hence, $0\le \mu _i^{(l,1)}\le 1$. Using the principle of mathematical induction, we suppose that $0 \le \mu _i^{(l,j)}\le 1$ is true and we will prove that $0 \le \mu _i^{(l,j+1)}\le 1$ is also true. Indeed, by substituting the term j by j-1 in Eq. (C4), then we have:

$$\begin{aligned}\mu _i^{(l,j+1)}\!=\!c_i\left( \!c_i \mu _i^{(l,j-1)}\!+(1\!-\!c_i) \mu _i^{*(j-1)}\!\right) +(1-c_i) \mu _i^{*(j)} \end{aligned}$$

so that

$$\begin{aligned} \mu _i^{(l,j+1)}=c_i^2 \mu _i^{(l,j-1)}+c_i(1-c_i) \mu _i^{*(j-1)}+(1-c_i) \mu _i^{*(j)} \end{aligned}$$

Next, by substituting the term j-1 by j-2, we have:

$$\begin{aligned} \mu _i^{(l,j+1)}=&\, c_i^2\left( c_i \mu _i^{(l,j-2)}+(1-c_i) \mu _i^{*(j-2)}\right) \\ {}&+c_i (1-c_i) \mu _i^{*(j-1)}+(1-c_i) \mu _i^{*(j)} \end{aligned}$$

So on, we continue substituting until the first term. We obtain afterward:

$$\begin{aligned} \mu _i^{(l,j+1)}\!=&\,c_i^j \mu _i^{(l,1)}\!+\!c_i^{j-1}(1\!-\!c_i) \mu _i^{*(1)}\!+\!c_i^{j-2}(1\!-\!c_i)\mu _i^{*(2)}\\&+\cdots +c_i (1\!-\!c_i) \mu _i^{*(j-1)}+(1\!-\!c_i) \mu _i^{*(j)} \end{aligned}$$

Accordingly, we write:

$$\begin{aligned} \mu _i^{(l,j+1)} = c_i^{j}\mu _i^{(l,1)}+(1-c_i) \sum _{k=0}^{j-1}c_i^{k} \mu _i^{*(j-k)} \end{aligned}$$

(C5)

Obviously, $\mu _i^{(l,j+1)}\!\ge \!0$ since $0\!\le \!\mu _i^{(l,1)}\!\le \!1$, $0\!<\!c_i\!<1\!$, and $0\!\le \!\mu _i^{*(j)}\!\le \! 1$ for all $j\in {\mathbb {N}}$.

Furthermore, from Eq. (C5) we have:

$\mu _i^{(l,j+1)}\!\le \! c_i^j \mu _i^{(l,1)}\!+\!(1\!-\!c_i) \displaystyle \!\sum _{k=0}^{j-1}\!c_i^k$ and $\displaystyle \sum _{k=0}^{j-1}c_i^k\!=\!\frac{(1\!-\!c_i^j)}{(1\!-\!c_i)}$.

This implies $\displaystyle \mu _i^{(l,j+1)}\le c_i^j \mu _i^{(l,1)}+(1-c_i)\frac{(1-c_i^j)}{(1-c_i)}$ and $\displaystyle \mu _i^{(l,j+1)}\le c_i^j \mu _i^{(l,1)}+1-c_i^j$,

i.e., $\displaystyle \mu _i^{(l,j+1)}\le c_i^j (\mu _i^{(l,1)}-1)+1$, since $0\le c_i^j \le 1$ and $\mu _i^{(l,1)}\le \!1\! \Rightarrow c_i^j (\mu _i^{(l,1)}-1)\!\le \! 0$. Hence, $0\!\le \! \mu _i^{(l,j+1)}\!\le \! 1$.

The first property of Theorem 1 is thus verified. $\mu _i^{(l,j)}$ is a bounded sequence.

Next, we demonstrate that $\mu _i^{(l,j)}$ is a monotonic sequence. We can see that $\mu _i^{(l,j)}$ is a recursive sequence that could be denoted as:

$$\begin{aligned} \mu _i^{(l,j+1)}=f(\mu _i^{(l,j)}), where f:[0,1]\longrightarrow [0,1] \end{aligned}$$

and

$$\begin{aligned} f(x)=c_i x+(1-c_i)\!\left( 1-(\!1-x)^{\lambda _l}\!\!\!\!\!\!\prod _{l^{'}=1,l^{'} \ne l}^{m}\!\!\!\!(1-\mu _i^{(l^{'},j) } )^{\lambda _{l^{'}}}\!\!\right) \nonumber \\ \end{aligned}$$

(C6)

To simplify the recursion function, we set

$p=\displaystyle \prod _{l^{'}=1,l^{'} \ne l}^{m}(1-\mu _i^{(l^{'},j) } )^{\lambda _{l^{'}}}$ where $p \in [0,1]$. Then, the function f(x) in Eq. (C6) could be written as:

$$\begin{aligned} f(x)=c_ix-(1-c_i )p(1-x)^{\lambda _l}+(1-c_i) \end{aligned}$$

(C7)

Next, we derive f(x)

$$\begin{aligned} \frac{\partial f(x)}{\partial x}=c_i+\lambda _l(1-c_i)p \end{aligned}$$

(C8)

Since $c_i> 0$ and $\lambda _l,p \in [0,1]$, we have: $\frac{\partial f(x)}{\partial x}>0$ for every $x\in [0,1]$. We deduce that f(x) is a strictly increasing function, and consequently, $\mu _i^{(l,j)}$ is a monotonic (increasing or decreasing) sequence. Its monotonicity depends on the sign of $\mu _i^{(l,2)}-\mu _i^{(l,1)}$. If $\mu _i^{(l,2)}-\mu _i^{(l,1)}>0$, then $\mu _i^{(l,j)}$ is a strictly increasing sequence. Otherwise, $\mu _i^{(l,j)}$ is a strictly decreasing sequence. From above, we proved that $\mu _i^{(l,j)}$ is a bounded monotonic sequence, and therefore, it converges. Now, it remains only to find its limit.

Based on the theorem of the fixed point: A fixed point l of a continuous recursion function f(x), which satisfies $f(l)=l$ and $\vert \frac{\partial f(x)}{\partial x}\vert <1$, is unique and limit of the recursive sequence expressed by f(x). By using Eq. (C7), we have:

$$\begin{aligned} f(\mu _i^{*(n_0)})=c_i \mu _i^{*(n_0)}-(1-c_i)p(1-\mu _i^{*(n_0)})^{\lambda _l}+(1-c_i) \end{aligned}$$

We have also: $p(1-\mu _i^{*(n_0)})^{\lambda _l}\!=\!\displaystyle \prod _{l=1}^m(1-\mu _i^{(l,n_0)})^{\lambda _l}$. Thereby

$$\begin{aligned} f(\mu _i^{*(n_0)})&\!=\!c_i \mu _i^{*(n_0)}\!-\!(1\!-\!c_i) \displaystyle \!\prod _{l=1}^{m}(1\!-\!\mu _i^{(l,n_0)})^{\lambda _l}\!+\!(1\!-\!c_i)\nonumber \\&\!=\!c_i \mu _i^{*(n_0)}+(1-c_i)(1-\displaystyle \prod _{l=1}^{m}(1-\mu _i^{(l,n_0)})^{\lambda _l})\nonumber \\&\!=\!c_i \mu _i^{*(n_0)}+(1-c_i) \mu _i^{*(n_0)}\nonumber \end{aligned}$$

We find that $f(\mu _i^{*(n_0)})=\mu _i^{*(n_0)}$. This means that $\mu ^{*(n_0)}$ is a fixed point of f(x). From Eq. (C8), by applying the triangular inequality, we have:

$$\begin{aligned} \vert \frac{\partial f(x)}{\partial x}\vert \le \vert c_i \vert + \lambda _l p\vert 1-c_i\vert <\vert c_i\vert +\vert 1-c_i \vert \end{aligned}$$

Since the only case that $\lambda _l p\vert 1-c_i\vert =\vert 1-c_i\vert$ is when $p=1$ and $\lambda _l=1$, which is absurd because if we admit that $\lambda _l=1$, then p=0. This means that $\vert \frac{\partial f(x)}{\partial x}\vert \le 1$. Now, recall that since f(x) is differentiable on the interval [0,1], then it is certainly continuous on this interval. From above, we proved that $\mu _i^{*(n_0)}$ is a unique fixed point of the continue function f(x) and furthermore the limit of its corresponding recursive sequence $\mu _i^{(l,j)}$. Hence, we can write:

$$\begin{aligned} \lim _{j\rightarrow +\infty }\mu _i^{(l,j)}=\mu _i^{*(n_0)} \end{aligned}$$

Likewise, we can demonstrate $\nu _i^{(l,j)}$ is a convergent sequence with

$$\begin{aligned} \lim _{j\rightarrow +\infty } \nu _i^{(l,j)}=\nu _i^{*(n_0)} \end{aligned}$$

This completes the proof of Theorem  1. Obviously, Theorem 1 is still valid despite the set D contains only one user story, simply we have to replace $c_i$ by $\lambda _l$. $\square$