Alessa, Nazek A.
Funding for this research was provided by:
This research was funded by the Deanship of Scientific Research at Princess Nourah bint Abdulrahman University through the Fast-track Research Funding Program.
Article History
Accepted: 5 April 2021
First Online: 7 May 2021
Declarations
:
: Disclosure of potential conflicts of interest: No conflict of interest.Research involving human participants and/or animals.
: This article does not contain any studies with human participants or animals.performed by any of the authors.
: No need it.The paper is novelty.
: Here we will consider an example clarifies Tri-level linear fractional programming problem (TL-LFPP) as:Equation removedEquation removedEquation removedEquation removedBy using a transformation method of Charnos and Cooper (Charnes and CooperCitationRef removed), which was illustrated in detail in (Subject and ShivCitationRef removed), to illustrate that TL-LFPP is equivalent to the following TL-LPP.Equation removedEquation removedEquation removedEquation removedWe choose <i>ν</i><sub>1</sub> = 10, <i>ν</i><sub>2</sub> = 6, <i>ν</i><sub>3</sub> = 2, <i>γ</i><sub>1</sub> = 0.95, <i>γ</i><sub>2</sub> = 0.85, [<i>σ</i><sub>1</sub><sup><i>L</i></sup>, <i>σ</i><sub>1</sub><sup><i>U</i></sup>] = [<i>σ</i><sub>2</sub><sup><i>L</i></sup>, <i>σ</i><sub>2</sub><sup><i>U</i></sup>] = [0.75, 0.9]. The solution of problem (15) is.<i>c</i> = (8.581, 6.1466, 9.96877), <i>θ</i> = 1.6, <i>ψ</i> = 0.58112.<i>X</i><sub><i>1</i></sub> = −516.22, <i>X</i><sub>2</sub> = –451.7, <i>X</i><sub>3</sub> = –371.435,<i>a</i><sub>1</sub>(<i>X</i><sub>1</sub>) = 0.804 < <i>γ</i><sub>1</sub> = 0.95. <i>a</i><sub>2</sub>(<i>X</i><sub>2</sub>) = 0.595 < <i>γ</i><sub>2</sub> = 0.85.<i>a</i><sub>3</sub>(<i>X</i><sub>3</sub>) = 0.581 < <i>σ</i><sub>1</sub> = 0.7412. <i>σ</i><sub>2</sub> = 0.9755.Both of DM<sub>1</sub> and DM<sub>2</sub> aren’t satisfied with the above solution, so if DM<sub>2</sub> changes <i>γ</i><sub><i>2</i></sub> = 0.85 to <i>γ</i><sup><i>'</i></sup><sub>2</sub> = 0.65. So the corresponding problem of (16) is formulated as:Equation removedThe solution of problem (19) is <i>c</i> = (8.791, 7.257, 8.916), <i>θ</i> = 2.1, <i>ψ</i> = 0.558775,<i>X</i><sub><i>1</i></sub>= −519.09, <i>X</i><sub><i>2</i></sub>= −453.12, <i>X</i><sub><i>3</i></sub> = −371.4, a<sub>1</sub>(<i>X</i><sub><i>1</i></sub>) = 0.865072 < <i>γ</i><sub><i>1</i></sub>= 0.95,<i>a</i><sub><i>2</i></sub>(<i>X</i><sub><i>2</i></sub>) = 0.65, <i>a</i><sub><i>3</i></sub>(<i>X</i><sub><i>3</i></sub>) = 0.558775, <i>σ</i><sub>1</sub> = 0.7514, <i>σ</i><sub>2</sub> = 0.8596.The termination condition (7) isn't satisfied the value of <i>a</i><sub>1</sub>(<i>X</i><sub>1</sub>), if DM<sub>1</sub> changes γ<sub>2</sub> = 0.95 to <i>γ</i><sup><i>'</i></sup><sub>1</sub> = 0.91 then solves the problem (20);Equation removedThe solution of problem (14) is.<i>c</i> = (8.873, 9.246, 7.917), <i>θ</i> = 1.513, <i>ψ</i> = 0.537895.<i>X</i><sub><i>1</i></sub> = −518.12, <i>X</i><sub>2</sub> = −450.36, <i>X</i><sub>3</sub> = −369.57,<i>a</i><sub><i>1</i></sub>(<i>X</i><sub>1</sub>) = 0.91, <i>a</i><sub><i>2</i></sub>(<i>X</i><sub>2</sub>) = 0.701, <i>a</i><sub><i>3</i></sub>(<i>X</i><sub>3</sub>) = 0.5379,<i>σ</i><sub>1</sub> = 0.778453, <i>σ</i><sub>2</sub> = 0.767755.By now, <i>a</i><sub><i>1</i></sub>(<i>X</i><sub>1</sub>) = 0.91 = <i>γ</i><sup><i>'</i></sup><sub>1</sub>, <i>a</i><sub><i>2</i></sub>(<i>X</i><sub>2</sub>) = 0.701 > 0.65 = <i>γ</i><sup><i>'</i></sup><sub>2</sub>,Moreover <i>σ</i><sub>1</sub> = 0.778453, <i>σ</i><sub>2</sub> = 0.767755 are all in [0.75, 0.9]. This is meaning that, all the proposed algorithm's termination conditions are satisfied, the DMs find the adequate solution.