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The range of optimal values in cost optimization models provides management with options for decision making. However, it can be quite challenging to achieve feasible range of optimality in Geometric programming (Gp) models having negative degrees of difficulty. In this paper, we conduct sensitivity analysis on the optimal solution of Geometric programming problem with negative degree of difficulty. Using imprest data, we determine the optimal objective function, dual decision variables, primal decision variables; the range of values, the cost coefficient and RHS constraint must lie for the solution to stay optimal. From the analysis, we established that incremental sensitivity analysis has the functional form .

In every optimization problem, it will be of interest not only to obtain an optimal solution of the problem but also to find how stable that optimal solution would be in the different coefficients of the problem [

We carry out sensitivity analysis or post optimality test in geometric programming on the cost coefficients of the objective function and the right-hand side (RHS) of the constraint equations to determine the effect of changes on the new optimal dual decision variables and the new optimal objective function. The range of values for the changes was determined; the new optimal objective function and the dual decision variables were also determined. Sensitivity analysis is the optimal dual decision variable for the problem. It is concerned with how small changes in the constraint or objective function affect the optimal objective value [

We observed that [

Our interest is to determine the range of values the change in the cost coefficients and the right-hand side will lie. We have illustrated the methodology by modeling a real life data (monthly imprests) obtained from the Department of Maritime Management Technology as a negative degree of difficulty geometric programming problem, and its optimal solution was determined before the sensitivity analysis carried out on it. We observed that [

y i ∗ = b i ( 0 ) + ∑ j = 1 n r j b i ( j ) (1)

The n-dimensional vectors b i ( 0 ) , b ( j ) ; i = 1 , ⋯ , n are constants; y_{i} is the dual decision variable and r_{j} is a constant. For positive degree of difficulty problem, k > 0, the optimal dual decision variables y i * are independent of the primal cost coefficients, c_{i} and the relationship among changes in the c_{i} and y_{i} are given by

d y i = ∑ j = 1 n { b i ( 0 ) ∑ i = 1 n [ J j l − 1 ( y ∗ ) ∑ i = 1 n b i ( i ) ( d c i / c i ) ] } (2)

where

J j l ( y ∗ ) = ∑ i = 1 n { b i ( 0 ) b i ( j ) / y i ∗ } (3)

where J_{ji} is the Jacobian matrix and

d v / v ∗ = ∑ y i ∗ d c i / c i (4)

For a change in the primal coefficient, d c i / c i we compute for the new solution as

v 1 = v ∗ + d v (5)

and the new dual variables becomes

y i 1 = y i ∗ + d s i (6)

The range of value for c i is

c i − y i ∗ D − 1 < c i 1 < c i + y i ∗ D − 1 (7)

where

D = ∑ j = 1 n { b i ( j ) ∑ i = 1 n [ J j l − 1 ( y ∗ ) b i ( i ) ] } (8)

For constraint equations, we compute for the range of values as follows:

g k ( x ) ≤ 1 (9)

Let the right-hand side be increased or decreased by R. Hence, we have

g k ( x ) ≤ 1 ± R

Let 1 ± R = Δ R

We have

g k ( x ) ≤ Δ R

Converting the constraint equation to standard form, we have

1 Δ R g k ( x ) ≤ 1 (10)

The new problem becomes:

Minimize f ( x ) (11)

Subject to 1 Δ R g k ( x ) ≤ 1 (12)

The effect of the change in the RHS on the cost coefficients of the constraint equation is then determined along side with its range of values.

The range of values for Δ R is

c R − y i ∗ D R − 1 < Δ R < c R + y i ∗ D R − 1 (13)

S/No | Items | Sept. (Amount) | Oct. (Amount) | Nov. (Amount) |
---|---|---|---|---|

1 | Snacks | 4000 | - | 3700 |

2 | FUTO Water | - | 7200 | - |

3 | Generator maintenance | 10,200 | - | 11,500 |

4 | Amstel malt | - | 12,800 | 4800 |

5 | Stationeries | 5800 | - | - |

Total | 20,000 | 20,000 | 20,000 |

Source: Dept. of Maritime Management Technology, FUTO, 2016.

_{i} to represent weights of various items the Department purchased during the period.

Let X_{1} = snacks; X_{2} = FUTO water; X_{3} = generator maintenance; X_{4} = Amstel malt and X_{5} = stationeries.

The total expenditure for the period is the sum of the individual monthly expenditure on these variables X i ; i = 1 , ⋯ , 5 . These sums are written as:

U i ( x ) = U 1 ( x ) + ⋯ + U n ( x ) (14)

where

U 1 ( x ) = 20000 x 1 x 3 x 5

U 2 ( x ) = 20000 x 2 x 4

U 3 ( x ) = 20000 x 1 x 3 x 4

x i > 0 ; C j > 0 and n = 3

Putting the U_{i}(x) in an unconstrained geometric programming form, we have

Minimize f ( x ) = 20000 x 1 x 3 x 5 + 20000 x 2 x 4 + 20000 x 1 x 3 x 4 (15)

Reformulating Equation (15) to constrained geometric programming problem, we have

Minimize f ( x ) = 20000 x 1 x 2 − 2 x 3 x 4 − 1 x 5 + 20000 x 1 − 1 x 2 2 x 3 − 2 x 4 x 5 − 1 + 20000 x 1 − 1 x 2 x 3 x 4 x 5 − 1 (16)

Subject to

x 1 2 x 2 2 x 5 x 4 + x 1 − 1 x 2 − 2 x 3 2 x 4 − 2 x 5 ≤ 1 (17)

Equations (16) and (17) are constrained geometric programming problem with negative one degree of difficulty (K = −1) given as K = n − ( m + 1 ) = − 1 ; where K is the degree of difficulty, n is the number of terms and m is the number of variables. Degree of difficulty is the measure of computational complexity of the problem.

Since the solution is not unique, we maximize the dual geometric program subject to linear constraint.

Maximize f ( y ) = ∏ k = 0 m ∏ j = 1 N k ( C k j y k j ∑ j = 1 N k y k j ) y k j (18)

Subject to

A y = B (19)

Forming the orthogonality and normality condition from the exponent matrix and writing it in the form of Equation (19), we have

y 1 − y 2 − y 3 + 2 y 4 − y 5 = 0

− 2 y 1 + 2 y 2 + y 3 + 2 y 4 − 2 y 5 = 0

y 1 − 2 y 2 + y 3 + 0 y 4 + 2 y 5 = 0

− y 1 + y 2 + y 3 + y 4 − 2 y 5 = 0

y 1 − y 2 − y 3 + y 4 + y 5 = 0

y 1 + y 2 + y 3 + 0 y 4 + 0 y 5 = 1

[ 1 − 1 − 1 2 − 1 − 2 2 1 2 − 2 1 − 2 1 0 2 − 1 1 1 1 − 2 1 − 1 − 1 1 1 1 1 1 0 0 ] A ( 6 × 5 ) [ y 1 y 2 y 3 y 4 y 5 ] y ( 5 × 1 ) = [ 0 0 0 0 0 1 ] B ( 6 × 1 )

>> A = [ 1 , − 1 , − 1 , 2 , − 1 ; ⋯ ; 1 , 1 , 1 , 0 , 0 ] ;

>> B = [ 0 ; 0 ; 0 ; 0 ; 0 ; 1 ] ;

y * = P i n v ( A ) ∗ B

y * = [ 0.4620 0.3662 0.1696 0.0530 0.0483 ] > 0

y * are the optimal weights of the dual decision variables and they satisfy the orthogonality and normality conditions.

From Equation (18), we correct for y * and compute the optimal objective function as follows:

f * ( y ) = f * ( x ) = ( ( 20000 0.4620 ) 0.4620 ) ∗ ( ( 20000 0.3662 ) 0.3662 ) ∗ ( ( 20000 0.1696 ) 0.1696 ) ∗ ( ( 1 0.0530 ) 0.0530 ) ∗ ( ( 1 0.0483 ) 0.0483 ) ∗ ( ( 0.1013 ) 0.1013 ) = 58534

The above is the global optimal objective function; that is, the optimal expenditure for the period should be 58,534 instead of 60,000. The contribution of each primal decision variable to the objective function is:

x i = exp ( w * ) = [ 12.0770 0.7094 0.6900 22.7982 1.8578 ] > 0

Allocation of New Values to the Items purchased

_{in} represents the initial allocation for each items; C_{j} represents the new average optimal allocation for each month and C_{jn} represents the final allocation for each items purchased.

Items | September (N) | October (N) | November (N) | ||||
---|---|---|---|---|---|---|---|

S/N | C_{in} | C_{jn} | C_{in} | C_{jn} | C_{in} | C_{jn} | |

1 | Snacks | 4000 | 3902.266 | 3700 | 3609.60 | ||

2 | FUTO Water | 7200 | 7024.0226 | ||||

3 | Gen. Maintenance | 10,200 | 9950.778 | 11,500 | 11,219.00 | ||

4 | Amstel Malt | 12,800 | 12,487.3074 | 4800 | 4682.73 | ||

5 | Stationery | 5800 | 5658.286 | ||||

Total | 20,000 | 19,511.33 | 20,000 | 19,511.33 | 20,000 | 19,511.33 |

A 10% increase in the September imprest will result in some changes in the optimal dual decision variables and the optimal objective function.

The optimal dual decision variable y i ∗ as presented in Equation (1) is computed as

y i ∗ = b i ( 0 ) + ∑ j = 1 n r j b i (j)

[ 0.0000 0.2505 0.11601 0.03625 0.03304 ] + r 1 [ 1.0000 0.2505 0.11601 0.03625 0.03304 ] ; where r 1 = 0.4620

We chose any value in the column of b^{(j}^{)} and let it be r_{1} and we make use of the relationship b 2 ( 1 + r ) = y i ∗ ⇒ b 2 ( 1 + r ) = 0.3662 b 2 = 0.3662 1 + 0.4620 = 0.2505 ; etc .

J 11 = ∑ i = 1 n b i ( i ) b j ( j ) y i ∗ = ∑ j = 1 n b j 2 y i ∗

J 11 = 1 0.4620 + 0.06275 0.3662 + 0.01346 0.1696 + 0.01314 0.0530 + 0.001092 0.0483 = 2.685753

Hence, if c_{1} = 20,000.00 is increased by 10%, making the total imprest for the month of September to be 22,000.00, we compute the changes in the dual decision variables as follows:

d y 5 = b 5 ( 1 ) J 11 − 1 b 1 ( 1 ) d c 1 / c 1

d y 5 = ( 0.03304 ) ( 1 / J 11 ) ( 1.0000 ) ( 1 / 10 ) = 0.001230

d y 4 = ( 0.03625 ) ( 1 / J 11 ) ( 1.0000 ) ( 1 / 10 ) = 0.00135

d y 3 = ( 0.11601 ) ( 1 / J 11 ) ( 1.0000 ) ( 1 / 10 ) = 0.00432

d y 2 = ( 0.2505 ) ( 1 / J 11 ) ( 1.0000 ) ( 1 / 10 ) = 0.00933

d y 1 = ( 1.0000 ) ( 1 / J 11 ) ( 1.0000 ) ( 1 / 10 ) = 0.03723

The new dual decision variables are:

y 1 = y i ∗ + d y i

y 1 = 0.4620 + 0.03723 = 0.49923

y 2 = 0.3662 + 0.00933 = 0.37553

y 3 = 0.1696 + 0.00432 = 0.17392

y 4 = 0.0530 + 0.00135 = 0.05435

y 5 = 0.0483 + 0.001230 = 0.04953

y i 1 = [ 0.49923 0.37553 0.17392 0.05435 0.04953 ] > 0

The new optimal dual decision variables satisfies orthogonality and normality conditions, the non-negativity condition, and we can observe some variations in the corresponding values of the dual decision variables.

The new optimal objective function is computed as follows:

f ∗ ( N e w ) ( x ) = ( ( 20000 0.49923 ) 0.49923 ) ∗ ( ( 20000 0.37553 ) 0.37553 ) ∗ ( ( 20000 0.17392 ) 0.17392 ) ∗ ( ( 1 0.05435 ) 0.05435 ) ∗ ( ( 1 0.04953 ) 0.04953 ) ∗ ( ( 0.10388 ) 0.10388 )

f ∗ ( N e w ) ( x ) = 96464

From the above sensitivity analysis, we can see that a 10% increase on September imprest will lead to increase in the optimal objective function values from 58,534 to 96,464 and the optimal dual decision variables changes from

y * = [ 0.4620 0.3662 0.1696 0.0530 0.0483 ] > 0 to y i 1 = [ 0.4992 0.3755 0.1739 0.05435 0.04953 ] > 0

The range of value for which the changes in September imprest lie is:

c − y i ∗ D − 1 < c 1 < c + y i ∗ D − 1

D − 1 = 2.685753

y i ∗ D − 1 = 0.4620 ∗ 2.685753 = 1.24

19998.76 ≤ c 1 ≤ 20001.24

So, if the cost coefficient (c_{1}), where c_{1} is the imprest for September, lies between 19,998.76 and 20,001.24, the optimal objective function will remain optimal, that is, will remain 58,534.

In this study, we have established that incremental sensitivity analysis on the negative degrees of difficulty geometric programming problems can be represented as

f ( x ) = Q x + c

where f ( x ) is the new optimal objective function; Q is the slope of the linear function and c is the initial global optimal solution and x is the percentage change in the cost coefficient; see

f ( x ) = 3793 x i + c i

f ( x ) = 3793 × 15 + 58534 = 115429naira

% change | f ( x ) ^{New} |
---|---|

10 | 96,464 |

9 | 91,733 |

8 | 84,500 |

5 | 75,165 |

4 | 71,468 |

2 | 64,727 |

1 | 61,502 |

0 | 58,534 |

Therefore, we can predict the new optimal solution without starting the calculations afresh.

If a constraint equation with a unit cost coefficient is added to the existing problem on impress, what happen to the optimal solution?

Solution:

Minimize f ( x ) = 20000 x 1 x 2 − 2 x 3 x 4 − 1 x 5 + 20000 x 1 − 1 x 2 2 x 3 − 2 x 4 x 5 − 1 + 20000 x 1 − 1 x 2 x 3 x 4 x 5 − 1

Subject to x 1 2 x 2 2 x 4 x 5 + x 1 − 1 x 2 − 2 x 3 2 x 4 − 2 x 5 ≤ 1

Subject to x 1 − 2 x 2 − 2 x 4 − 1 x 5 − 1 ≤ 1

Forming orthogonality and normality conditions, we have

y 1 − y 2 − y 3 + 2 y 4 − y 5 − 2 y 6 = 0

− 2 y 1 + 2 y 2 + y 3 + 2 y 4 − 2 y 5 − 2 y 6 = 0

y 1 − 2 y 2 + y 3 + 0 y 4 + 2 y 5 + 0 y 6 = 0

− y 1 + y 2 + y 3 + y 4 − 2 y 5 − y 6 = 0

y 1 − y 2 − y 3 + y 4 + y 5 − y 6 = 0

y 1 + y 2 + y 3 + 0 y 4 + 0 y 5 + 0 y 6 = 1

[ 1 − 1 − 1 2 − 1 − 2 − 2 2 1 2 − 2 − 2 1 − 2 1 0 2 0 − 1 1 1 1 − 2 − 1 1 − 1 − 1 1 1 − 1 1 1 1 0 0 0 ] A ( 6 × 6 ) [ y 1 y 2 y 3 y 4 y 5 y 6 ] y ( 6 × 1 ) = [ 0 0 0 0 0 1 ] B ( 6 × 1 )

>> A = [ 1 , − 1 , − 1 , 2 , − 1 , − 2 ; ⋯ ; 1 , 1 , 1 , 0 , 0 , 0 ] ;

>> B = [ 0 ; 0 ; 0 ; 0 ; 0 ; 1 ] ;

y * = P i n v ( A ) ∗ B = 0

and f * ( x ) = 0 .

Hence, we conclude that addition of constraint equations in this problem will lead to degenerate and infeasible optimal solution.

Since the expenditure of monthly imprest can vary from time to time depending on the Departmental needs, therefore, we advise as follows (see

1) The Department could spend N3902.266 instead of N4000 and N3609.60 instead of N3700 respectively on snacks for the month of September and November, 2016.

2) The Department could spend N7024.0226 instead of N7200 on FUTO water for the month of October, 2016.

3) The Department could spend N9950.778 instead of N10,200 and N11,219.00 instead of N11,500 respectively on generator maintenance for the month of September and November, 2016.

4) The Department could spend N12,487.3074 instead of N12,800 and N4682.73 instead of N4800 respectively on Amstel Malt for the month of October and November, 2016.

5) The Department could spend N5658.286 instead of N5800 on stationery for the month of September, 2016, if she had applied our model.

In addition, we conducted sensitivity analysis on the modeled problem, and observed that the incremental sensitivity analysis is of the form f ( x ) = Q x i + c i . When there is 15% increase in the September allocation (expenditure), the optimal objective function will be f ( x ) = 3793 × 15 + 58534 = 115429 . Sensitivity analysis helps us to determine the boundary of adjustment of the cost coefficients so that the optimal objective function will not be violated. We found such point to be: 19998.76 ≤ c 1 ≤ 20001.24 . Moreover, these results can guide the Department on how to allocate her scarce resources in particular and to the university in general.

We acknowledge the authors, researchers and publishers whose materials we have used. We also acknowledge the Department of Maritime Management Technology, Federal University of Technology Owerri (FUTO) for providing us with data for this paper.

The authors declare no conflicts of interest regarding the publication of this paper.

Amuji, H.O., Olewuezi, N.P., Onwuegbuchunam, D.E. and Igboanusi, C. (2020) Sensitivity Analysis on the Negative Degree of Difficulty Geometric Programming Problem. American Journal of Operations Research, 10, 13-23. https://doi.org/10.4236/ajor.2020.101002