Management of Deterministic and Stochastic Projects

Description

Management of Deterministic and Stochastic Projects

Details

Management problems of deterministic and stochastic projects. It obtains the duration of a project and the appropriate slack for each activity in a deterministic context. In addition it obtains a schedule of activities' time (Castro, Gómez & Tejada (2007) <doi:10.1016/j.orl.2007.01.003>). It also allows the management of resources. When the project is done, and the actual duration for each activity is known, then it can know how long the project is delayed and make a fair delivery of the delay between each activity (Bergantiños, Valencia-Toledo & Vidal-Puga (2018) <doi:10.1016/j.dam.2017.08.012>). In a stochastic context it can estimate the average duration of the project and plot the density of this duration, as well as, the density of the early and last times of the chosen activities. As in the deterministic case, it can make a distribution of the delay generated by observing the project already carried out.

DAG plot

Description

This function plots a directed acyclic graph (DAG).

Usage

dag.plot(
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  critical.activities = NULL
)

Arguments

Value

A plot.

Examples

prec1and2<-matrix(c(0,1,0,2,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0),nrow=5,ncol=5,byrow=TRUE)
prec3and4<-matrix(0,nrow=5,ncol=5)
prec3and4[3,1]<-3
dag.plot(prec1and2,prec3and4)

Problems of distribution of delay in deterministic projects

Description

This function calculates the delay of a project once it has been completed. In addition, it also calculates the distribution of the delay between the different activities with the proportional, truncated proportional and Shapley rule.

Usage

delay.pert(
  duration,
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  observed.duration,
  delta = NULL,
  cost.function = NULL
)

Arguments

Details

Given a problem of sharing delays in a project (N,\prec,\{\bar{X}_i\}_{i\in N},\{x_i\}_{i\in N}), such that \{\bar{X}_i\}_{i\in N} is the expected value of activities' duration and \{x_i\}_{i\in N} the observed value. If D(N,\prec,\{\bar{X}_i\}_{i\in N}) is the expected project time and D(N,\prec,\{x_i\}_{i\in N}) is the observed project time, it has to d=D(N,\prec,\{\bar{X}_i\}_{i\in N})-\delta is the delay, where \delta can be any arbitrary value greater than zero. The following rules distribute the delay costs among the different activities.

The proportional rule, from Brânzei et al. (2002), distributes the delay, d, proportionally. So that each activity receives a payment of:

\phi_i=\frac{\displaystyle x_{i}-\bar{X}_{i}}{\displaystyle \sum_{j\in N}\max\{x_{j}-\bar{X}_{j},0\}}\cdot C(D(N,\prec,\{\bar{X}_i\}_{i\in N})).

The truncated proportional rule, from Brânzei et al. (2002), distributes the delay, d, proportionally, where the individual delay of each player is reduced to d if if is larger. So that each activity receives a payment of:

\bar{\phi}_i=\frac{\displaystyle \min\{x_{i}-\bar{X}_{i},C(D(N,\prec,\{\bar{X}_i\}_{i\in N}))\}}{\displaystyle \sum_{j\in N} \max\{\min\{x_{j}-\bar{X}_{j},C(D(N,\prec,\{\bar{X}_i\}_{i\in N}))\},0\}}\cdot C(D(N,\prec,\{\bar{X}_i\}_{i\in N})).

These values are only well defined when the sum of the individual delays is different from zero.

Shapley rule distributes the delay, d, based on the Shapley value for TU games, see Bergantiños et al. (2018). Given a project problem with delays (N,\prec,\{\bar{X}_i\}_{i\in N},\{x_i\}_{i\in N}), its associated TU game, (N,v), is v(S)=C(D(N,\prec,(\{\bar{X}_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))) for all S\subseteq N, where C is the costs function (by default C(D(N,\prec,y))=D(N,\prec,y)-\delta. If the number of activities is greater than ten, the Shapley value, of the game (N,v), is estimated using a unique sampling process for all players, see Castro et al. (2009).

Value

The delay value and a solution matrix.

References

berg

Bergantiños, G., Valencia-Toledo, A., & Vidal-Puga, J. (2018). Hart and Mas-Colell consistency in PERT problems. Discrete Applied Mathematics, 243, 11-20.

bran

Brânzei, R., Ferrari, G., Fragnelli, V., & Tijs, S. (2002). Two approaches to the problem of sharing delay costs in joint projects. Annals of Operations Research, 109(1-4), 359-374.

castro

Castro, J., Gómez, D., & Tejada, J. (2009). Polynomial calculation of the Shapley value based on sampling. Computers & Operations Research, 36(5), 1726-1730.

Examples

prec1and2<-matrix(c(0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration<-c(2,1,1,4,2)
observed.duration<-c(2.5,1.25,2,4.5,3)
delta<-6
delay.pert(duration,prec1and2=prec1and2,observed.duration=observed.duration,
delta=delta,cost.function=NULL)

Problems of distribution of delay in deterministic projects with unions a priori

Description

This function calculates the delay of a project with unions a priori once it has been completed. In addition, it also calculates the distribution of the delay between the different activities with the proportional, truncated proportional and Owen rule.

Usage

delay.pert.unions(
  duration,
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  union,
  observed.duration,
  delta = NULL,
  cost.function = NULL
)

Arguments

Details

Given a problem of sharing delays in a project (N,\prec,\{\bar{X}_i\}_{i\in N},\{x_i\}_{i\in N}), such that \{\bar{X}_i\}_{i\in N} is the expected value of activities' duration and \{x_i\}_{i\in N} the observed value. If D(N,\prec,\{\bar{X}_i\}_{i\in N}) is the expected project time and D(N,\prec,\{x_i\}_{i\in N}) is the observed project time, it has to d=D(N,\prec,\{\bar{X}_i\}_{i\in N})-\delta is the delay, where \delta can be any arbitrary value greater than zero. The following rules distribute the delay costs among the different activities.

The proportional rule, from Brânzei et al. (2002), distributes the delay, d, proportionally. So that each activity receives a payment of:

\phi_i=\frac{\displaystyle x_{i}-\bar{X}_{i}}{\displaystyle \sum_{j\in N}\max\{x_{j}-\bar{X}_{j},0\}}\cdot C(D(N,\prec,\{\bar{X}_i\}_{i\in N})).

The truncated proportional rule, from Brânzei et al. (2002), distributes the delay, d, proportionally, where the individual delay of each player is reduced to d if if is larger. So that each activity receives a payment of:

\bar{\phi}_i=\frac{\displaystyle \min\{x_{i}-\bar{X}_{i},C(D(N,\prec,\{\bar{X}_i\}_{i\in N}))\}}{\displaystyle \sum_{j\in N} \max\{\min\{x_{j}-\bar{X}_{j},C(D(N,\prec,\{\bar{X}_i\}_{i\in N}))\},0\}}\cdot C(D(N,\prec,\{\bar{X}_i\}_{i\in N})).

These values are only well defined when the sum of the individual delays is different from zero.

Owen rule distributes the delay, d, based on the Owen value for TU games with a priori unions. Given a project problem with delays and unions (N,\prec,P,\{\bar{X}_i\}_{i\in N},\{x_i\}_{i\in N}), its associated TU game with a priori unions, (N,v,P), is v(S)=C(D(N,\prec,(\{\bar{X}_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))) for all S\subseteq N, where C is the costs function (by default C(D(N,\prec,y))=D(N,\prec,y)-\delta. If the number of activities is greater than ten, the Owen value, of the game (N,v,P), is estimated using a unique sampling process for all players.

Value

The delay value and a solution matrix.

Examples

prec1and2<-matrix(c(0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration<-c(2,1,1,4,2)
observed.duration<-c(2.5,1.25,2,4.5,3)
delta<-6
union<-list(c(1,2),c(3,4),c(5))
delay.pert.unions(duration,prec1and2=prec1and2,union=union,observed.duration=observed.duration,
delta=delta,cost.function=NULL)

Problems of distribution of delay in stochastic projects

Description

This function calculates the delay of a stochastic project, once it has been carried out. In addition, it also calculates the distribution of the delay on the different activities with the Stochastic Shapley rule.

Usage

delay.stochastic.pert(
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  distribution,
  values,
  observed.duration,
  percentile = NULL,
  delta = NULL,
  cost.function = NULL,
  compilations = 1000
)

Arguments

Details

Given a problem of sharing delays in a stochastic project (N,\prec,\{X_i\}_{i\in N},\{x_i\}_{i\in N}), such that \{X_i\}_{i\in N} is the random variable of activities' durations and \{x_i\}_{i\in N} the observed value. It is defined as E(D(N,\prec,\{X_i\}_{i\in N})) the expected project time, where E is the mathematical expectation, and D(N,\prec,\{x_i\}_{i\in N}) the observed project time, then d=D(N,\prec,\{X_i\}_{i\in N})-\delta, with \delta>0, normally \delta>E(D(N,\prec,\{X_i\}_{i\in N})), is the delay. The proportional and truncated proportional rule, see delay.pert function, can be adapted to this context by using the mean of the random variables.

The Stochastic Shapley, Gonçalves-Dosantos et al. (2020), rule is based on the Shapley value for the TU game (N,v) where v(S)=E(C(D(N,\prec,(\{X_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))), for all S\subseteq N, where C is the costs function (by default C(y)=D(N,\prec,y)-\delta). If the number of activities is greater than ten, the Shapley value, of the game (N,v), is estimated using a unique sampling process for all players, see Castro et al. (2009).

The Stochastic Shapley rule 2 is based on the sum of the Shapley values for the TU games (N,v) and (N,w) where v(S)=E(C(D(N,\prec,(\{X_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))))-E(C(D(N,\prec,(\{X_i\}_{i\in N})))) and w(S)=E(C(D(N,\prec,(\{0_i\}_{i\in N\backslash S},\{X_i\}_{i\in S})))), for all S\subseteq N, 0_N denotes the vector in R^N whose components are equal to zero and where C is the costs function (by default C(y)=D(N,\prec,y)-\delta).

Value

A delay value and solution vector.

References

castro

Castro, J., Gómez, D., & Tejada, J. (2009). Polynomial calculation of the Shapley value based on sampling. Computers & Operations Research, 36(5), 1726-1730.

gon

Gonçalves-Dosantos, J.C., García-Jurado, I., Costa, J. (2020) Sharing delay costs in Stochastic scheduling problems with delays. 4OR, 18(4), 457-476

Examples

prec1and2<-matrix(c(0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
distribution<-c("TRIANGLE","TRIANGLE","TRIANGLE","TRIANGLE","EXPONENTIAL")
values<-matrix(c(1,3,2,1/2,3/2,1,1/4,9/4,1/2,3,5,4,1/2,0,0),nrow=5,byrow=TRUE)
observed.duration<-c(2.5,1.25,2,4.5,3)
percentile<-NULL
delta<-6.5
delay.stochastic.pert(prec1and2=prec1and2,distribution=distribution,values=values,
observed.duration=observed.duration,percentile=percentile,delta=delta,
cost.function=NULL,compilations=1000)

Problems of distribution of delay in stochastic projects

Description

This function calculates the delay of a stochastic project, once it has been carried out. In addition, it also calculates the distribution of the delay on the different activities with the Stochastic Shapley rule.

Usage

delay.stochastic.pert.unions(
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  union,
  distribution,
  values,
  observed.duration,
  percentile = NULL,
  delta = NULL,
  cost.function = NULL,
  compilations = 1000
)

Arguments

Details

Given a problem of sharing delays in a stochastic project with unions (N,P,\prec,\{X_i\}_{i\in N},\{x_i\}_{i\in N}), such that \{X_i\}_{i\in N} is the random variable of activities' durations and \{x_i\}_{i\in N} the observed value. It is defined as E(D(N,\prec,\{X_i\}_{i\in N})) the expected project time, where E is the mathematical expectation, and D(N,\prec,\{x_i\}_{i\in N}) the observed project time, then d=D(N,\prec,\{X_i\}_{i\in N})-\delta, with \delta>0, normally \delta>E(D(N,\prec,\{X_i\}_{i\in N})), is the delay. The proportional and truncated proportional rule, see delay.pert function, can be adapted to this context by using the mean of the random variables.

The Stochastic Owen rule rule is based on the Owen value for the TU game (N,v,P) where v(S)=E(C(D(N,\prec,(\{X_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))), for all S\subseteq N, where C is the costs function (by default C(y)=D(N,\prec,y)-\delta). If the number of activities is greater than ten, the Owen value, of the game (N,v,P), is estimated using a unique sampling process for all players.

The Stochastic Owen rule 2 is based on the sum of the Owen values for the TU games (N,v,P) and (N,w,P) where v(S)=E(C(D(N,\prec,(\{X_i\}_{i\in N\backslash S},\{x_i\}_{i\in S}))))-E(C(D(N,\prec,(\{X_i\}_{i\in N})))) and w(S)=E(C(D(N,\prec,(\{0_i\}_{i\in N\backslash S},\{X_i\}_{i\in S})))), for all S\subseteq N, 0_N denotes the vector in R^N whose components are equal to zero and where C is the costs function (by default C(y)=D(N,\prec,y)-\delta).

Value

A delay value and solution vector.

Examples

prec1and2<-matrix(c(0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
distribution<-c("TRIANGLE","TRIANGLE","TRIANGLE","TRIANGLE","EXPONENTIAL")
values<-matrix(c(1,3,2,1/2,3/2,1,1/4,9/4,1/2,3,5,4,1/2,0,0),nrow=5,byrow=TRUE)
observed.duration<-c(2.5,1.25,2,4.5,3)
percentile<-NULL
delta<-6.5
union<-list(c(1,2),c(3,4),c(5))
delay.stochastic.pert.unions(prec1and2=prec1and2,union=union,distribution=distribution,
values=values,observed.duration=observed.duration,percentile=percentile,delta=delta,
cost.function=NULL,compilations=1000)

Early time for a deterministic projects

Description

This function calculates the early time for one project.

Usage

early.time(prec1and2 = matrix(0), prec3and4 = matrix(0), duration)

Arguments

Value

Early time vector.

References

burke

Burke, R. (2013). Project management: planning and control techniques. New Jersey, USA.

Examples

prec1and2<-matrix(c(0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration<-c(3,2,1,1.5,4.2)
early.time(prec1and2,duration=duration)

Last time for a deterministic projects

Description

This function calculates the last time for one project.

Usage

last.time(prec1and2 = matrix(0), prec3and4 = matrix(0), duration, early.times)

Arguments

Value

Last time vector.

References

bur

Burke, R. (2013). Project management: planning and control techniques. New Jersey, USA.

Examples

prec1and2<-matrix(c(0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration<-c(3,2,1,1.5,4.2)
early.times<-c(0,0,3.5,2,0)
last.time(prec1and2,duration=duration,early.times=early.times)

Project resource levelling

Description

This function calculates the schedule of the project so that the consumption of resources is as uniform as possible.

Usage

levelling.resources(
  duration,
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  resources,
  int = 1
)

Arguments

Details

The problem of leveling resources takes into account that in order for activities to be carried out in the estimated time, a certain level of resources must be used. The problem is to find a schedule that allows to execute the project in the estimated time so that the temporary consumption of resources is as level as possible.

Value

A solution matrices.

References

heg

Hegazy, T. (1999). Optimization of resource allocation and leveling using genetic algorithms. Journal of construction engineering and management, 125(3), 167-175.

Examples

duration<-c(3,4,2,1)
resources<-c(4,1,3,3)
prec1and2<-matrix(c(0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0),nrow=4,ncol=4,byrow=TRUE)

levelling.resources(duration,prec1and2,prec3and4=matrix(0),resources,int=1)

Build a precedence matrix

Description

This function calculates the costs per activity to accelerate the project.

Usage

mce(
  duration,
  minimum.durations,
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  activities.costs,
  duration.project = NULL
)

Arguments

Details

The MCE method (Minimal Cost Expediting) tries to speed up the project at minimum cost. It considers that the duration of some project activities could be reduced by increasing the resources allocated to them (and thus increasing their implementation costs).

Value

A solution matrices.

References

kelley

Kelley Jr, J. E. (1961). Critical-path planning and scheduling: Mathematical basis. Operations research, 9(3), 296-320.

Examples

duration<-c(5,4,5,2,2)
minimum.durations<-c(3,2,3,1,1)
activities.costs<-c(1,1,1,1,1)
prec1and2<-matrix(c(0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration.project<-6

mce(duration,minimum.durations,prec1and2,prec3and4=matrix(0),activities.costs,duration.project)

Organize project activities

Description

This function organizes the activities of a project, in such a way that if i precedes j then i is less strict than j.

Usage

organize(prec1and2 = matrix(0), prec3and4 = matrix(0))

Arguments

Value

A list containing:

• Precedence: ordered precedence matrix.

• Order: new activities values.

Examples

prec1and2<-matrix(c(0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
organize(prec1and2)

Build a precedence matrix

Description

This function builds a unique type 1 precedence matrix given any kind of precedence.

Usage

rebuild(prec1and2 = matrix(0), prec3and4 = matrix(0))

Arguments

Details

There are four types of precedence between two activities i,j: Type 1: the activity j cannot start until activity i has finished. Type 2: the activity j cannot start until activity i has started. Type 3: the activity j cannot end until activity i has ended. Type 4: the activity j cannot end until activity i has started.

All these precedences can be written only as type 1. It should be noted that precedence type 1 implies type 2, and type 2 implies type 4. On the other hand, precedence type 1 implies type 3, and type 3 implies type 4.

Value

A list containing:

• Precedence: precedence matrix.

• Type 2: activities related to type 2 precedence.

• Type 3: activities related to type 3 precedence.

• Type 4: activities related to type 4 precedence.

Examples

prec1and2<-matrix(c(0,1,0,2,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,2,0),nrow=5,ncol=5,byrow=TRUE)
prec3and4<-matrix(0,nrow=5,ncol=5)
prec3and4[3,1]<-3
rebuild(prec1and2,prec3and4)

Project resource allocation

Description

This function calculates the project schedule so that resource consumption does not exceed the maximum available per time period..

Usage

resource.allocation(
  duration,
  prec1and2,
  prec3and4 = matrix(0),
  resources,
  max.resources,
  int = 1
)

Arguments

Details

The problem of resource allocation takes into account that in order for activities to be carried out in the estimated time, a certain level of resources must be used. The problem is that the level of resources available in each period is limited. The aim is to obtain the minimum time and a schedule for the execution of the project taking into account this new restriction.

Value

A solution matrices.

References

hega

Hegazy, T. (1999). Optimization of resource allocation and leveling using genetic algorithms. Journal of construction engineering and management, 125(3), 167-175.

Examples

duration<-c(3,4,2,1)
prec1and2<-matrix(c(0,1,0,0,0,0,0,0,0,0,0,1,0,0,0,0),nrow=4,ncol=4,byrow=TRUE)
resources<-c(4,1,3,3)
max.resources<-4

resource.allocation(duration,prec1and2,prec3and4=matrix(0),resources,max.resources,int=1)

Schedule for deterministic projects

Description

This function calculates the duration of the project, the slacks for each activity, as well as the schedule of each activity.

Usage

schedule.pert(
  duration,
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  PRINT = TRUE
)

Arguments

Value

A list of a project schedule and if PRINT=TRUE a plot of schedule.

References

burk

Burke, R. (2013). Project management: planning and control techniques. New Jersey, USA.

Examples

prec1and2<-matrix(c(0,0,1,0,0,0,0,0,1,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
duration<-c(3,2,1,1.5,4.2)
schedule.pert(duration,prec1and2)

Stochastic projects

Description

This function calculates the average duration time for a stochastic project and the activities criticality index. It also plots the estimate density of the project duration, as well as the estimate density of the early and last times.

Usage

stochastic.pert(
  prec1and2 = matrix(0),
  prec3and4 = matrix(0),
  distribution,
  values,
  percentile = 0.95,
  plot.activities.times = NULL,
  compilations = 1000
)

Arguments

Value

Two values, average duration time and the maximum time allowed, a critically index vector and a durations histogram.

Examples

prec1and2<-matrix(c(0,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0),nrow=5,ncol=5,byrow=TRUE)
distribution<-c("TRIANGLE","TRIANGLE","TRIANGLE","TRIANGLE","EXPONENTIAL")
values<-matrix(c(1,3,2,1/2,3/2,1,1/4,9/4,1/2,3,5,4,1/2,0,0),nrow=5,byrow=TRUE)
percentile<-0.95
plot.activities.times<-c(1,4)
stochastic.pert(prec1and2=prec1and2,distribution=distribution,values=values,
percentile=percentile,plot.activities.times=plot.activities.times)