# Instalando o pacote Lpsolve<\/span><\/span>\ninstall.packages<\/span> (<\/span>"<\/span>lpSolve<\/span>"<\/span>)<\/span><\/span>\n<\/span>\n# Instalando o pacote LpSolveAPI<\/span><\/span>\ninstall<\/span> .packages <\/span>(<\/span>"<\/span>lpSolveAPI<\/span>"<\/span>)<\/span><\/span><\/code><\/pre><\/div>\n\n\n\n
Declarando o problema<\/h2>\n\n\n\n
Para ficar mais f\u00e1cil o entendimento da Programa\u00e7\u00e3o Linear, vamos utilizar o exemplo de otimiza\u00e7\u00e3o linear encontrado no livro Modeling and Solving Linear Programming with R. Sendo um problema de produ\u00e7\u00e3o de duas cadeiras, segue o enunciado do problema: <\/p>\n\n\n\n
Uma empresa produz dois modelos de cadeiras: 4P e 3P. O modelo 4P precisa de 4 pernas, 1 assento e 1 encosto. Por outro lado, o modelo 3P precisa de 3 pernas e 1 assento. A empresa tem um estoque inicial de 200 pernas, 500 assentos e 100 encostos. Se a empresa precisar de mais pernas, assentos e encostos, pode comprar blocos de madeira padr\u00e3o, cujo custo \u00e9 de 80 reais por bloco.<\/em><\/p>\n\n\n\n
 A empresa pode produzir 10 assentos, 20 pernas e 2 encostos de um bloco de madeira padr\u00e3o. O custo de produ\u00e7\u00e3o do modelo 4P \u00e9 de 30 reais por cadeira, enquanto o custo do modelo 3P \u00e9 de 40 reais por cadeira. Por fim, a empresa informou que a quantidade m\u00ednima de cadeiras a serem produzidas \u00e9 de 1000 unidades por m\u00eas. Sendo necess\u00e1rio definir um modelo de programa\u00e7\u00e3o linear, que minimize o custo total (os custos de produ\u00e7\u00e3o das duas cadeiras, mais a compra de novos blocos de madeira).<\/em><\/p>\n\n\n\n
Caracter\u00edsticas das Cadeiras<\/h3>\n\n\n\n
Modelo Cadeira<\/td>
N\u00famero de Pernas<\/td>
N\u00famero de assentos<\/td>
N\u00famero de Encostos<\/td>
Custo da Produ\u00e7\u00e3o<\/td><\/tr>
4P<\/td>
4<\/td>
1<\/td>
1<\/td>
R$30<\/td><\/tr>
3P<\/td>
3<\/td>
1<\/td>
0<\/td>
R$40<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
Estoque inicial<\/h3>\n\n\n\n
N\u00famero de pernas<\/td>
N\u00famero de assentos<\/td>
N\u00famero de encostos<\/td><\/tr>
200<\/td>
500<\/td>
100<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
1 bloco de madeira: custo de 80 reais<\/h3>\n\n\n\n
Produz – N\u00ba de pernas<\/td>
Produz – N\u00ba de assentos<\/td>
Produz – N\u00ba de encostos<\/td><\/tr>
20<\/td>
10<\/td>
2<\/td><\/tr><\/tbody><\/table><\/figure>\n\n\n\n
Agora que temos um problema a ser resolvido, precisamos identificar os 3 componentes principais.<\/p>\n\n\n\n
1\u00ba) Vari\u00e1veis de Decis\u00e3o: S\u00e3o as vari\u00e1veis utilizadas no modelo.<\/p>\n\n\n\n
\n
modelo 4P: \u20ac 30 (Custo de produ\u00e7\u00e3o unid. do modelo 4P)<\/li>\n\n\n\n
modelo 3P: \u20ac 40 (Custo de produ\u00e7\u00e3o unid. do modelo 3P)<\/li>\n\n\n\n
bloco madeira: \u20ac 80 (Custo da unid. bloco de madeira)<\/li>\n<\/ul>\n\n\n\n
2\u00ba) Fun\u00e7\u00e3o objetivo: \u00c9 uma fun\u00e7\u00e3o matem\u00e1tica que representa o principal objetivo do tomador de decis\u00e3o. Ela pode ser de dois tipos: miniza\u00e7\u00e3o ou maximiza\u00e7\u00e3o<\/p>\n\n\n\n
custo = 30 + modeloAP + 40 + modelo3P + 80 * blocoMadeira = MIN(custo)<\/em><\/p>\n\n\n\n
3\u00ba) Restri\u00e7\u00f5es: S\u00e3o as regras que dizem o que podemos (ou n\u00e3o) fazer e\/ou quais s\u00e3o as limita\u00e7\u00f5es dos recursos ou
atividades associadas ao modelo.<\/p>\n\n\n\n
\n
modelo_4P + modelo_3P <= 500estoque<\/em> + 10um bloco de madeira <\/em>(N\u00ba de acentos)<\/li>\n\n\n\n
4 * modelo_4P + 3 * modelo_3P <= 200estoque<\/em> + 20um bloco de madeira<\/em> (N\u00ba de pernas)<\/li>\n\n\n\n
modelo_4P <= 100estoque<\/em> + 2um bloco de madeira<\/em> (N\u00ba de Encostos)<\/li>\n\n\n\n
modelo_4P + modelo_3P >= 1000 (Quantidade m\u00ednima de cadeiras)<\/li>\n<\/ul>\n\n\n\n
Implementando a solu\u00e7\u00e3o no R<\/h2>\n\n\n\n
Vamos usar o pacote lpSolve<\/strong> para encontrar a solu\u00e7\u00e3o para o nosso problema<\/p>\n\n\n\n
# Carregando o pacote LpSoLlve<\/span><\/span>\nlibrary<\/span> (<\/span>lpSolve<\/span>)<\/span><\/span>\n<\/span>\n# Definindo as constantes da fun\u00e7\u00e3o objetivo<\/span><\/span>\nconstantes_objetivo<\/span> <\/span><-<\/span> <\/span>c<\/span>(<\/span>30<\/span>, <\/span>40<\/span>, <\/span>80<\/span>)<\/span><\/span>\n<\/span>\n# Criando uma matriz com as restri\u00e7\u00f5es<\/span><\/span>\nrestricoes<\/span> <\/span><-<\/span> <\/span>matrix<\/span>(<\/span>c<\/span>(<\/span><\/span>\n<\/span>\n1<\/span>, <\/span>1<\/span>, <\/span>-<\/span>10<\/span>, <\/span># N\u00ba de ocentos<\/span><\/span>\n<\/span>\n4<\/span>, <\/span>3<\/span>, <\/span>-<\/span>20<\/span>, <\/span># N\u00ba de pernas<\/span><\/span>\n<\/span>\n1<\/span>, <\/span>0<\/span>, <\/span>-<\/span>2<\/span>, <\/span># N\u00ba de Encostos<\/span><\/span>\n<\/span>\n1<\/span>, <\/span>1<\/span>, <\/span>0<\/span> <\/span># Quant\u00eddade m\u00edntma de cadeiras<\/span><\/span>\n), <\/span>nrow<\/span> <\/span>=<\/span> <\/span>4<\/span> , <\/span>byrow<\/span> <\/span>=<\/span> <\/span>TRUE<\/span> )<\/span><\/span>\n<\/span>\n# Valores que ficam do lado direito da restr\u00ed\u00e7\u00e3o<\/span><\/span>\nvalores_restricao<\/span> <\/span><-<\/span> <\/span>c<\/span>(<\/span>500<\/span>, <\/span>200<\/span>, <\/span>100<\/span>, <\/span>1000<\/span>)<\/span><\/span>\n<\/span>\n# Definindo as equa\u00e7\u00f5es ou inequa\u00e7\u00f5es da restri\u00e7\u00e3o<\/span><\/span>\ndirecao_restricao<\/span> <\/span><-<\/span> <\/span>c<\/span>(<\/span>"<\/span><=\u201d, \u201c<=<\/span>"<\/span>, \u201c<\/span><=<\/span>\u201d, <\/span>"<\/span>>=<\/span>"<\/span>)<\/span><\/span>\n<\/span>\n# Encontrando a solu\u00e7\u00e3o \u00f3tima<\/span><\/span>\nsolucao_otima<\/span> <\/span><-<\/span> lp<\/span>(<\/span><\/span>\ndirection<\/span> <\/span>=<\/span> <\/span>"<\/span>min<\/span>"<\/span>,<\/span> <\/span># Objetiva<\/span><\/span>\nobjective.in<\/span> <\/span>=<\/span> <\/span>constantes_objetivo<\/span>,<\/span><\/span>\nconst.mat<\/span> <\/span>=<\/span> <\/span>restricoes<\/span>,<\/span><\/span>\nconst.dir<\/span> <\/span>=<\/span> <\/span>direcao_restricao<\/span>,<\/span><\/span>\nconst.rhs<\/span> <\/span>=<\/span> <\/span>valores_restricao<\/span>,<\/span><\/span>\nall.int<\/span> <\/span>=<\/span> <\/span>T<\/span><\/span>\n)<\/span><\/span>\n<\/span>\n# Status da sotu\u00e7\u00e3o<\/span><\/span>\n<\/span>\nstatus_solucao<\/span> <\/span><-<\/span> <\/span>solucao_otima<\/span>$<\/span>status<\/span><\/span>\n# Solu\u00e7\u00e3o do problema<\/span><\/span>\nsolucao_problema<\/span> <\/span><-<\/span> <\/span>solucao_otima<\/span>$<\/span>solution<\/span><\/span>\nnames<\/span>(<\/span>solucao_problema<\/span>) <\/span><-<\/span> <\/span>c<\/span>(\u201c<\/span>Cadeira<\/span> <\/span>4P<\/span>\u201d, \u201c<\/span>Cadeira<\/span> <\/span>3P<\/span>"<\/span>, \u201cBloco Madeira\u201d)<\/span><\/span>\n# Custo total <\/span><\/span>\ncusto_total <- solucao_otima$objval<\/span><\/span><\/code><\/pre><\/div>\n\n\n\n
Visualizando a solu\u00e7\u00e3o<\/h2>\n\n\n\n
# Status da solu\u00e7\u00e3o<\/span><\/span>\n# 0 = sucesso, 2 = nenhuma solu\u00e7\u00e3o vi\u00e1vel<\/span><\/span>\nprint<\/span>(<\/span>paste<\/span>(<\/span>"<\/span>Status da solu\u00e7\u00e3o:<\/span>"<\/span>, <\/span>ifelse<\/span>(<\/span>status_solucao<\/span> <\/span>==<\/span> <\/span>0<\/span>, \u201c<\/span>Sucesso<\/span>.\u201d, \u201c<\/span>Nenhuma<\/span> <\/span>Solu\u00e7\u00e3o<\/span> <\/span>V\u00ed\u00e1vel<\/span>.\u201d<\/span>)<\/span>))<\/span><\/span>\n<\/span>\n## [1] "Status da solu\u00e7\u00e3o: Sucesso.\u201d<\/span><\/span>\n<\/span>\n# Solu\u00e7\u00e3o do problema<\/span><\/span>\nprint<\/span>(<\/span>solucao_problema<\/span>)<\/span><\/span>\n<\/span>\n##  Cadeira 4P   Cadeira 3P  Bloco Madeira<\/span><\/span>\n##              420               580                   161<\/span><\/span>\n<\/span>\n# Verificando o custo total<\/span><\/span>\nprint<\/span>(<\/span>paste<\/span>(<\/span>"<\/span>Custo total:<\/span>"<\/span>, <\/span>custo<\/span> <\/span>total<\/span>))<\/span><\/span>\n<\/span>\n## [1] "Custo total: 48680\u201d<\/span><\/span><\/code><\/pre><\/div>\n\n\n\n
Outro pacote dispon\u00edvel para linguagem R \u00e9 o IpSolveAPI<\/strong> que permite visualizar um relat\u00f3rio de resolu\u00e7\u00e3o do problema de Programa\u00e7\u00e3o Linear, segue abaixo o c\u00f3digo da resolu\u00e7\u00e3o do problema utilizando esse pacote:<\/p>\n\n\n\n
# Carregando o pacote LpSolveAPI<\/span><\/span>\n<\/span>\nlibrary<\/span>(<\/span>lpsolveAPI<\/span>)<\/span><\/span>\n<\/span>\n# Definindo que o problema ter\u00e1 4 restri\u00e7\u00f5es com 3 vari\u00e1veis<\/span><\/span>\nproblema<\/span> <\/span><-<\/span> make.lp<\/span>(<\/span>nrow<\/span> <\/span>=<\/span> <\/span>4<\/span>,<\/span> <\/span>ncol<\/span> <\/span>=<\/span> <\/span>3<\/span>)<\/span><\/span>\n<\/span>\n# Definindo o tipo de problema que ser\u00e1 resolvido 'minimiza\u00e7\u00e3o'<\/span><\/span>\nlp.control<\/span>(<\/span>problema<\/span>,<\/span> <\/span>sense<\/span>=<\/span>"<\/span>min<\/span>"<\/span>)<\/span><\/span>\n<\/span>\n## $anti.degen<\/span><\/span>\n## [1] \u201cfixedvars\u201d \u201cstalling"<\/span><\/span>\n##<\/span><\/span>\n## $basis.crash<\/span><\/span>\n## [1] "none\u201d<\/span><\/span>\n##<\/span><\/span>\n## $bb.depthlimit<\/span><\/span>\n## [1] -50<\/span><\/span>\n##<\/span><\/span>\n## $bb. floorfirst<\/span><\/span>\n## [1] \u201cautomatic\u201d<\/span><\/span>\n##<\/span><\/span>\n## $bb.rule<\/span><\/span>\n## [1] "pseudononint\u201d \u201cgreedy" \u201cdynanic" "rcostfixing"<\/span><\/span>\n##<\/span><\/span>\n## $break.at.first<\/span><\/span>\n## [1] FALSE<\/span><\/span>\n##<\/span><\/span>\n## $break.at.value<\/span><\/span>\n## [1] -1e+30<\/span><\/span>\n##<\/span><\/span>\n## $epsilon<\/span><\/span>\n## epsb  epsd  epsel  epsint epsperturb  epspivot<\/span><\/span>\n## 1e-10 1e-09 1e-12 1e-07 1e-05 2e-07<\/span><\/span>\n##<\/span><\/span>\n## $improve<\/span><\/span>\n## [1] "dualfeas" "thetagap"<\/span><\/span>\n##<\/span><\/span>\n## $infinite<\/span><\/span>\n## [1] 1e+30<\/span><\/span>\n##<\/span><\/span>\n## $maxpivot<\/span><\/span>\n## [1] 250<\/span><\/span>\n##<\/span><\/span>\n## $mip.gap<\/span><\/span>\n## absolute relative<\/span><\/span>\n1e-11<\/span><\/span>\n1e-11<\/span><\/span>\n##<\/span><\/span>\n## $negrange<\/span><\/span>\n## [1] -1e+06<\/span><\/span>\n##<\/span><\/span>\n## $obj.in.basis<\/span><\/span>\n## [1] TRUE<\/span><\/span>\n##<\/span><\/span>\n## $pivoting<\/span><\/span>\n## [1] "devex" "adaptive"<\/span><\/span>\n##<\/span><\/span>\n## $presolve<\/span><\/span>\n## [1] "none"<\/span><\/span>\n##<\/span><\/span>\n## $scalelimit<\/span><\/span>\n## [1] 5<\/span><\/span>\n##<\/span><\/span>\n## $scaling<\/span><\/span>\n## [1] "geometric" "equilibrate" "integers"<\/span><\/span>\n##<\/span><\/span>\n## $sense<\/span><\/span>\n## [1] "minimize"<\/span><\/span>\n##<\/span><\/span>\n## $simplextype<\/span><\/span>\n## [1] "dual" "primal"<\/span><\/span>\n\ufeff<\/span>##<\/span><\/span>\n## $timeout<\/span><\/span>\n## [1] @<\/span><\/span>\n##<\/span><\/span>\n## $verbose<\/span><\/span>\n## [1] "neutral"<\/span><\/span>\n<\/span>\n# Definindo o tipo das vari\u00e1veis de decis\u00e3o <\/span><\/span>\nset.type<\/span>(<\/span>problema<\/span>,<\/span> <\/span>1<\/span>:<\/span>3<\/span>,<\/span> <\/span>type<\/span>=<\/span>c<\/span>(<\/span>"<\/span>integer<\/span>"<\/span>)<\/span>)<\/span><\/span>\n<\/span>\n# Definindo as constantes da fun\u00e7\u00e3o objetivo <\/span><\/span>\nconstantes<\/span> <\/span>objetivo<\/span> <\/span><-<\/span> <\/span>c<\/span>(<\/span>38<\/span>, <\/span>40<\/span>, <\/span>30<\/span>) <\/span><\/span>\nset.objfn<\/span>(<\/span>problema<\/span>,<\/span> <\/span>constantes_objetivo<\/span>)<\/span><\/span>\n<\/span>\n# Adicionando as restri\u00e7\u00f5es<\/span><\/span>\nadd.constraint<\/span>(<\/span>problema<\/span>,<\/span> <\/span>c<\/span>(<\/span>1<\/span>,<\/span> <\/span>1<\/span>,<\/span> <\/span>-<\/span>10<\/span>)<\/span>, <\/span>"<\/span><=<\/span>"<\/span>, <\/span>500<\/span>) add.constraint<\/span>(<\/span>problema<\/span>,<\/span> <\/span>c<\/span>(<\/span>4<\/span>,<\/span> <\/span>3<\/span>,<\/span> <\/span>-<\/span>20<\/span>)<\/span>, <\/span>"<\/span><<\/span>"<\/span>, <\/span>200<\/span>) add.constraint <\/span>(<\/span>problema<\/span>,<\/span> <\/span>c<\/span>(<\/span>1<\/span>,<\/span> <\/span>0<\/span>,<\/span> <\/span>-<\/span>2<\/span>)<\/span>, <\/span>"<\/span><<\/span>"<\/span>, <\/span>100<\/span>) add.constraint<\/span>(<\/span>problema<\/span>,<\/span> <\/span>c<\/span>(<\/span>1<\/span>,<\/span> <\/span>1<\/span>,<\/span> <\/span>0<\/span>)<\/span>, <\/span>"<\/span>><\/span>"<\/span>, <\/span>1000<\/span>)<\/span><\/span>\n<\/span>\n# Visualizando o relat\u00f3rio do problema print(problema)<\/span><\/span>\n## Model name:<\/span><\/span>\n##             C1 C2 C3<\/span><\/span>\n## Minimize    30 40 80<\/span><\/span>\n## R1           0  0  0 free 0<\/span><\/span>\n## R2           0  0  0 free 0<\/span><\/span>\n## R3           0  0  0 free 0<\/span><\/span>\n## R4           0  0  0 free 0<\/span><\/span>\n## R5           1  1-10 <= 500<\/span><\/span>\n## R6           4  3-20 <= 200<\/span><\/span>\n## R7           1  0 -2 <= 100<\/span><\/span>\n## R8           1  1  0 >=1000<\/span><\/span>\n## Kind        Std Std Std<\/span><\/span>\n## Type        Int Int Int<\/span><\/span>\n## Upper       Inf Inf Inf<\/span><\/span>\n## Lowen        0  0  0<\/span><\/span>\n<\/span>\n#Resolvendo o problema<\/span><\/span>\nsolve<\/span>(<\/span>problema<\/span>) <\/span># Status da solu\u00e7\u00e3o: 0 = sucesso, 2 nenhuma solu\u00e7\u00e3o vi\u00e1vel<\/span><\/span>\n<\/span>\n## [1] 0<\/span><\/span>\n<\/span>\n# Obtendo os valores das vari\u00e1veis<\/span><\/span>\nget.variables<\/span>(<\/span>problema<\/span>)<\/span><\/span>\n <\/span><\/span>\n## [1] 420 580 161<\/span><\/span>\n<\/span>\n# Obtendo o valor da fun\u00e7\u00e3o objetivo get.objective (problema) # Custo total<\/span><\/span>\n<\/span>\n## [1] 48680 <\/span><\/span><\/code><\/pre><\/div>\n\n\n\n
Por fim, o resultado foi completamente resolvido utilizando a Linguagem R, das duas formas que apresentamos. Ainda restou alguma d\u00favida? Deixe aqui nos coment\u00e1rios. At\u00e9 a pr\u00f3xima! <\/p>\n","protected":false},"excerpt":{"rendered":"
Descubra o que \u00e9 Pesquisa Operacional e como a programa\u00e7\u00e3o linear est\u00e1 inserida dentro deste conceito t\u00e3o importante da estat\u00edstica.<\/p>\n","protected":false},"author":14,"featured_media":27653,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"","site-content-layout":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","spay_email":"","footnotes":""},"categories":[445,442],"tags":[38,46,455,176,383,384,381,382,350],"class_list":["post-27638","post","type-post","status-publish","format-standard","has-post-thumbnail","hentry","category-ferramentas-e-tecnologias","category-fundamentos-de-estatistica","tag-data-science","tag-estatistica-2","tag-ferramentas","tag-linguagem-r","tag-lpsolve","tag-lpsolve-api","tag-pesquisa-operacional","tag-r","tag-statplace-2"],"yoast_head":"\n