Optimize Heavy Equipment Rental Costs with Linear Programming

School

Sebelas Maret University**We aren't endorsed by this school

Course

SOLO 3

Subject

Industrial Engineering

Date

Dec 12, 2024

Pages

Uploaded by UltraFlower635

Nabila Nur Amalina Luthfi F0222129 Assignment 5 1.A heavy equipment company “Mbah Bejo” is a company that rents out heavy equipment such as excavators (backhoe), fork-lifts, palette-jacks, mini-cranes, etc. This company has 14 heavy equipment located in Jakarta totalling 6 units and in Surabaya 8 units. Heavy equipment will be rented by a road construction company for the construction of Trans-Java toll roads which will be used in 6 cities, namely Tasikmalaya 2, Cirebon 1, Jogja 4, Solo 3, Madiun 2, and Jember 2. Due to road conditions, transportation cannot go directly from the origin city to the destination city and must go through transit cities, namely Bandung, Semarang and Malang. The flow of goods delivery and the cost of transporting a heavy equipment is shown in the following table. Table of transportation cost/unit from origin city to transit city. Bandung Semarang Malang Jakarta 10 17 25 Surabaya 22 15 12 Table of transportation cost/unit from transit city to destination city. Tasikmalaya Cirebon Jogjakarta Solo Madiun Jember Bandung 8 12 Semarang 15 12 10 Malang 23 18 7 12 a.Draw the supply/availability of equipment, the demand/need of equipment and the transshipment cost/unit from origin city to transit city, then from transit city to destination city. b.Formulate the problem into a linear programming to determine the minimum cost of transshipment faced by the company. c.Solve the problem using POM-QM for Windows. d.Interpret and explain the route of transshipment from origin city, transit city to destination city. e.How much minimum transshipment cost should be incurred by the company to ship the heavy equipment from origin city to destination city. f.Draw the route and allocation of heavy equipment rented by the construction company from origin city to destination city which have the minimum shipment cost. 2.Four automobiles have entered Bubba’s Repair Shop for various types of work, ranging from a transmission overhaul to a brake job. The experience level of the mechanics is quite varied, and Bubba would like to minimizethe time required to complete all of the jobs. He has estimated the time in minutes for each mechanic to complete each job. Billy can complete job 1 in 400 minutes, job 2 in 90 minutes, job 3 in 60 minutes, and job 4 in 120 minutes. Taylor will finish job 1 in 650 minutes, job 2 in 120 minutes, job 3 in 90 minutes, and job 4 in 180 minutes. Mark will finish job 1 in 480 minutes, job 2 in 120 minutes, job 3 in 80 minutes, and job 4 in 180 minutes. John will complete job 1 in 500 minutes, job 2 in 110 minutes, job 3 in 90 minutes, and job 4 in 150 minutes. Each mechanic should be assigned to just one of these jobs. What is the minimum total time required to finish the four jobs? Who should be assigned to each job?

*** 1. a. To From TasikmalayaCirebonYogyakartaSoloMadiunJemberBandung (T)Semarang (T)Malang (T)SupplyJakartaX X X X X X 10 17 25 6 SurabayaX X X X X X 22 15 12 8 Bandung (T)8 12 X X X X X X X 14 Semarang (T)X 15 12 10 X X X X X 14 Malang (T)X X 23 18 7 12 X X X 14 Demmand2 1 4 3 2 2 14 14 14 To From Tasikmalaya Cirebon Yogyakarta Solo Madiun Jember Bandung (T) Semarang (T) Malang (T) SupplJakarta 18 22 29 27 30 37 10 17 25 6 Surabaya 30 30 27 25 19 24 22 15 12 8 Bandung (T) 8 12 X X X X X X X 14 Semarang (T) X 15 12 10 X X X X X 14 Malang (T) X X 23 18 7 12 X X X 14 Demmand 2 1 4 3 2 2 14 14 14 b. Minimum Transportation Cost= 10XAC + 17XAD + 25XAE + 22XBC + 15XBD + 12 XBE + 8XCF + 12XCG + 15XDG + 12XDH + 10XDI + 18XEI + 7XEJ + 12 XEK. Constraint Function = XAC + XAD + XAE = 6 ( Transporation function from Jakarta to the transit point (Bandung, Semarang, Malang)) XBC + XBD + XBE = 8 (Transportation function from Surabaya to the transit point (Bandung, Semarang, Malang)) XAC + XBC – XCF – XCG = 0 (To make sure all the goods are out of the Bandung transit area.) XAD + XBD – XDG – XDH – XDH – XDI = 0 (To make sure all the goods are out of the Semarang transit area.) XAE + XBE – XEI – XEJ – XEK = 0 (To make sure all the goods are out of the Malang transit area)

XCF = 2 (Bandung (T) ke Tasikmalaya) XCG + XDG = 1 (From Bandung (T) and Semarang (T) to Cirebon) XDH + XEH = 4 (From Semarang (T) and Malang (T) to Yogyakarta) XDI + XEI = 3 (From Semarang (T) and Malang (T) to Solo) XEJ = 2 ( From Malang (T) to Madiun) XEK = 2 (From Malang (T) to Jember) c. d. From that QM solve, knowing: -There are 6 Excavator from Jakarta transit to Bandung -There are 8 Excavator fom Surabaya transit to Malang -From Bandung (Transit) the goods will be sent to Jogjakarta (4 unit), Solo (3 unit), Madiun (2 unit), and Jember (2 unit) -From Semarang (Transit) the goods will be sent to Tasikmalaya (2 unit), and Semarang (5 Unit) -From Malang (Transit) the goods will be sent to Cirebon (1 unit) and Semarang (7 Unit) e. The minimum transhipment cost that will be borne by the company is as follows: = (6 x 10 + 8 x 12) = 60 + 96 = $156 f. Delivery routes and machine allocation from Jakarta and Surabaya to destination locations:

2. Job Worker’s Name Job 1 Job 2 Job 3 Job 4 Billy 400 90 60 120 Taylor 650 120 90 180 Mark 480 120 80 180 John 500 110 90 150 Objective Function: minimize employee assignment cost Z= 400XA1 + 90XA2 + 60XA3 + 120 XA4 + 650 XB1 + 120XB2 + 90XB3 + 180XB4 + 480XC1 + 120XC2 + 80XC3 + 180XC4 + 500XD1 + 110XD2 + 90XD3 + 150XD4 Constraint Function = XA1 + XA2 + XA3 + XA4 ≤1 (constraint for Billy) XB1 + XB2 + XB3 + XB4 ≤1 constraint for Taylor) XC1 + XC2 +XC3 + XC4 ≤1 (constraint for Mark) XD1 + XD2 + XD3 + XD4 ≤1 (constraint for John) XA1 + XB1 + XC1 + XD1 ≤1 constraint for Job 1) XA2 + XB2 + XC2 + XD2 ≤1(constraint for Job 2) XA3 + XB3 + XC3 + XD3 ≤1(constraint for Job 3) XA4 + XB4 + XC4 + XD4 ≤1 (constraint for Job 4)

Job Worker’s Name Job 1 Job 2 Job 3 Job 4 Billy 340 30 0 60 Taylor 560 30 0 90 Mark 400 40 0 100 John 410 20 0 60 Job Worker’s Name Job 1 Job 2 Job 3 Job 4 Billy 0 10 0 0 Taylor 220 10 0 30 Mark 60 20 0 70 John 70 0 0 0 Job Worker’s Name Job 1 Job 2 Job 3 Job 4 Billy 0 10 10 0 Taylor 210 0 0 20 Mark 50 10 0 60 John 70 0 10 0 Workers Assigned Work Minimum cost incurred Mark 3 80 Taylor 2 120 John 4 150 Billy 1 400 Total 750