IE 416 Extra Credit

Page 371 Problem 4

Group 1Vanessa Condie

Lucia HortaGeoffrey Cheung

October 4, 2011

Industrial and Manufacturing Engineering DepartmentCalifornia State Polytechnic University, Pomona

Table of Contents

Problem Statement:..........................................................................................................................2

Calculate Plant Capacity..................................................................................................................2

Initial Graphical Representation......................................................................................................3

Dummy Demand Point....................................................................................................................3

Updated Graphical Representation..................................................................................................4

Transportation Tableau....................................................................................................................4

Mathematical Representation..........................................................................................................4

Initial Basic Feasible Solution (BFS) using Northwest Corner Method (NWC)............................5

Solving Using WinQSB...................................................................................................................5

Sensitivity Analysis.........................................................................................................................7

Report to Manager.........................................................................................................................10

Part B.............................................................................................................................................14

1

Problem Statement:

Steelco manufactures three types of steel at different plants. The time required to manufacture 1 ton of steel (regardless of type) and the costs at each plant are shown in Table 8. Each week, 100 tons of each type of steel (1, 2, and 3) must be produced. Each plant is open 40 hours per week.a) Formulate a balanced transportation problem to minimize the cost of meeting Steelco’s

weekly requirements.b) Suppose the time required to produce 1 ton of steel depends on the type of steel as well as on

the plant at which it is produced (see Table 9, page 372). Could a transportation problem still be formulated?

Table 8Table 8 Cost ($) Plant Steel 1 Steel 2 Steel 3 Time (minutes)

1 60 40 28 202 50 30 30 163 43 20 20 15

Table 9Table 9 Time (minutes)Plant Steel 1 Steel 2 Steel 3

1 15 12 152 15 15 203 10 10 15

Calculate Plant Capacity

We are given demand for each type of steel in units of tons. The capability of each plant is given in minutes. We must convert capability of each plant to units of tons so that we can solve.

40 hour work week100 tons of each steel type per week

40 hrs∗60minshr

=2400mins forwork

Plant 12400mins

20mins/ ton=120 tons

Plant 2

2

2400mins16mins /ton

=150 tons

Plant 32400mins

15mins/ ton=160 tons

Plant Supply Steel Type SupplyPlant 1 120 tons Steel 1 100 tonsPlant 2 150 tons Steel 2 100 tonsPlant 3 160 tons Steel 3 100 tonsTotal 430 tons Total 300 tons

At this point it does not matter which plant produces which type of steel.

Total Supply: 430 tonsTotal Demand: 300 tons

Initial Graphical Representation

Dummy Demand Point

∑ si≥∑ d iAdd Dummy Demand point

Capacity: ∑ si−∑ d i ;Cij=0130 tons excess supply

3

4

Updated Graphical Representation

Transportation Tableau

Steel1 Steel 2 Steel 3 Dummy

Plant 1 60 40 28 0

120

Plant 2 50 30 30 0

150

Plant 3 20 20 20 0

160

100 100 100 130 430

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Mathematical Representation

min z=60 x11+40x12+28 x13+50 x21+30 x22+30 x23+43 x31+20 x32+20x33

Supply constraintsx11+x12+x13+x14≤120x21+x22+x23+x24≤150x31+x32+ x33+x34≤160

Demand constraintsx11+x21+x31≥100x12+x22+x32≥100x13+x23+x33≥100x14+x24+x34≥130 (We do not formulate the dummy.)

Xij = # sent from I to j

Initial Basic Feasible Solution (BFS) using Northwest Corner Method (NWC)

Steel1 Steel 2 Steel 3 Dummy Capacit

y

Plant 1 60 40 28 0

120100 20

Plant 2 50 30 30 0

150 80 70

Plant 3 20 20 20 0

160 30 130

Capacity 100 100 100 130 430

X11=100X12=20X22=80X23=70X33=30X34=130

Total Cost = 11900

6

Solving Using WinQSB

Input data into WinQSB table.

WinQSB Solution have complete solution with NBV

Our optimal solution is 9320.

Graphical Solution

Summary TableFrom To Amount Unit Cost Total CostPlant 1 Steel 3 40 28 1120Plant 2 Steel 1 100 50 5000Plant 3 Steel 2 100 20 2000Plant 3 Steel 3 60 20 1200      Total 9320

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Our optimal solution is 9320.

Sensitivity Analysis

We are interested in performing sensitivity analysis from plant 1 to steel 3 because it contains the second highest cost and has a reduction price of $8 per unit increase. Not clear and may be not a good one ?????

Range of Optimality

Range of Feasibility

8

Sensitivity Analysis from $26 to $31 unit cost for Plant 1 to Steel 3. Explain motivation for this range

Results of this analysis are below:

Based on the results, we should not manufacture send steel 3 at from plant 1 if the unit cost is higher than $30.

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Sensitivity analysis from 150 to 170 for capacity of Plant 3. Change heading

Plant 3We are interested in the production capacity of Plant 3. If we purchase new faster machines we can increase capacity. If we purchase new cheaper machines, they will be slower and our capacity will decrease. Rewrite motivation

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We can see that as we increase capacity of Plant 3, the total cost will continue to decrease.

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Report to Manager

Our recommended plan to meet demand has a minimum total weekly cost of $9320.

A summary plan is shown below:

Fromproduce

Amount Unit Cost Total Cost

Plant 1 Steel 3 40 28 1120Plant 2 Steel 1 100 50 5000Plant 3 Steel 2 100 20 2000Plant 3 Steel 3 60 20 1200Plant 1 – Excess 80Plant 2 – Excess 50      Total 9320

From Plant 1 we should send manufacture 40 tons of Steel 3. From Plant 2 to Steel 1 we should send 100 tons of steel. From Plant 3 to Steel 2 we should send 100 tons of steel and from Plant 3 to Steel 3 we should send 60 tons of steel. There will be extra capacity (or unused capacity) at Plant 1 of 80 tons of steel and at Plant 2 of 50 tons of steel. All the capacity at Plant 3 will be used. There is not alternate solution (not for manager).

The table below is not suitable for a manager report.

Based on the table above, we can increase the cost of production shipment from Plant 1 to Steel 3 up to $30, and still have the same amount of steel processed at this plant. However, any change of shipping cost will affect the final total cost. It is important that while the minimum cost of shipping from Plant 3 to Steel 2 is listed as -$30, this is not feasible as it would require us being paid to ship steel instead of paying to ship steel (not for manager). We can manufacture the same amount of Steel 2 at plant 3 for any unit cost of manufacturing that is below 21.

If we want to send steel from Plant 1 to Steel 1 we need to reduce the unit cost of $60. The cost should be reduced below $50 in order to have shipment from Plant 1 to Steel 1. rewrite

12

The graph above shows the effect of changes in cost of shipping from Plant 1 to Steel 3 on total cost. The detailed shipping plan changes depending on the cost and if we want such detailed information we need to solve the problem again at the desired point. The graph indicates that if the cost of shipment goes above $30, we will not ship any steel from Plant 1 to Steel 3.

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Based on the table above, no amount of capacity change for Plant 1 and 2 will affect total cost. (Refer to the ranges above, your sentence is not true of all possible values.) For every 1 increase in capacity in Plant 3, we can reduce total cost by $8. We can increase our capacity in Plant 3 up to 200 tons of steel. Needs rewrite

The graph above shows the effect of changes in supply of Plant 3 on total cost. The detailed shipping plan changes depending on the cost and if we want such detailed information we need to solve the problem again at the desired point. The graph indicates that as we continue to increase capacity, total cost will continue to decrease.

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Part B

Note: This page will be updated with a more detailed solution once we have learned more linear programming techniques.

No you cannot create balanced transportation problem because each type of steel has different production times. This means that the capacity will constantly change.

In order to solve this problem, use Linear Programming (to be taught in class later).

Brief explanation:Setup decision variables and then setup objective function and constraints. There are 15 constraints because each constraint needs to take into account supply, demand, and time available to produce.

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