<p> Nikolaos Kapetanios IE 416 HW 2</p><p>Parisay’s comments are in red. Problem 4 Page 71 Source of the problem: page 71, W.L. Winston, “operations Research, Application and Algorithms”, 4th Edition, Thomson Learning, 2004, ISBN: 0-534-38058-1</p><p>By: B. Bajaj</p><p>Goldilocks needs to find at least 12 lb of gold and at least 18 lb of silver to pay the monthly rent. There are two mines in which Goldilocks can find the gold and silver. Each day that Goldilocks spends in mine 1, she finds 2 lb of gold and 2 lb of silver. Each day that Goldilocks spends in mine 2, she finds 1 lb of gold and 3 lb of silver. Formulate an LP to help Goldilocks meet her requirements while spending as little time as possible in the mines.</p><p>Summary by Patrick: Material Mine 1 Mine 2 Rent Gold 2 1 > 12 Silver 2 3 > 18</p><p>Here Goldilocks is looking for a minimum gold and silver output from each mine to pay her rent, the only variable is the amount of days she spends in each mine, therefore the following decision variables are defined.</p><p>Let:</p><p>X1 = Number of Days spent in Mine 1</p><p>X 2 = Number of Days spent in Mine 2</p><p>Therefore Goldilocks would like to minimize the following Objective function</p><p>Z = X1 + X 2 {Minimum}</p><p>Subject to (S.T.):</p><p>2X1+ X2 ≥ 12 {Gold Constraint}</p><p>1 2X1 + X2 ≥ 18 {Silver Constraint} These equations first need to be defined in WinQSB. Two variables and two constraints along with an objective function that is to be minimized were defined in the software. </p><p>Next the values of the objective function and constraints were entered in the software:</p><p>Figure 1: Initial data input in WinQSB</p><p>WinQSB then provided a combined report of the output displaying the optimal solution given the constraints of the problem.</p><p>Figure 3: Combined value output.</p><p>Software states that in order to achieve the maximum solution goldilocks will have to spend 4.5 days in mine one and 3 days in mine 2 making the minimum days spent in both</p><p>2 mines to be 7.5 days. Upon review these two values uphold the two constraints (do not use word “constraints” in report to the manager, be more specific) defined in the problem and are reasonable.</p><p>Report to Management: by Mr. Yi</p><p>Based on the information that we have, it has been determined that we will be able to meet Goldilocks’ requirements in 7.5 days. By spending 4.5 days in Mine</p><p>1 and 3 days in Mine 2, a total of 12 lb of gold and 18 lb of silver will be collected.</p><p>Goldilocks will then be able to afford to pay her rent with those 7.5 days of work.</p><p>Sensitivity analysis: (By Mr. David Chavez) case1: Select one basic variable of your choice.</p><p>The basic variable that I have chosen is X1= Number of Days spent in Mine 1</p><p>3 case 2: Select one nonbasic variable of your choice. If you do not have any nonbasic variable you will skip this case.</p><p>Skip case 3: Select one binding constraint of your choice. </p><p>The binding constraint that I have choose is C1 = Gold Constraint</p><p>4 case 4: Select one nonbinding constraint of your choice. If you do not have any nonbinding constraint you will skip this case.</p><p>Skip b) Write a report of your sensitivity graph and analysis for a hypothetical manager. Use a language understandable to most people. Do not use mathematical abbreviations. (Read the file on report writing for blending problem.) </p><p>The following sensitivity analysis report or report to the manager is not suitable or is wrong. Can you figure out why? </p><p>The following pertains to the minimum number of days in mines one and two, to collect at least 12 and 18 lbs of gold and silver to meet the cost of monthly rent. The total minimum number days required to meet our demand in both mines are 7.5 days. In mine 1 we will be spending 4.5 days and in mine 2 we will be spending 3 days. It will </p><p>5 only take 7.5 days to pay for the rent. Considering that in mine 1 we collect 2 lbs of silver and 2 lbs of gold which is more then what we collect in mine 2, I saw an opportunity to make this mine more efficient. Therefore I performed an analysis for mine 1 that incorporated a “what if” situation. If we wanted to know the minimum amount of gold and silver to collect in mine #1 that would contribute to minimizing the days we spend in the mines then I recommend increasing our gold intake from 12 to 18 lbs, this will reduce the amount of time spent in mine 1 by 30 %, we only spend .67% of a day oppose from a full 100% of a day.</p><p>Sensitivity analysis report for coefficient of X1 (by Parisay):</p><p>Notice coefficient for X1 is “1”. What does this coefficient mean? It means if </p><p>Goldiloc spends one day in Mine 1 she will obtain one day worth of 2 lb gold and 2 lb silver. One way to interpret this coefficient is that it represents efficiency (or productivity). That is Goldiloc will obtain 2 lb gold and 2 lb silver in each day if her efficiency is 100%. Is there any other way to interpret it? I do not know! Email me if you know.</p><p>For example, if efficiency is less than 100% (i.e. Goldilock is not feeling well) we will expect to obtain less of the gold and silver in one day. Therefore, she should work longer hours per day to be able to obtain the same gold and silver in one day. As a numerical example, consider the efficiency is 90% and not 100%. Therefore she should work about</p><p>11% more in one day to be able to get the same amount of gold and silver. (w*0.9=1day </p><p>>> w=1.11 days, or instead of 8 hours she had to work 8.88 hours) Which means the coefficient of X1 in objective function should be 1.11 instead of 1. Now the sensitivity graph will show that we need to spend more total number of days to pay the rent, which makes sense. </p><p>6 Similarly, if efficiency is more than 100% we will expect to obtain more of the gold and silver in one day. Efficiency can be translated (calculated) to a coefficient that is less than 1. For example, assume that efficiency is 110% instead of 100%. Therefore she can work about 9% less in one day to be able to get the same amount of gold and silver. </p><p>(w*1.1=1day >> w= 0.91 days, or instead of 8 hours she can work 7.27 hours.) This means the coefficient of X1 in objective function should be 0.91 instead of 1. Now the sensitivity graph will show that we need to spend less total number of days to pay the rent, which makes sense. </p><p>Now having good understanding (interpretation) of this coefficient, we can discuss that our motivation for selecting this coefficient for sensitivity analysis is that we predict that Goldilock can be more or less efficient when digging </p><p>Mine 1 and we would like to see the effect of her efficiency on the solution.</p><p>The next question can be whether to select coefficient of X1 or X2 for sensitivity analysis. In general, if no other practical information is available, we select a variable that has higher value as solution. In here, X1=4.5 and X2=3, therefore I will select coefficient of X1. Can you tell why? Because any small change in the coefficient of X1 will be multiplied by 4.5 (higher solution) and will have more impact on objective function value. However, in practical cases, you need to check which one of coefficients is more unstable (or uncertain) or your manager has interest in.</p><p>Is there any other way to incorporate efficiency in our analysis? Yes. We can incorporate efficiency in technological coefficients, that is, coefficients of constraints in left hand side. For example, If Goldiloc has 90% efficiency at Mine 1 then instead of 2 </p><p>7 lb she will obtain 2*90% =1.8 lb of gold and silver each day. Therefore the related constraints will be replaced by:</p><p>1.8X1+ X2 ≥ 12 {Gold Constraint}</p><p>1.8X1 + X2 ≥ 18 {Silver Constraint} Using this new problem should provide the same answer as the discussion above. Try it. </p><p>Then which way of approach is more suitable to incorporate efficiency? Why? First student who emails me the correct answer will get extra credit!</p><p>In the case of selecting a constraint for rhs (right hand side) sensitivity analysis, if no other practical information is available, we select a constraint with the highest shadow price. Because any small change in the value of rhs will be multiplied by</p><p>SP (higher shadow price) and will have more impact on objective function value. In this problem it does not make any difference which constraint we select.</p><p>The final report should put together all the pieces of information and reports.</p><p>8</p>