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Vera_Pavlovna [14]
3 years ago
5

Kalyan Singhal Corp. makes three products, and it has three machines available as resources as given in the following LP problem

:Maximize contribution = 6X1 + 8X2 + 4X3Subject to:1X1 + 7X2 + 4X3 <= 72(hours on machine 1)2X1 + 1X2 + 7X3 <= 76 (hours on machine 2)8X1 + 4X2 + 1X3 <= 72 (hours on machine 3)X1, X2, X3 >=0(a) Determine the optimal solution using LP software. the optimal achieved isX1=X2-X3=contribution =(b) Is there unused time available on any of the machines with the optimal solution?(c) What would it be worth to the firm to make an additional hour of time available on the third machine?(d) How much would the firm’s profit increase if an extra 15 hours of time were made available on the second machine at no extra cost?

Business
1 answer:
alukav5142 [94]3 years ago
4 0

Answer:

a) X1=5.64

X2=4.57

X3= 8.59

Optimal solution=104.77

b) Additional hour on machine 3 = 105.42 – 104.77 = 0.64

c) Additional 15 hours on machine 2 = 105.37-104.77 = 0.6

Explanation:

Please consider the data provided by the exercise. If you have any question please write me back. All the exercises are solved in a single sheet with the formulas indications.  SOLVED WITH EXCEL SOLVER.

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Answer:

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11 months ago
Which of the following best describes equilibrium?
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Answer:

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3 years ago
Sheffield Laboratories holds a valuable patent (No. 758-6002-1A) on a precipitator that prevents certain types of air pollution.
vlabodo [156]

Answer:

                        SHEFFIELD LABORATORY

                            PATENT (NO. 78-6002-1A)

Carrying value as at Dec 31

                                      2011                     2015                   2018

Cost                            $182,300           $349,000           $385,000

Amortization             <u> (10,724)   </u>             <u>(79,433)  </u>           <u>(142,403)</u>

                                 <u> 171,576     </u>         <u>   269,567  </u>           <u>242,597</u>

Cost

As at 31 Dec 2011

Design and construction of a prototype                                     $89,000

Testing of models                                                                           40,600

Fees paid engineers and lawyers to prepare application          <u> 52,700</u>

                                                                                                       <u>$182,300</u>

As at 31 Dec 2012

Cost as at Jan 1, 2012                                                                $182,300

Additional cost during the year:

Engineering activity necessary to advance.                           <u> $84,500  </u>

                                                                                                    <u>$266,800</u>

As at 31 Dec 2013

Cost as at Jan 1, 2013                                                                $266,800

Additional cost during the year:

legal fee paid                                                                              <u>$40600  </u>

                                                                                                 <u>   $307,400</u>

As at 31 Dec 2014

Cost as at Jan 1, 2013                                                                $307,400

Additional cost during the year:

Research aimed at modifying the design                                <u>$41,600 </u>

                                                                                                   <u> $349,000</u>

As at 31 Dec 2018

Cost as at Jan 1, 2018                                                                $349,000

Additional cost during the year:

legal fee paid in unseccesful patent infrigement.                   <u>  $36,000  </u>

                                                                                                    <u>$385,000</u>

Amortization for the year    

Dec 31 2011         $182,300/17 =  $10,724

Dec 31 2012

182,300/17                                          10,724

84,500/0                                          <u>      -       </u>

                                                         <u>  10,724</u>

<u />

Dec 31 , 2013 :

  $182,300/17 =  $10,724            

  84,500/16    =      5,281

40,600/0     =     <u>    -  </u>

                         <u>   16,005</u>

Dec 31 2014  =  

$182,300/17 =  $10,724            

  84,500/16    =      5,281

40,600/16     =       2,538

41,600/17      =       <u> 2,447</u>

                         <u>   20,990</u>

Dec 31 2018  =  

$182,300/17 =  $10,724            

  84,500/16    =      5,281

40,600/16     =       2,538

41,600/17      =        2,447

36,000/0     =    <u>       -</u>

                         <u>   20,990</u>

Explanation:

5 0
3 years ago
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