













                          Linear Programming Examples
                                      and
                                 Documentation
                                      for
                                AS-EASY-AS v5.70





                                 Copyright 1996
                                  TRIUS, Inc.
                                  P.O. Box 249
                           N. Andover, MA 01845-0249





                               Tel: 508-794-9377
                               Fax: 507-688-6312
                               BBS: 508-794-0762
                          Web: http://www.triusinc.com
                                 CIS: GO TRIUS



        All material herein is provided on an as-is basis. Neither TRIUS
       nor any of its employees and/or agents provide any warranty as to
         the applicability of this information to specific situations.
        Any such applicability has to be determined solely by the user.


       AS-EASY-AS v5.7 & LINEAR PROGRAMMING
       ====================================
       A number of people have contacted us, since the release of AS-
       EASY-AS v5.7, telling us how impressed they are with the new
       features, in particular the powerful linear programming option,
       but they don't know exactly how or when to use it.

       Unfortunately, linear programming can be a fairly complex topic,
       and we don't have universal answers on when/how to use it.  In
       response to all those comments, however, we've put together these
       very detailed examples of linear programming situations and
       solutions, hoping that they will give some more insight to this
       unique, powerful AS-EASY-AS feature.




       Statement of problem (1)
       ------------------------
       A designer of expensive leather jackets, created two new jacket
       designs for the new season, a long one and a short one.  Each
       short leather jacket requires 1 labor-hour from the cutting
       department and 3 labor-hours from the sewing department. Each
       long leather jacket requires 2 labor hours from the cutting
       department and 4 labor-hours from the sewing department.  This
       designer is sharing cutting and sewing services with other
       designers, and as such, there are only 32 labor-hours per week
       available in the cutting department and 84 labor-hours per week
       available in the sewing department for him.  In addition, because
       of the limited appeal of long leather jackets, the distributor
       cannot take any more that 12 long leather jackets per week.  If
       the designer makes $50 profit on each short jacket and $80 on
       each long one, how many jackets of each type should he have
       manufactured per week in order to maximize his profit?




       Objective Function
       ------------------
       If we let,  X1 be the number of short, and X2 be the number of
       long leather jackets produced per week, so that the designer's
       profit is maximized, then the profit, per week, would be,

                P = 50*X1 + 80*X2

       Looking at this equation from a mathematical standpoint, it
       looks like the profit (P) can be made as large as we like by
       simply producing more and more leather jackets.  But, here is
       where the various real-life limitations come in.  Any
       manufacturer, no matter how small or how large, has
       manufacturing limits imposed by available materials, available
       manpower, demand, available plant capacity, etc.   These
       limits are usually referred to as constraints.



                                           Linear Programming   Pg: 1/2

       Here is what these constraints look like in our situation
       (again, assuming X1 short jackets and X2 long jackets are
       produced).




       Cutting department constraint
       -----------------------------
                                                                
        Cutting time     Cutting time   Equal   Maximum Labor   
        for short     +  for long       to or   hours           
        jackets          jackets        Less    available in    
        per week         per week       than    cutting dept.   
                                                                

           1*X1         +      2*X2          =<          32




       Sewing department constraint
       ------------------------------
                                                               
        Sewing time     Sewing time   Equal   Maximum Labor    
        for short    +  for long      to or   hours            
        jackets         jackets       Less    available in     
        per week        per week      than    Assembly dept.   
                                                               

           3*X1        +        4*X2        =<          84




       Demand constraint
       -----------------
       As stated in the problem, the distributor cannot take any more
       than 12 long leather jackets per week, so

          X2 =< 12




       Non-Negative Constraints
       ------------------------
       Many times we overlook these constraints, and luckily, in most
       situations they have no bearing, but they are very important
       and we should always include them if we want to guarantee the
       correct solution every time.

          X1 >= 0
          X2 >= 0



                                           Linear Programming   Pg: 1/3

       Mathematical Model
       ------------------
       Now, let us group all our data and equations together, in
       preparation to enter them into AS-EASY-AS.

        We want to maximize:  P = 50*X1 + 80*X2   <- Objective function

               Subject to:    X1 + 2*X2  =< 32  
                            3*X1 + 4*X2  =< 84  
                                     X2  =< 12   <- Constraints
                                     X1  >=  0  
                                     X2  >=  0  




       AS-EASY-AS Model
       ----------------
       A portion of the AS-EASY-AS screen, showing the entry for this
       problem, is shown below (Note that all instructions given
       further down are based on the assumption that the data entry
       in AS-EASY-AS corresponds to the cells shown below).

         [  A ]....A...B...C...D....E...
            1
            2        1    2  LE     32      <- 1st constraint
            3        3    4  LE     84      <- 2nd constraint
            4        0    1  LE     12      <- 3rd constraint
            5        1    0  GE      0      <- 4th constraint
            6        0    1  GE      0      <- 5th constraint
            7       50   80  EQ   MAX       <- Objective function


       After you've entered the data in your AS-EASY-AS worksheet, we
       are ready to solve this problem.

            1.  Locate the cursor in cell B2, and press the "/" key to
                invoke the main AS-EASY-AS menu.

            2.  Press the letter "A" to select Array.

            3.  Press the letter "L" to select Linear.  The prompt
                on the first line of the screen should display,
                "SIMPLEX/ First Range: B2..B2"

            4.  Use the cursor keys to highlight the cell range
                B2..E7, and press ENTER to accept the range. The prompt
                on the first line of the screen should display,
                "Indicate OUTPUT range: B2"

            5.  Move the cursor to D9 and press ENTER.






                                           Linear Programming   Pg: 1/4

       The program will solve the problem for you and the screen
       should look like the one shown below.

         [  A ]....A...B...C...D....E...
            1
            2        1    2  LE     32
            3        3    4  LE     84
            4             1  LE     12
            5        1    0  GE      0
            6        0    1  GE      0
            7       50   80  EQ   MAX
            8
            9                MAX  1480
           10                 X1    20
           11                 X2     6
           12



       The results in cells D9..E11 indicate that this was a
       maximization problem.  The maximum value of the objective
       function is 1480, and the corresponding X1 and X2 values are
       20 and 6.  Or, in terms of our real life problem, given the
       constraints we have, the designer can maximize his weekly
       profit ($1,480), by having 20 short and 6 long jackets
       manufactured per week.



       Observations
       ------------
       In order to provide a better understanding into the powerful
       process of linear programming, here we present some common
       questions that might come up.


       Q1. Well, since we know that long leather jackets make
           significantly more profit for the designer, why not
           manufacture just long jackets and skip the short ones
           altogether?

       A1. Because for starters, according to the stated constraints,
           the distributor can only take 12 long jackets a week.  so
           the maximum profit that could be achieved would be 12 * 80
           = 960 which is lower than that produced by our solution,
           i.e., it's not the 'maximum'.


       Q2. If Q1 is true, then why not produce the max 12 long
           jackets and then use the remaining resources for short
           jackets?  It might generate more profits!






                                           Linear Programming   Pg: 1/5

       A2. Production of 12 long jackets would use up 12 * 2 = 24
           hours of the 32 available in the cutting department.
           There would only be 8 labor-hours left, which would result
           in a maximum of 8 short jackets produced.  This would
           result in an additional profit of 8 * 50 = 400.  Added to
           the 960 profit realized by the 12 long jackets (see A1),
           it would give us a total of 1360, which is still less than
           the amount given by the optimized solution!


       Q3. The cutting time required for short jackets is 1 labor-
           hour, and that for long jackets is 2 labor-hours, i.e.,
           twice as long.  However, in terms of profit, long jackets
           produce only about 60% more (less than a factor of 2).
           Wouldn't it make sense to produce ONLY short jackets and
           maximize the profit, especially where there is no
           distributor limit as in Q1 above?

       A3. The cutting dept. could produce a maximum of

             32 labor hours per week
             -----------------------  =  32 short jackets in a week
             1 labor-hour per jacket

           whereas the sewing dept could complete a maximum of

             84 labor hours per week
             -----------------------  =  28 short jackets in a week
             3 labor-hour per jacket

           The 28 short jackets (maximum possible per week), would
           generate a profit of $1250, which is less than the profit
           generated by the combination suggested by the AS-EASY-AS
           solution.























                                           Linear Programming   Pg: 1/6

       Statement of problem (2)
       ------------------------
       A cattle rancher uses three types of cattle food, Type(1),
       Type(2), and Type(3).  The cost per pound of each is:
       Type(1)=$1.5, Type(2)=$3.5, and Type(3)=$2.0. The rancher
       wants to meet published nutritional minimum daily requirements
       (MDR), in milligrams of Vitamins A, B and C per animal.  The
       MDR and nutritional content for the three types of cattle food,
       in mg/lb, is shown in the table below.

             Vitamin   MDR   Type(1)  Type(2)  Type(3)
             -----------------------------------------
               A       120     8        2       20
               B       180     9       11        5
               C       100     1       10       20

       Because of protein content, however, an animal cannot eat more
       than 15 lbs of Type(1), 10 lbs of Type(2) and 5 lbs of Type(3)
       of cattle food per day.  How many pounds of each type of
       cattle food should the rancher purchase per day in order to
       minimize his cost, and still meet the MDR?




       Objective Function
       ------------------
       If we let X1, X2 and X3 be the number of pounds of each type
       of food the rancher buys per day, for each animal, then the
       total cost per day (the function that needs to be minimized),
       is:

         P = 1.5*X1 + 3.5*X2 + 2.0*X3

       Looking at this equation from a mathematical standpoint, it
       looks like the expense can be continuously minimized by simply
       reducing the pounds of each type of cattle food purchased, or
       by purchasing only Type(1) feed which is the least expensive
       one (shown as X1 in the above equation). But, here is where the
       various real-life limitations come in.  There are minimum
       nutritional requirements, minimum feed per day per animal, etc.
       These limits are usually referred to as constraints.

       Here is what these constraints look like in our situation.
       (Note that all the quantities below are per day, per animal).












                                                  Linear Programming   Pg: 2/7

       Vitamin A Nutritional Constraints
       ---------------------------------
                                                             
        Vitamin A     Vitamin A     Vitamin A      Minimum   
        obtained   +  obtained   +  obtained   >=  Vitamin A 
        from food     from food     from food      required  
        Type(1)       Type(2)       Type(3)        per day   
                                                             
           8*X1      +     2*X2      +     20*X3     >=    120



       Vitamin B Nutritional Constraints
       ---------------------------------
                                                             
        Vitamin B     Vitamin B     Vitamin B      Minimum   
        obtained   +  obtained   +  obtained   >=  Vitamin B 
        from food     from food     from food      required  
        Type(1)       Type(2)       Type(3)        per day   
                                                             
           9*X1      +     11*X2      +     5*X3     >=    180



       Vitamin C Nutritional Constraints
       ---------------------------------
                                                             
        Vitamin C     Vitamin C     Vitamin C      Minimum   
        obtained   +  obtained   +  obtained   >=  Vitamin C 
        from food     from food     from food      required  
        Type(1)       Type(2)       Type(3)        per day   
                                                             
           1*X1      +    10*X2      +     20*X3     >=    100



       Total Animal Feed Constraints
       -----------------------------
                         
         Total Type(1)   
         Food an animal   =< 15
         can eat per day 
                         
               X1           =< 15


                         
         Total Type(2)   
         Food an animal   =< 10
         can eat per day 
                         
               X2           =< 10





                                                  Linear Programming   Pg: 2/8

                         
         Total Type(3)   
         Food an animal   =< 5
         can eat per day 
                         
               X3           =< 5



       Non-Negative Constraints
       ------------------------
       Many times we overlook these constraints, and luckily, in most
       situations they have no bearing, but they are very important
       and we should always include them if we want to guarantee the
       correct solution every time.

          X1 >= 0
          X2 >= 0
          X3 >= 0



       Mathematical Model
       ------------------
       Now, let us group all our data and equations together, in
       preparation to enter them into AS-EASY-AS.

       Minimize:  P = 1.5*X1 + 3.5*X2 + 2*X3 <- Objective function

          Subject to: 8*X1 +  2*X2 + 20*X3 >= 120 
                      9*X1 + 11*X2 +  5*X3 >= 180 
                      1*X1 + 10*X2 + 20*X3 >= 100 
                                        X1 =< 15   <- Constraints
                                        X2 =< 10  
                                        X3 =<  5  
                                        X1 >=  0  
                                        X2 >=  0  
                                        X3 >=  0  




       AS-EASY-AS Model
       ----------------
       A portion of the AS-EASY-AS screen, showing the entry for this
       problem, is shown below (Note that all instructions given
       further down are based on the assumption that the data entry
       in AS-EASY-AS corresponds to the cells shown below).









                                                  Linear Programming   Pg: 2/9

         [  A ]....A...B...C...D...E....F......
            1
            2    8    2   20  GE    120   <- 1st constraint
            3    9   11    5  GE    180   <- 2nd constraint
            4    1   10   20  GE    100   <- 3rd constraint
            5    1    0    0  LE     15   <- 4th constraint
            6    0    1    0  LE     10   <- 5th constraint
            7    0    0    1  LE      5   <- 6th constraint
            8    1    0    0  GE      0   <- 7th constraint
            9    0    1    0  GE      0   <- 8th constraint
           10    0    0    1  GE      0   <- 9th constraint
           11  1.5  3.5  2.0  EQ  MIN     <- Objective Function

       After you've entered the data in your AS-EASY-AS worksheet, we
       are ready to solve this problem.

            1.  Locate the cursor in cell A2, and press the "/" key to
                invoke the main AS-EASY-AS menu.

            2.  Press the letter "A" to select Array.

            3.  Press the letter "L" to select Linear.  The prompt
                on the first line of the screen should display,
                "SIMPLEX/ First Range: A2..A2"

            4.  Use the cursor keys to highlight the cell range
                A2..E11, and press ENTER to accept the range. The prompt
                on the first line of the screen should display,
                "Indicate OUTPUT range: A2"

            5.  Move the cursor to A14 and press ENTER.

       The program will solve the problem for you and the screen
       should look like the one shown below.

         [  A ]....A...B...C...D...E....F......
            1
            2    8    2   20  GE    120
            3    9   11    5  GE    180
            4    1   10   20  GE    100
            5    1    0    0  LE     15
            6    0    1    0  LE     10
            7    0    0    1  LE      5
            8    1    0    0  GE      0
            9    0    1    0  GE      0
           10    0    0    1  GE      0
           11  1.5  3.5  2.0  EQ  MIN
           12
           13
           14 MIN    38
           15  X1    15
           16  X2   2.8
           17  X3   2.9
           18



                                                 Linear Programming   Pg: 2/10

       The results in cells A14..B17 indicate that this was a
       minimization problem. The minimum value of the objective
       function is 38, and the corresponding X1, X2 and X3 values are
       15, 2.8, and 2.9.  Or, in terms of our real life problem,
       given the constraints we have, the rancher can minimize his
       cost to $38 per day per animal,  and still meet all his
       nutritional requirements by buying 15 lbs of Type(1), 2.8 lbs of
       Type(2), and 2.9 lbs of Type(3) cattle food.

















































                                                 Linear Programming   Pg: 2/11

       Statement of problem (3)
       ------------------------
       A nutritionist recommends to a colleague the following
       minimum daily requirements of vitamin B and vitamin C:

          400 units of vitamin B
          800 units of vitamin C

       The local pharmacy supplies two different vitamin tablets, brand Y
       and brand Z.  Vitamin tablet Y contains 75 units of vitamin B and
       100 units of vitamin C and costs $0.05.  Vitamin tablet Z contains
       50 units of vitamin B and 200 units of vitamin C and costs $0.04.
       How many vitamin tablets of each brand should be consumed to
       satisfy the daily requirements at a minimal cost?



       Objective Function
       ------------------
       If we let X1 be the number of brand Y tablets and X2 be the
       number of brand Z tablets, then the total cost per day (the
       function that needs to be minimized), is:

          P = .05*X1 + .04*X2

       Looking at this equation from a mathematical standpoint, it
       looks like the expense can be minimized by simply purchasing
       brand Z tablets ($0.04).  However, in real life, like in the
       previous examples, there are limitations.  There are minimal
       nutritional daily requirements to be met.  Such limitations
       limits are usually referred to as constraints.

       Here is what these constraints look like in our situation:


       Vitamin B Nutritional Constraints
       ---------------------------------
                                                  
        Vitamin B     Vitamin B      Minimum units
        obtained   +  obtained   >=  Vitamin B    
        from          from           required     
        Brand Y       Brand Z        per day      
                                                  
           75*X1     +     50*X2     >=     400


       Vitamin C Nutritional Constraints
       ---------------------------------
                                                            
        Vitamin C        Vitamin C         Minimum units    
        obtained      +  obtained      >=  Vitamin C        
        from Brand Y     from Brand Z      required per day 
                                                            
          100*X1        +    200*X2        >=     800



                                                 Linear Programming   Pg: 3/12

       Non-Negative Constraints
       ------------------------
       Many times we overlook these constraints. In most situations
       they have no bearing to the solution of the problem. However,
       they are very important and we should always include them if we
       want to guarantee the correct solution every time.

          X1 >= 0
          X2 >= 0



       Mathematical Model
       ------------------
       Now, let us group all our data and equations together, in
       preparation to enter them into AS-EASY-AS.

       Minimize:  P = .05*X1 + .04*X2   <- Objective function

          Subject to:  75*X1 +  50*X2 >= 400 
                      100*X1 + 200*X2 >= 800  <- Constraints
                                   X1 >=  0  
                                   X2 >=  0  



       AS-EASY-AS Model
       ----------------
       A portion of the AS-EASY-AS screen, showing the entry for this
       problem, is shown below (Note that all instructions given
       further down are based on the assumption that the data entry
       in AS-EASY-AS corresponds to the cells shown below).


         [  A ]....A...B...C...D....E...
            1
            2       75   50  GE    400      <- 1st constraint
            3      100  200  GE    800      <- 2nd constraint
            4        1    0  GE      0      <- 3rd constraint
            5        0    1  GE      0      <- 4th constraint
            6      .05  .04  EQ   MIN       <- Objective function


       After you've entered the data in your AS-EASY-AS worksheet, we
       are ready to solve this problem.

            1.  Locate the cursor in cell B2, and press the "/" key to
                invoke the main AS-EASY-AS menu.

            2.  Press the letter "A" to select Array.

            3.  Press the letter "L" to select Linear.  The prompt
                on the first line of the screen should display,
                "SIMPLEX/ First Range: B2..B2"



                                                 Linear Programming   Pg: 3/13

            4.  Use the cursor keys to highlight the cell range
                B2..E6, and press ENTER to accept the range. The prompt
                on the first line of the screen should display,
                "Indicate OUTPUT range: B2"

            5.  Move the cursor to D9 and press ENTER.

       The program will solve the problem for you and the screen
       should look like the one shown below.

         [  A ]....A...B...C...D....E...
            1
            2       75   50  GE    400
            3      100  200  GE    800
            4        1    0  GE      0
            5        0    1  GE      0
            6      .05  .04  EQ   MIN
            7
            8
            9                MIN   .28
           10                 X1     4
           11                 X2     2
           12



       The results in cells D9..E11 indicate that this was a
       minimization problem.  The minimum value of the objective
       function is .28, and the corresponding X1 and X2 values are 4
       and 2.  Or, in terms of our real life problem, given the
       constraints we have, the individual can minimize the cost to
       $0.28 per day and still meet all nutritional requirements by
       purchasing 4 tablets of brand Y and 2 tablets of brand Z.
























                                                 Linear Programming   Pg: 3/14


       Statement of problem (4)
       ------------------------
       A savings & loan company has $3 million in funds to lend.  Local
       and State laws require that at least 50% of all monies loaned
       for mortgages must be for first mortgages and that at least 30%
       of the total amount loaned must be for either first or second
       mortgages.  Company policy requires that signature and
       automobile loans cannot exceed 25% of the total amount loaned
       and that signature loans cannot exceed 15% of the total amount
       loaned.  How much money should be allocated to each type of loan
       in order to maximize the company's return?

       The types of loans and annual returns for each type are
       given below:

         -------------------------------------
         Type of Loan            Annual Return
         -------------------------------------
         Signature               18%
         First Mortgage          12%
         Second Mortgage         14%
         Automobile              16%
         -------------------------------------




       Objective Function
       ------------------
       If we let,

           X1 represent Signature loans,
           X2 represent First Mortgages,
           X3 represent Second Mortgages, and
           X4 represent Automobile loans,

       then the company's return (the function that needs to be
       maximized), can be expressed as:

          P = .18*X1 + .12*X2 + .14*X3 + .16*X4

       Looking at this equation from a mathematical standpoint, it
       looks like the annual return can be maximized by simply
       providing Signature loans.  However, in real life (and as
       indicated in this example), we have to take into account State and
       Federal regulations as well as Company guidelines and limitations.
       These limitations will become the problems constraints.









                                                 Linear Programming   Pg: 4/15

       Fund Constraints
       ----------------
                                                         
        Sign.     First      Second     Auto       Total 
        Loan   +  Mortg.  +  Mortg.  +  Loan   <=  Loan  
        Amount    Amount     Amount     Amount     Amount
                                                         
           X1    +      X2    +     X3     +     X4    <=  3000000



       State & Federal Law Constraints
       -------------------------------
                                                   
        First          Second          30% of      
        Mortgage   +   Mortgage    >=  Total Loan  
        Amount         Amount          Amount      
                                                   
           X2        +      X3         >=     900000

                                           Assures First and Second
        First          Second              Mortgages are at least
        Mortgage   -   Mortgage    >=  0   equal, to  meet "50% of
        Amount         Amount              Total Mortgage Funds
                                           loaned must be for First
           X2        -      X3         >=   0    Mortgages" constraint.



       Company Guidelines Constraints
       ------------------------------
                                                   
        Signature      Automobile      25% of      
        Loan       +   Loan        <=  Total Loan  
        Amount         Amount          Amount      
                                                   
           X1        +      X4         <=     750000

                                  
        Signature      15% of     
        Loan       <=  Total Loan 
        Amount         Amount     
                                  
           X1        <=     450000



       Non-Negative Constraints
       ------------------------
       Many times we overlook these constraints, and luckily, in most
       situations they have no bearing, but they are very important
       and we should always include them if we want to guarantee the
       correct solution every time.




                                                 Linear Programming   Pg: 4/16

          X1 >= 0
          X2 >= 0
          X3 >= 0
          X4 >= 0


       Mathematical Model
       ------------------
       Now, let us group all our data and equations together, in
       preparation to enter them into AS-EASY-AS.

       Minimize:  P = .18*X1 + .12*X2 + .14*X3 + .16*X4   <- Objective function

          Subject to: X1 + X2 + X3 + X4 <= 300000 
                                X2 + X3 >= 900000 
                                X2 - X3 >=     0  
                                X1 + X4 <= 750000 
                                     X1 <= 450000  <- Constraints
                                     X1 >=     0  
                                     X2 >=     0  
                                     X3 >=     0  
                                     X4 >=     0  



       AS-EASY-AS Model
       ----------------
       A portion of the AS-EASY-AS screen, showing the entry for this
       problem, is shown below (Note that all instructions given
       further down are based on the assumption that the data entry
       in AS-EASY-AS corresponds to the cells shown below).


         [  A ]....A...B...C...D....E........F...G
            1
            2   1    1    1    1   LE    3000000   <- 1st constraint
            3   0    1    1    0   GE     900000   <- 2nd constraint
            4   0    1   -1    0   GE          0   <- 3rd constraint
            5   1    0    0    1   LE     750000   <- 4th constraint
            6   1    0    0    0   LE     450000   <- 5th constraint
            7   1    0    0    0   GE          0   <- 6th constraint
            8   0    1    0    0   GE          0   <- 7th constraint
            9   0    0    1    0   GE          0   <- 8th constraint
           10   0    0    0    1   GE          0   <- 9th constraint
           11 .18  .12  .14  .16   EQ     MAX      <- Objective Function

       After you've entered the data in your AS-EASY-AS worksheet, we
       are ready to solve this problem.

            1.  Locate the cursor in cell A2, and press the "/" key to
                invoke the main AS-EASY-AS menu.

            2.  Press the letter "A" to select Array.




                                                 Linear Programming   Pg: 4/17

            3.  Press the letter "L" to select Linear.  The prompt
                on the first line of the screen should display,
                "SIMPLEX/ First Range: A2..A2"

            4.  Use the cursor keys to highlight the cell range
                A2..F11, and press ENTER to accept the range. The prompt
                on the first line of the screen should display,
                "Indicate OUTPUT range: A2"

            5.  Move the cursor to B13 and press ENTER.

       The program will solve the problem for you and the screen
       should look like the one shown below.

         [  A ]....A...B...C...D....E........F...G
            1
            2   1    1    1    1   LE    3000000
            3   0    1    1    0   GE     900000
            4   0    1   -1    0   GE          0
            5   1    0    0    1   LE     750000
            6   1    0    0    0   LE     450000
            7   1    0    0    0   GE          0
            8   0    1    0    0   GE          0
            9   0    0    1    0   GE          0
           10   0    0    0    1   GE          0
           11 .18  .12  .14  .16   EQ     MAX
           12
           13      max   423559.6
           14      X1      450000
           15      X2     1125000
           16      X3     1125000
           17      X4      300000


       The results in cells B13..C17 indicate that this was a maximization
       problem.  The maximum value of the objective function is 423,559.6,
       and the corresponding X1, X2, X3, and X4 values are 450,000,
       1,125,000, 1,125,000, and 300,000.  Or, putting it in terms that
       the loan company's officers would understand, given the constraints
       we have, the bank can maximize their returns and still meet all
       State/Federal law and company guidelines requirements by lending an
       amount equal to $450,000 for Signature loans, $1,125,000 for First
       mortgages, $1,125,000 for Second mortgages, and $300,000 for
       Automobile loans. The maximum return, under those conditions, would
       be: $423,559.60

       --- End










                                                 Linear Programming   Pg: 4/18

