Real-World Applications
3.12 Multiperiod Work Scheduling
7 Suppose we are borrowing $1,000 at 12% annual interest with 60 monthly payments. Assume equal payments are made at the end of month 1, month 2, . . . month 60. We know that entering into Excel the function
PMT(.01, 60, 1,000) would yield the monthly payment ($22.24).
It is instructive to use LP to determine the montly pay-ment. Let p be the (unknown) monthly paypay-ment. Each month we owe .01 (our current unpaid balance) in interest. The remainder of our monthly payment is used to reduce the un-paid balance. For example, suppose we un-paid $30 each month.
At the beginning of month 1, our unpaid balance is $1,000.
Of our month 1 payment, $10 goes to interest and $20 to paying off the unpaid balance. Then we would begin month 2 with an unpaid balance of $980. The trick is to use LP to determine the monthly payment that will pay off the loan at the end of month 60.
8 You are a CFA (chartered financial analyst). Madonna has come to you because she needs help paying off her credit card bills. She owes the amounts on her credit cards shown in Table 43. Madonna is willing to allocate up to $5,000 per month to pay off these credit cards. All cards must be paid off within 36 months. Madonna’s goal is to minimize the total of all her payments. To solve this problem, you must understand how interest on a loan works. To illustrate, suppose Madonna pays $5,000 on Saks during month 1.
Then her Saks balance at the beginning of month 2 is 20,000 (5,000 .005(20,000))
This follows because during month 1 Madonna incurs .005(20,000) in interest charges on her Saks card. Help Madonna solve her problems!
9 Winstonco is considering investing in three projects. If we fully invest in a project, the realized cash flows (in millions of dollars) will be as shown in Table 44. For example, project 1 requires cash outflow of $3 million today
and returns $5.5 million 3 years from now. Today we have
$2 million in cash. At each time point (0, .5, 1, 1.5, 2, and 2.5 years from today) we may, if desired, borrow up to $2 million at 3.5% (per 6 months) interest. Leftover cash earns 3% (per 6 months) interest. For example, if after borrowing and investing at time 0 we have $1 million we would receive
$30,000 in interest at time .5 years. Winstonco’s goal is to maximize cash on hand after it accounts for time 3 cash flows. What investment and borrowing strategy should be used? Remember that we may invest in a fraction of a project. For example, if we invest in .5 of project 3, then we have cash outflows of $1 million at time 0 and .5.
T A B L E 43
Card Balance ($) Monthly Rate (%)
Saks Fifth Avenue 20,000 .5
Bloomingdale’s 50,000 1
Macys 40,000 1.5
T A B L E 44
Cash Flow
Time (Years) Project 1 Project 2 Project 3
0 3.0 2 2.0
0.5 1.0 .5 2.0
1 1.8 1.5 1.8
1.5 1.4 1.5 1
2 1.8 1.5 1
2.5 1.8 1.2 1
3 5.5 1 6
At the beginning of January, 50 skilled technicians work for CSL. Each skilled technician can work up to 160 hours per month. To meet future demands, new technicians must be trained. It takes one month to train a new technician. During the month of training, a trainee must be supervised for 50 hours by an experienced technician. Each experienced technician is paid $2,000 a month (even if he or she does not work the full 160 hours).
During the month of training, a trainee is paid $1,000 a month. At the end of each month, 5% of CSL’s experienced technicians quit to join Plum Computers. Formulate an LP whose solution will enable CSL to minimize the labor cost incurred in meeting the ser-vice requirements for the next five months.
Solution CSL must determine the number of technicians who should be trained during month t (t 1, 2, 3, 4, 5). Thus, we define
xt number of technicians trained during month t (t 1, 2, 3, 4, 5) CSL wants to minimize total labor cost during the next five months. Note that
Total labor cost cost of paying trainees cost of paying experienced technicians To express the cost of paying experienced technicians, we need to define, for t 1, 2, 3, 4, 5,
yt number of experienced technicians at the beginning of month t Then
Total labor cost (1,000x1 1,000x2 1,000x3 1,000x4 1,000x5)
(2,000y1 2000y2 2,000y3 2,000y4 2,000y5) Thus, CSL’s objective function is
min z 1,000x1 1,000x2 1,000x3 1,000x4 1,000x5
2,000y1 2,000y2 2,000y3 2,000y4 2,000y5
What constraints does CSL face? Note that we are given y1 50, and that for t 1, 2, 3, 4, 5, CSL must ensure that
Number of available technician hours during month t
Number of technician hours required during month t (72) Because each trainee requires 50 hours of experienced technician time, and each skilled technician is available for 160 hours per month,
Number of available technician hours during month t 160yt 50xt
Now (72) yields the following five constraints:
160y1 50x1 6,000 (month 1 constraint) 160y2 50x2 7,000 (month 2 constraint) 160y3 50x3 8,000 (month 3 constraint) 160y4 50x4 9,500 (month 4 constraint) 160y5 50x5 11,000 (month 5 constraint)
As in the other multiperiod formulations, we need constraints that relate variables from different periods. In the CSL problem, it is important to realize that the number of skilled technicians available at the beginning of any month is determined by the number of skilled technicians available during the previous month and the number of technicians trained during the previous month:
(73)
technicians trained during month (t 1)
experienced technicians who quit during month (t 1)
For example, for February, (73) yields
y2 y1 x1 0.05y1 or y2 0.95y1 x1
Similarly, for March, (73) yields
y3 0.95y2 x2
and for April,
y4 0.95y3 x3
and for May,
y5 0.95y4 x4
Adding the sign restrictions xt 0 and yt 0 (t 1, 2, 3, 4, 5), we obtain the follow-ing LP:
min z 1,000x1 1,000x2 1,000x3 1,000x4 1,000x5
2,000y1 2,000y2 2,000y3 2,000y4 2,000y5
s.t. 160y1 50x1 6,000 y1 50 160y2 50x2 7,000 0.95y1 x1 y2
160y3 50x3 8,000 0.95y2 x2 y3
160y4 50x4 9,500 0.95y3 x3 y4
160y5 50x5 11,000 0.95y4 x4 y5
xt, yt 0 (t 1, 2, 3, 4, 5)
The optimal solution is z 593,777; x1 0; x2 8.45; x3 11.45; x4 9.52; x5 0;
y1 50; y2 47.5; y3 53.58; y4 62.34; and y5 68.75.
In reality, the yt’s must be integers, so our solution is difficult to interpret. The prob-lem with our formulation is that assuming that exactly 5% of the employees quit each month can cause the number of employees to change from an integer during one month to a fraction during the next month. We might want to assume that the number of em-ployees quitting each month is the integer closest to 5% of the total workforce, but then we do not have a linear programming problem!
P R O B L E M S
Group A
Experienced technicians available at beginning of month (t 1) Experienced technicians available
at beginning of month t
1 If y1 38, then what would be the optimal solution to CSL’s problem?
2 An insurance company believes that it will require the following numbers of personal computers during the next six months: January, 9; February, 5; March, 7; April, 9;
May, 10; June, 5. Computers can be rented for a period of one, two, or three months at the following unit rates: one-month rate, $200; two-one-month rate, $350; three-one-month rate,
$450. Formulate an LP that can be used to minimize the
cost of renting the required computers. You may assume that if a machine is rented for a period of time extending beyond June, the cost of the rental should be prorated. For example, if a computer is rented for three months at the beginning of May, then a rental fee of 2
3(450) $300, not $450, should be assessed in the objective function.
3 The IRS has determined that during each of the next 12 months it will need the number of supercomputers given in Table 45. To meet these requirements, the IRS rents
supercomputers for a period of one, two, or three months.
It costs $100 to rent a supercomputer for one month, $180 for two months, and $250 for three months. At the beginning of month 1, the IRS has no supercomputers. Determine the rental plan that meets the next 12 months’ requirements at minimum cost. Note: You may assume that fractional rentals are okay, so if your solution says to rent 140.6 computers for one month we can round this up or down (to 141 or 140) without having much effect on the total cost.
Group B
4 You own a wheat warehouse with a capacity of 20,000 bushels. At the beginning of month 1, you have 6,000 bushels
T A B L E 45
Month Computer Requirements
1 800
2 1,000
3 600
4 500
5 1,200
6 400
7 800
8 600
9 400
10 500
11 800
12 600
of wheat. Each month, wheat can be bought and sold at the price per 1000 bushels given in Table 46.
The sequence of events during each month is as follows:
a You observe your initial stock of wheat.
b You can sell any amount of wheat up to your initial stock at the current month’s selling price.
c You can buy (at the current month’s buying price) as much wheat as you want, subject to the warehouse size limitation.
Your goal is to formulate an LP that can be used to deter-mine how to maximize the profit earned over the next 10 months.
T A B L E 46
Month Selling Price ($) Purchase Price ($)
1 3 8
2 6 8
3 7 2
4 1 3
5 4 4
6 5 3
7 5 3
8 1 2
9 3 5
10 2 5
S U M M A R Y Linear Programming Definitions
A linear programming problem (LP) consists of three parts:
1 A linear function (the objective function) of decision variables (say, x1, x2, . . . , xn) that is to be maximized or minimized.
2 A set of constraints (each of which must be a linear equality or linear inequality) that restrict the values that may be assumed by the decision variables.
3 The sign restrictions, which specify for each decision variable xjeither (1) variable xjmust be nonnegative—xj 0; or (2) variable xjmay be positive, zero, or negative—xj
is unrestricted in sign (urs).
The coefficient of a variable in the objective function is the variable’s objective func-tion coefficient. The coefficient of a variable in a constraint is a technological coefficient.
The right-hand side of each constraint is called a right-hand side (rhs).
A point is simply a specification of the values of each decision variable. The feasible region of an LP consists of all points satisfying the LP’s constraints and sign restrictions.
Any point in the feasible region that has the largest z-value of all points in the feasible re-gion (for a max problem) is an optimal solution to the LP. An LP may have no optimal solution, one optimal solution, or an infinite number of optimal solutions.
A constraint in an LP is binding if the left-hand side and the right-hand side are equal when the values of the variables in the optimal solution are substituted into the constraint.
Graphical Solution of Linear Programming Problems
The feasible region for any LP is a convex set. If an LP has an optimal solution, there is an extreme (or corner) point of the feasible region that is an optimal solution to the LP.
We may graphically solve an LP (max problem) with two decision variables as follows:
Step 1 Graph the feasible region.
Step 2 Draw an isoprofit line.
Step 3 Move parallel to the isoprofit line in the direction of increasing z. The last point in the feasible region that contacts an isoprofit line is an optimal solution to the LP.