Chapter 3, Problem 56RP

Explanation of Solution

LP to minimize the cost:

• Assume the following for shifts,
  • Consider the time from midnight to 6 A.M. as shift “1”.
  • Consider the time from 6 A.M. to noon as shift “2”.
  • Consider the time from noon to 6 P.M. as shift “3”.
  • Consider the time from 6 P.M. to midnight as shift “4”.
• Let “${\text{X}}_{\text{1}}$” be the number of employees who works both shifts “1” and “2”.
• Let “${\text{X}}_2$” be the number of employees who works both shifts “2” and “3”.
• Let “${\text{X}}_3$” be the number of employees who works both shifts “3” and “4”.
• Let “${\text{X}}_4$” be the number of employees who works both shifts “4” and “1”.
• Let “${\text{C}}_{\text{m}}$” be the number of customers left after end of shift “m”.
• Let “${\text{S}}_{\text{m}}$” be the number of customers served during shift “m”.
• The Linear Programming (LP) to minimize the sum of impatience costs of bank’s worker and customer, is as follows,

$\text{Min}{\text{z=120(X}}_{\text{1}}{\text{+X}}_2+{\text{X}}_3+{\text{X}}_4){\text{+5(C}}_1{\text{+C}}_2+{\text{C}}_3+{\text{C}}_4)$

• The above function subject to the following conditions,
  • “${\text{C}}_1=100-{\text{S}}_{\text{1}}$” denotes the number of customers, who are left after end of shift “1”