# Replacement Models : Staffing Problem

In the previous sections, we discussed about replacement problems, which were not related to human resources working in an organization.

The replacement models can also be used to solve the problems of staff replacement. This section focuses on the problem of replacing staff in an organization. Staff replacement is essential due to the following factors:

• Inefficiency
• Resignation
• Retirement
• Unexpected events (like accident, death, etc.)

"Companies that sensibly manage their investment in people will prosper in the long run." - Tom DeMarco & Tim Lister

#### Assumption

• The life distribution for the service of staff in an organization is predetermined.

#### Example 1

A team of software developers at www.universalteacher.com is planned to rise to a strength of 50 persons, and then to remain at that level. Consider the following data:

Year Total % who have left upto the end of the year
1 5
2 30
3 50
4 60
5 70
6 75
7 80
8 85
9 90
10 100

On the basis of above information, determine:

What is the recruitment per year necessary to maintain the strength? There are 8 senior posts for which the length of service is the main criterion. What is the average length of service after which new entrant can expect his promotion to one of these posts?

Solution.

##### Calculating values for table 1
• The values in column (c) are obtained by subtracting the corresponding elements of column (b) from 100.
• The values in column (d) are obtained by dividing the corresponding elements in column (b) by 100.
• The values in column (e) are obtained by dividing the corresponding elements in column (c) by 100.

#### Table 1

On small screens, use horizontal scrollbar to view full calculation

Year No.of persons
who leave at the end of the year
No. of persons in service at the end of year Prob. of leaving at the end of the year Prob. of in service at the end of the year
a b c d e
0 0 100 0 1.00
1 5 95 0.05 0.95
2 30 70 0.30 0.70
3 50 50 0.50 0.50
4 60 40 0.60 0.40
5 70 30 0.70 0.30
6 75 25 0.75 0.25
7 80 20 0.80 0.20
8 85 15 0.85 0.15
9 90 10 0.90 0.10
10 100 0 1.00 0
Total   455

From table 1, we find that with a recruitment policy of 100 persons every year, the total number of persons serving in the organization would have been 455. Hence, if we want to maintain a strength of 50 persons then we should recruit

 100 x 50 ---------- 455 = 1000 ------- 91 = 10.98

= 11 persons/year

Every year 11 persons should be recruited to maintain a strength of 50. Number of survivals after each year can be obtained by multiplying the various values of column (e) by 11.

"New people are recruited to replace the old ones." - Vinay Chhabra & Manish Dewan

#### Table 2

Year Number of persons in service
0 11
1 10
2 8
3 6
4 4
5 3
6 3
7 2
8 2
9 1
10 0

Now there are 8 senior posts. From table 2, it can seen that there are 3 persons in service during the sixth year, 2 in seventh year, 2 in eighth year, and 1 in ninth year. Hence, promotions of new recruits will start by the end of sixth year and will continue upto seventh year.

#### Example 2

The Railway Ministry requires 200 private assistants, 300 private secretaries, and 50 section officers. Persons are recruited at the age of 21, if still in service, retire at the age of 60. Given the following life table, determine

• How many persons should be recruited every year ?
• At what age promotions should take place ?

Use horizontal scrollbar to view full calculation

 Age No. in service Age No. in service Age No. in service Age No. in service 21 22 23 24 25 26 27 28 1000 600 400 380 311 260 229 206 29 30 31 32 33 34 35 36 190 180 174 166 162 155 150 146 37 38 39 40 41 42 43 44 145 135 131 125 120 112 105 100 45 46 47 48 49 50 51 52 94 86 80 73 65 60 53 46 53 54 55 56 57 58 59 60 40 32 26 23 19 13 11 0

Solution.

If a policy of recruiting 1000 persons every year is followed, then the total number of employees in service between the age 21 to 59 years will be equal to 6403. But the requirement of organization is 550 (200 + 300 + 50) employees.

Therefore, to maintain a strength of 550 employees, the organization should recruit:
(1000 X 550)/6403 = 86 (approx.) persons every year.

##### Private assistants

Out of a strength of 550, there are 200 private assistants. Hence, out of a strength of 1000 there will be
(200 X 1000)/550 = 364 private assistants.
From the above life table, 364 is available upto 24 years. Therefore, the promotion of private assistants will take place in 25th year.

##### Private secretaries

Out of a strength of 1000 there will be
(300 X 1000)/550 = 545 private secretaries.

##### Section officers

Number of section officers = 1000 - (364 + 545) = 91.
From the above life table, we find that at the age of 46 only 86 will survive. Therefore, promotion of private secretaries will take place in 46th year.

In this chapter, you were introduced to the problem of replacement that is experienced in systems (both machines and men) where efficiency of items worsen over their life span or sometimes fail completely. The chapter focussed on the following categories:

• Replacement of items whose efficiency deteriorates with time.
• Replacement of items that fail suddenly.
• Staffing problem.