Download presentation

Presentation is loading. Please wait.

Published byAugust Mason Modified about 1 year ago

1
ISEN 315 Spring 2011 Dr. Gary Gaukler

2
Introduction to Aggregate Planning Goal: To plan gross work force levels and set firm-wide production plans, based on predicted demand for aggregate units.

3
Hierarchy of Planning Forecast of aggregate demand over time horizon Aggregate Production Plan: determine aggregate production and workforce levels over time horizon Master Production Schedule: Disaggregate the aggregate plan and determine per-item production levels Materials Requirements Planning: Detailed schedule for production/replenishment activities

4
Prototype Aggregate Planning Example The washing machine plant is interested in determining work force and production levels for the next 8 months. Forecasted aggregate demands for Jan-Aug. are: 420, 280, 460, 190, 310, 145, 110, 125. Starting inventory at the end of December is 200 and the firm would like to have 100 units on hand at the end of August. Find monthly production levels.

5
Step 1: Determine “net” demand. MonthNet PredictedCum. Net Demand Demand 1(Jan) (Feb) (Mar) (Apr) (May) (June) (July) (Aug)

6
Constant Work Force Plan Suppose that we are interested in determining a production plan that doesn’t change the size of the workforce over the planning horizon. How would we do that?

7
Monthly Production = 1940/8 = or rounded to 243/month. But: there are stockouts.

8
How can we have a constant work force plan with no stockouts? Using the graph, find a straight line that lies completely above the cumulative net demand curve:

9
From the previous graph, we see that the cum. net demand curve is crossed at period 3, so that monthly production is 960/3 = 320. Ending inventory each month is found from: Month Cum. Net. Dem. Cum. Prod. Invent. 1(Jan) (Feb) (Mar) (Apr.) (May) (June) (July) (Aug)

10
But - may not be realistic for several reasons: It may not be possible to achieve the production level of 320 unit/mo with an integer number of workers Since all months do not have the same number of workdays, a constant production level may not translate to the same number of workers each month.

11
To overcome these shortcomings: Assume number of workdays per month is given K factor computed where K = # of aggregate units produced by one worker in one day Suppose that we are told that over a period of 40 days, the plant had 38 workers who produced 520 units. It follows that: K= 520/(38*40) =.3421 = average number of units produced by one worker in one day.

12
Computing Constant Work Force Assume we are given the following # working days per month: 22, 16, 23, 20, 21, 17, 18, 10. March is still critical month. Cum. net demand thru March = 960. Cum # working days = = 61. Find 960/61 = units/day implies /.3421 = 46 workers required.

13
Constant Work Force Production Plan Mo # wk days Prod. Cum Cum Nt End Inv Level Prod Dem Jan Feb Mar Apr May Jun Jul Aug

14
Addition of Costs Holding Cost (per unit per month): $8.50 Hiring Cost per worker: $800 Firing Cost per worker: $1,250 Payroll Cost: $75/worker/day Shortage Cost: $50 unit short/month

15
Cost Evaluation of Constant Work Force Plan Assume that the work force at end of Dec was 40. Cost to hire 6 workers: 6*800 = $4800 Inventory Cost: accumulate ending inventory: ( ) = Add in 100 units netted out in Aug = Hence Inv. Cost = 2193*8.5=$18, Payroll cost: ($75/worker/day) (46 workers) (167days) = $576,150 Cost of plan: $576,150 + $18, $4800 = $599,590.50

16
Cost Reduction in Constant Work Force Plan

17
Zero Inventory Plan (Chase Strategy) Change the workforce each month in order to reduce ending inventory to nearly zero by matching the workforce with monthly demand as closely as possible. This is accomplished by computing the # units produced by one worker each month (by multiplying K by #days per mo.) and then taking net demand each month and dividing by this quantity. The resulting ratio is rounded up and possibly adjusted downward.

18
Chase vs. Constant

19
Linear Programming Class of optimization problems Linear objective function Linear constraints Real variables Efficiently solved

20
Aggregate Planning LP Parameters : –c H, c F –c I –c R –c O, c U, c S

21
Aggregate Planning LP Parameters : –n t –K –I 0,W 0 –D t

22
Aggregate Planning LP Decision variables : –W t –P t –I t –H t, F t

23
Aggregate Planning LP Decision variables : –O t –U t –S t

24
Aggregate Planning LP

25
Constraints:

26
Aggregate Planning LP Constraints:

27
Aggregate Planning LP Objective function:

Similar presentations

© 2016 SlidePlayer.com Inc.

All rights reserved.

Ads by Google